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#41
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[Ohms Law] Watts and Impedance?
Hello,
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#42
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[Ohms Law] Watts and Impedance?
Hello,
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#43
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[Ohms Law] Watts and Impedance?
Hello,
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#44
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[Ohms Law] Watts and Impedance?
Subject:
[Ohms Law] Watts and Impedance? Date: Sat, 07 Feb 2004 13:01:36 GMT From: Computer Prog Organization: Optimum Online Newsgroups: rec.audio.tech References: 1 , 2 , 3 , 4 Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#45
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[Ohms Law] Watts and Impedance?
Subject:
[Ohms Law] Watts and Impedance? Date: Sat, 07 Feb 2004 13:01:36 GMT From: Computer Prog Organization: Optimum Online Newsgroups: rec.audio.tech References: 1 , 2 , 3 , 4 Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#46
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[Ohms Law] Watts and Impedance?
Subject:
[Ohms Law] Watts and Impedance? Date: Sat, 07 Feb 2004 13:01:36 GMT From: Computer Prog Organization: Optimum Online Newsgroups: rec.audio.tech References: 1 , 2 , 3 , 4 Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#47
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[Ohms Law] Watts and Impedance?
Subject:
[Ohms Law] Watts and Impedance? Date: Sat, 07 Feb 2004 13:01:36 GMT From: Computer Prog Organization: Optimum Online Newsgroups: rec.audio.tech References: 1 , 2 , 3 , 4 Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US Thanks. Knowing that the magic box is only going to take 200 watts out of the wall because that's all the speaker wants, we can plug in the power and the voltage the magic box needs to do that (into the formula) and solve for the current: I = 200W/120V = 1.666...A ~ 1.7A OK, so now what happens if we plug the magic box into a 14V supply and the speaker still wants 200 watts? Well, the magic box has no choice but to get whatever current it needs from the 14V supply to make that happen, so if we use our I = P/E formula again it comes out: I = P/E = 200W/14V = 14.286A ~ 14.3A So, if you've got a 200 watt load, it doesn't make any difference what its impedance is, the primary power source has to supply that power, so the current it'll have to supply will depend on its voltage. |
#48
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[Ohms Law] Watts and Impedance?
Computer Prog,
To some extent you are asking "how heavy is a ton of bricks?" Are they heavier if I stack them in a single brick column, in a one high row, or some combination? The obvious answer is that they always weigh a ton, but one stacking configuration may be more convenient to truck around than another. In your amplifier situation, unless you have some sort of exotic process running inside (be sure to market it), an audio amplifier cannot create or destroy energy. It can only transform the energy from one form to another. And --- there is always a "handling charge" for this transformation. In general, only 30-50% of the amplifier's input energy from the battery is available for running the speakers. --- With respect to your voltage issues. A common trick in automotive amplifiers is to use two amplifiers per channel. One amplifier runs 0 to +14 and the other is turned on it's head to run 0 to -14 volts. The speaker, connected between both amplifiers, doesn't know or care about this trick and thinks the amplifier is actually a 28 volt unit. Other possibilities include switching power supplies that will step the battery voltage up before applying it to the amplifier. --- At your level, assuming that the amplifier has reasonable design and build qualities and was fairly rated by some known standard, the major issues a does the unit have enough power for your application; can your battery charging system come up with the required power; and can the vehicle wiring support the required current? The rest is your judgement about how the system looks, operates, and sounds. Don't worry too much about what is inside the box. A common audio marketing strategy is to convince the customer that a unit benefits from some exclusive circuit innovation. Checking around, you'll find that no other product has this innovation. Once you are convinced about the innovation's benefit, you'll have to purchase that product. I rememberer one product line that touted an exclusive design approach. They were very decent products from an excellent manufacturer, but it was mostly a (well executed) standard textbook design with a few impressive sounding terms thrown into the consumer literature. From my point of view, the only novelty was that they published some graphs in the consumer literature. Very similar graphs had been published almost a decade prior by another manufacturer and before that in my textbooks. From a marketing standpoint the program was a success because it generated some excitement for the sales staff who were more likely to mention the product to customers. ----------------------------------------------------------- spam: wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15 13 (Barry Mann) [sorry about the puzzle, spammers are ruining my mailbox] ----------------------------------------------------------- |
#49
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[Ohms Law] Watts and Impedance?
Computer Prog,
To some extent you are asking "how heavy is a ton of bricks?" Are they heavier if I stack them in a single brick column, in a one high row, or some combination? The obvious answer is that they always weigh a ton, but one stacking configuration may be more convenient to truck around than another. In your amplifier situation, unless you have some sort of exotic process running inside (be sure to market it), an audio amplifier cannot create or destroy energy. It can only transform the energy from one form to another. And --- there is always a "handling charge" for this transformation. In general, only 30-50% of the amplifier's input energy from the battery is available for running the speakers. --- With respect to your voltage issues. A common trick in automotive amplifiers is to use two amplifiers per channel. One amplifier runs 0 to +14 and the other is turned on it's head to run 0 to -14 volts. The speaker, connected between both amplifiers, doesn't know or care about this trick and thinks the amplifier is actually a 28 volt unit. Other possibilities include switching power supplies that will step the battery voltage up before applying it to the amplifier. --- At your level, assuming that the amplifier has reasonable design and build qualities and was fairly rated by some known standard, the major issues a does the unit have enough power for your application; can your battery charging system come up with the required power; and can the vehicle wiring support the required current? The rest is your judgement about how the system looks, operates, and sounds. Don't worry too much about what is inside the box. A common audio marketing strategy is to convince the customer that a unit benefits from some exclusive circuit innovation. Checking around, you'll find that no other product has this innovation. Once you are convinced about the innovation's benefit, you'll have to purchase that product. I rememberer one product line that touted an exclusive design approach. They were very decent products from an excellent manufacturer, but it was mostly a (well executed) standard textbook design with a few impressive sounding terms thrown into the consumer literature. From my point of view, the only novelty was that they published some graphs in the consumer literature. Very similar graphs had been published almost a decade prior by another manufacturer and before that in my textbooks. From a marketing standpoint the program was a success because it generated some excitement for the sales staff who were more likely to mention the product to customers. ----------------------------------------------------------- spam: wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15 13 (Barry Mann) [sorry about the puzzle, spammers are ruining my mailbox] ----------------------------------------------------------- |
#50
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[Ohms Law] Watts and Impedance?
Computer Prog,
To some extent you are asking "how heavy is a ton of bricks?" Are they heavier if I stack them in a single brick column, in a one high row, or some combination? The obvious answer is that they always weigh a ton, but one stacking configuration may be more convenient to truck around than another. In your amplifier situation, unless you have some sort of exotic process running inside (be sure to market it), an audio amplifier cannot create or destroy energy. It can only transform the energy from one form to another. And --- there is always a "handling charge" for this transformation. In general, only 30-50% of the amplifier's input energy from the battery is available for running the speakers. --- With respect to your voltage issues. A common trick in automotive amplifiers is to use two amplifiers per channel. One amplifier runs 0 to +14 and the other is turned on it's head to run 0 to -14 volts. The speaker, connected between both amplifiers, doesn't know or care about this trick and thinks the amplifier is actually a 28 volt unit. Other possibilities include switching power supplies that will step the battery voltage up before applying it to the amplifier. --- At your level, assuming that the amplifier has reasonable design and build qualities and was fairly rated by some known standard, the major issues a does the unit have enough power for your application; can your battery charging system come up with the required power; and can the vehicle wiring support the required current? The rest is your judgement about how the system looks, operates, and sounds. Don't worry too much about what is inside the box. A common audio marketing strategy is to convince the customer that a unit benefits from some exclusive circuit innovation. Checking around, you'll find that no other product has this innovation. Once you are convinced about the innovation's benefit, you'll have to purchase that product. I rememberer one product line that touted an exclusive design approach. They were very decent products from an excellent manufacturer, but it was mostly a (well executed) standard textbook design with a few impressive sounding terms thrown into the consumer literature. From my point of view, the only novelty was that they published some graphs in the consumer literature. Very similar graphs had been published almost a decade prior by another manufacturer and before that in my textbooks. From a marketing standpoint the program was a success because it generated some excitement for the sales staff who were more likely to mention the product to customers. ----------------------------------------------------------- spam: wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15 13 (Barry Mann) [sorry about the puzzle, spammers are ruining my mailbox] ----------------------------------------------------------- |
#51
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[Ohms Law] Watts and Impedance?
Computer Prog,
To some extent you are asking "how heavy is a ton of bricks?" Are they heavier if I stack them in a single brick column, in a one high row, or some combination? The obvious answer is that they always weigh a ton, but one stacking configuration may be more convenient to truck around than another. In your amplifier situation, unless you have some sort of exotic process running inside (be sure to market it), an audio amplifier cannot create or destroy energy. It can only transform the energy from one form to another. And --- there is always a "handling charge" for this transformation. In general, only 30-50% of the amplifier's input energy from the battery is available for running the speakers. --- With respect to your voltage issues. A common trick in automotive amplifiers is to use two amplifiers per channel. One amplifier runs 0 to +14 and the other is turned on it's head to run 0 to -14 volts. The speaker, connected between both amplifiers, doesn't know or care about this trick and thinks the amplifier is actually a 28 volt unit. Other possibilities include switching power supplies that will step the battery voltage up before applying it to the amplifier. --- At your level, assuming that the amplifier has reasonable design and build qualities and was fairly rated by some known standard, the major issues a does the unit have enough power for your application; can your battery charging system come up with the required power; and can the vehicle wiring support the required current? The rest is your judgement about how the system looks, operates, and sounds. Don't worry too much about what is inside the box. A common audio marketing strategy is to convince the customer that a unit benefits from some exclusive circuit innovation. Checking around, you'll find that no other product has this innovation. Once you are convinced about the innovation's benefit, you'll have to purchase that product. I rememberer one product line that touted an exclusive design approach. They were very decent products from an excellent manufacturer, but it was mostly a (well executed) standard textbook design with a few impressive sounding terms thrown into the consumer literature. From my point of view, the only novelty was that they published some graphs in the consumer literature. Very similar graphs had been published almost a decade prior by another manufacturer and before that in my textbooks. From a marketing standpoint the program was a success because it generated some excitement for the sales staff who were more likely to mention the product to customers. ----------------------------------------------------------- spam: wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15 13 (Barry Mann) [sorry about the puzzle, spammers are ruining my mailbox] ----------------------------------------------------------- |
#52
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote: Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? --- If both amps are capable of delivering 200W into their respective loads then, yes, for the same power output they will draw identical currents from the same supply voltage. --- My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US --- Just looking at the link reveals that neither amp is capable of putting out 200 watts, so, yeah, if that's the part the salesman's talking about he's wrong (to put it politely) to start with. But, you bring up a good point, which is: if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply, how much power can you pump into the speaker? Assuming that the amp has an "H" bridge output stage, it can make the louspeaker think it's got 28 Volts, peak-to-peak, across it, which is 9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its impedance, we can say: P = E²/R = 9.9VRMS²/8R = 12.25W So the best you can do with a 14V supply is to put about 12W into an 8 ohm speaker. Since there are four speakers total, if we assume two per channel and assume that they'll be connected in parallel, then we can say: P = E²/R = 98/4 = 24.5W and since the amp is a stereo amp, if we assume 24.5 watts per channel we come out with about 50 watts, somewhat higher than the basic amp is rated for. To find the voltage we need to drive a 4 ohm load to 39 watts, we can say: E = sqrt(PR) ~ 12.5V, which is right in the ballpark for a 12V automotive-type application. But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... -- John Fields |
#53
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote: Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? --- If both amps are capable of delivering 200W into their respective loads then, yes, for the same power output they will draw identical currents from the same supply voltage. --- My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US --- Just looking at the link reveals that neither amp is capable of putting out 200 watts, so, yeah, if that's the part the salesman's talking about he's wrong (to put it politely) to start with. But, you bring up a good point, which is: if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply, how much power can you pump into the speaker? Assuming that the amp has an "H" bridge output stage, it can make the louspeaker think it's got 28 Volts, peak-to-peak, across it, which is 9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its impedance, we can say: P = E²/R = 9.9VRMS²/8R = 12.25W So the best you can do with a 14V supply is to put about 12W into an 8 ohm speaker. Since there are four speakers total, if we assume two per channel and assume that they'll be connected in parallel, then we can say: P = E²/R = 98/4 = 24.5W and since the amp is a stereo amp, if we assume 24.5 watts per channel we come out with about 50 watts, somewhat higher than the basic amp is rated for. To find the voltage we need to drive a 4 ohm load to 39 watts, we can say: E = sqrt(PR) ~ 12.5V, which is right in the ballpark for a 12V automotive-type application. But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote: Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? --- If both amps are capable of delivering 200W into their respective loads then, yes, for the same power output they will draw identical currents from the same supply voltage. --- My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US --- Just looking at the link reveals that neither amp is capable of putting out 200 watts, so, yeah, if that's the part the salesman's talking about he's wrong (to put it politely) to start with. But, you bring up a good point, which is: if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply, how much power can you pump into the speaker? Assuming that the amp has an "H" bridge output stage, it can make the louspeaker think it's got 28 Volts, peak-to-peak, across it, which is 9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its impedance, we can say: P = E²/R = 9.9VRMS²/8R = 12.25W So the best you can do with a 14V supply is to put about 12W into an 8 ohm speaker. Since there are four speakers total, if we assume two per channel and assume that they'll be connected in parallel, then we can say: P = E²/R = 98/4 = 24.5W and since the amp is a stereo amp, if we assume 24.5 watts per channel we come out with about 50 watts, somewhat higher than the basic amp is rated for. To find the voltage we need to drive a 4 ohm load to 39 watts, we can say: E = sqrt(PR) ~ 12.5V, which is right in the ballpark for a 12V automotive-type application. But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote: Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? --- If both amps are capable of delivering 200W into their respective loads then, yes, for the same power output they will draw identical currents from the same supply voltage. --- My reason for asking this question in the first place was because my brother just purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms. This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated @ 8-ohms since that would require more voltage at each speaker and we are only starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V (input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage) amp. I think the answer is no. Here is a link to the Harley part: http://www.harley-davidson.com/gma/g...bmLocale=en_US --- Just looking at the link reveals that neither amp is capable of putting out 200 watts, so, yeah, if that's the part the salesman's talking about he's wrong (to put it politely) to start with. But, you bring up a good point, which is: if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply, how much power can you pump into the speaker? Assuming that the amp has an "H" bridge output stage, it can make the louspeaker think it's got 28 Volts, peak-to-peak, across it, which is 9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its impedance, we can say: P = E²/R = 9.9VRMS²/8R = 12.25W So the best you can do with a 14V supply is to put about 12W into an 8 ohm speaker. Since there are four speakers total, if we assume two per channel and assume that they'll be connected in parallel, then we can say: P = E²/R = 98/4 = 24.5W and since the amp is a stereo amp, if we assume 24.5 watts per channel we come out with about 50 watts, somewhat higher than the basic amp is rated for. To find the voltage we need to drive a 4 ohm load to 39 watts, we can say: E = sqrt(PR) ~ 12.5V, which is right in the ballpark for a 12V automotive-type application. But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... -- John Fields |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote:
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms of course it will go much lower when you only use the battery voltage of 12V, then you get 18/36W for any higher wattage the amplifier must make use of a voltage inverter i.e. 12V 40V for 200W/8 Ohms or 28V for 200W/4 Ohms The current drawn from the battery/generator will be about the same whatever the speaker impedance is when driven by the same power of 200W, around 60A at full power stereo 2x200W RMS... Peter |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote:
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms of course it will go much lower when you only use the battery voltage of 12V, then you get 18/36W for any higher wattage the amplifier must make use of a voltage inverter i.e. 12V 40V for 200W/8 Ohms or 28V for 200W/4 Ohms The current drawn from the battery/generator will be about the same whatever the speaker impedance is when driven by the same power of 200W, around 60A at full power stereo 2x200W RMS... Peter |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote:
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms of course it will go much lower when you only use the battery voltage of 12V, then you get 18/36W for any higher wattage the amplifier must make use of a voltage inverter i.e. 12V 40V for 200W/8 Ohms or 28V for 200W/4 Ohms The current drawn from the battery/generator will be about the same whatever the speaker impedance is when driven by the same power of 200W, around 60A at full power stereo 2x200W RMS... Peter |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote:
Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms of course it will go much lower when you only use the battery voltage of 12V, then you get 18/36W for any higher wattage the amplifier must make use of a voltage inverter i.e. 12V 40V for 200W/8 Ohms or 28V for 200W/4 Ohms The current drawn from the battery/generator will be about the same whatever the speaker impedance is when driven by the same power of 200W, around 60A at full power stereo 2x200W RMS... Peter |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote: But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's more like it! -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote: But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's more like it! -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote: But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's more like it! -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote: But how about that 70 watt amp? It's not likely they've got a DC to DC converter in there to boost the supply voltage, and the best you can do with a 14V supply and two 4 ohm loads is 50 watts, so they must be playing with the speaker impedance. For a nomonal 12.5V supply, then, let's look at what the speakers need to look like to get us 70 watts out. Since: P = E²/R we can rearrange that to: R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms Which, for a stereo rig means two, 2.2 ohm speakers in parallel per channel. Hmmm.... Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's more like it! -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote: On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. --- Another error, dammit! with a 14.4V supply and no bridge, the maximum signal swing into the load will be 14.4VPP, which is ~ 5VRMS Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4 ohm load, 25/4 ~ 6.25 watts. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote: On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. --- Another error, dammit! with a 14.4V supply and no bridge, the maximum signal swing into the load will be 14.4VPP, which is ~ 5VRMS Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4 ohm load, 25/4 ~ 6.25 watts. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote: On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. --- Another error, dammit! with a 14.4V supply and no bridge, the maximum signal swing into the load will be 14.4VPP, which is ~ 5VRMS Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4 ohm load, 25/4 ~ 6.25 watts. -- John Fields |
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[Ohms Law] Watts and Impedance?
On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote: On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel wrote: Computer Prog wrote: Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms the maximum you can get out of 14,4V is: 26W into 8 Ohms and 52W into 4 Ohms --- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is (14.4V/sqrt(2))²/8 ~ 13W and in a 4 ohm load (14.4V/sqrt(2))²/4 ~ 26W Assuming, of course, that the loudspeaker is AC coupled to the amp and that the voltage swing into the loudspeaker is +/- 7.2V. --- Another error, dammit! with a 14.4V supply and no bridge, the maximum signal swing into the load will be 14.4VPP, which is ~ 5VRMS Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4 ohm load, 25/4 ~ 6.25 watts. -- John Fields |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote in message ...
Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? Yes. In both cases a voltage supply of 14 volts to the amplifier, even if it is bridged. So in the first case the 14 volt has to be stepped up to U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case to 40 volts, plus losses in the design. Stepping up voltage is commonly done in high(er) power car audio power amps. There are mainly three classes of amplifiers for car audio (or more?). -The first uses a single output stage directly connected to the speaker, max output (assuming 14 volt supply and 4 ohm speakers) = (14/2/sqrt(2))^2/4=6 watts. -The second uses two output stages in a bridge and can deliver (14/sqrt(2))^2/4=24 watts -The third class uses step-up of the voltage supply and can in principle deliver any power. In rare cases loudspeaker with lower impedance is used, but not very often. |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote in message ...
Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? Yes. In both cases a voltage supply of 14 volts to the amplifier, even if it is bridged. So in the first case the 14 volt has to be stepped up to U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case to 40 volts, plus losses in the design. Stepping up voltage is commonly done in high(er) power car audio power amps. There are mainly three classes of amplifiers for car audio (or more?). -The first uses a single output stage directly connected to the speaker, max output (assuming 14 volt supply and 4 ohm speakers) = (14/2/sqrt(2))^2/4=6 watts. -The second uses two output stages in a bridge and can deliver (14/sqrt(2))^2/4=24 watts -The third class uses step-up of the voltage supply and can in principle deliver any power. In rare cases loudspeaker with lower impedance is used, but not very often. |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote in message ...
Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? Yes. In both cases a voltage supply of 14 volts to the amplifier, even if it is bridged. So in the first case the 14 volt has to be stepped up to U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case to 40 volts, plus losses in the design. Stepping up voltage is commonly done in high(er) power car audio power amps. There are mainly three classes of amplifiers for car audio (or more?). -The first uses a single output stage directly connected to the speaker, max output (assuming 14 volt supply and 4 ohm speakers) = (14/2/sqrt(2))^2/4=6 watts. -The second uses two output stages in a bridge and can deliver (14/sqrt(2))^2/4=24 watts -The third class uses step-up of the voltage supply and can in principle deliver any power. In rare cases loudspeaker with lower impedance is used, but not very often. |
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[Ohms Law] Watts and Impedance?
Computer Prog wrote in message ...
Hello, Thanks for the reply, but I think you misinterpreted my question. In my example I wanted both amps to run off the same input voltage, like a car audio application. Amp1 - 200W @ 14.4V @ 8-ohms Amp1 - 200W @ 14.4V @ 4-ohms Assume that both amps are connect to their proper impedance load. Will both amps draw the same current from the car alternator when they are producing the 200W off the 14V input voltage? Yes. In both cases a voltage supply of 14 volts to the amplifier, even if it is bridged. So in the first case the 14 volt has to be stepped up to U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case to 40 volts, plus losses in the design. Stepping up voltage is commonly done in high(er) power car audio power amps. There are mainly three classes of amplifiers for car audio (or more?). -The first uses a single output stage directly connected to the speaker, max output (assuming 14 volt supply and 4 ohm speakers) = (14/2/sqrt(2))^2/4=6 watts. -The second uses two output stages in a bridge and can deliver (14/sqrt(2))^2/4=24 watts -The third class uses step-up of the voltage supply and can in principle deliver any power. In rare cases loudspeaker with lower impedance is used, but not very often. |
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[Ohms Law] Watts and Impedance?
On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote: My main question was whether the two 200W amps would draw the same current from the power source if driven by the same input voltage. For example lets say they are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at their rated impedance load. I think both amps will draw the same current from the car alternator/batter when running at 200W? The difference is the 8-ohm amp will have to send more voltage to the speaker terminals, while the 4-ohm amp will have to send more current to the speaker terminals. You're muddying the water by making them car amps :-) The basic power supply is a nominal 12 volts. Basic physics tells us that there isn't 200 watts to be had into a 4ohm load. So the amp must include some sort of voltage convertor between the battery and the power amp's supply rail. All you can do with an amplifier is present a voltage to the load. The load will determine how much current is drawn. Until it tries to draw more than the amp can provide. Which is by no means a theoretical limit. |
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[Ohms Law] Watts and Impedance?
On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote: My main question was whether the two 200W amps would draw the same current from the power source if driven by the same input voltage. For example lets say they are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at their rated impedance load. I think both amps will draw the same current from the car alternator/batter when running at 200W? The difference is the 8-ohm amp will have to send more voltage to the speaker terminals, while the 4-ohm amp will have to send more current to the speaker terminals. You're muddying the water by making them car amps :-) The basic power supply is a nominal 12 volts. Basic physics tells us that there isn't 200 watts to be had into a 4ohm load. So the amp must include some sort of voltage convertor between the battery and the power amp's supply rail. All you can do with an amplifier is present a voltage to the load. The load will determine how much current is drawn. Until it tries to draw more than the amp can provide. Which is by no means a theoretical limit. |
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[Ohms Law] Watts and Impedance?
On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote: My main question was whether the two 200W amps would draw the same current from the power source if driven by the same input voltage. For example lets say they are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at their rated impedance load. I think both amps will draw the same current from the car alternator/batter when running at 200W? The difference is the 8-ohm amp will have to send more voltage to the speaker terminals, while the 4-ohm amp will have to send more current to the speaker terminals. You're muddying the water by making them car amps :-) The basic power supply is a nominal 12 volts. Basic physics tells us that there isn't 200 watts to be had into a 4ohm load. So the amp must include some sort of voltage convertor between the battery and the power amp's supply rail. All you can do with an amplifier is present a voltage to the load. The load will determine how much current is drawn. Until it tries to draw more than the amp can provide. Which is by no means a theoretical limit. |
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[Ohms Law] Watts and Impedance?
On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote: My main question was whether the two 200W amps would draw the same current from the power source if driven by the same input voltage. For example lets say they are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at their rated impedance load. I think both amps will draw the same current from the car alternator/batter when running at 200W? The difference is the 8-ohm amp will have to send more voltage to the speaker terminals, while the 4-ohm amp will have to send more current to the speaker terminals. You're muddying the water by making them car amps :-) The basic power supply is a nominal 12 volts. Basic physics tells us that there isn't 200 watts to be had into a 4ohm load. So the amp must include some sort of voltage convertor between the battery and the power amp's supply rail. All you can do with an amplifier is present a voltage to the load. The load will determine how much current is drawn. Until it tries to draw more than the amp can provide. Which is by no means a theoretical limit. |
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[Ohms Law] Watts and Impedance?
John Fields wrote:
--- Without using a bridging configuration, the maximum power you can dissipate in an 8 ohm load from a 14.4VDC source is=20 =20 (14.4V/sqrt(2))=B2/8 ~ 13W Actually, the maximum power you can dissipate in an 8-ohm load from a=20 14.4 VDC source is simply 14.4**2/8 =3D 26 W ! This is at DC though, of course. |
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