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#1
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"Fixed" 8 volt output o-loads eq
My deck (Eclipse 8053) has biamp outputs - provided the "rear" channels are
switched to become "mid" and the front channels become the "high" outputs. So, in biamp mode I have no rear channels---but the deck also has a "Fixed Out" that runs at full volume only. I ran this fixed out to an old alpine 10-band that I am using to control the rear speakers. Fuzztone. On the EQ, there is an led array that shows total overload unless I lower all the bands to minus 12 db (or -15 db, maybe--I forget) and it's still slightly distorted at those settings. My plan is to try a 10k resistor (guessing on the value here) inline with the tip of the interconnect rca before the EQ. Is this plan valid??? Another idea, better, would be to able to sum the mids & highs for the back speakers *without having them be summed to the front speakers. That way, fronts and rear would be controlled by the head unit's volume control. I don't know how to build this circuit, other than buying a little mixer. |
#2
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"Fixed" 8 volt output o-loads eq
My plan is to try a 10k resistor (guessing on the value here) inline with
the tip of the interconnect rca before the EQ. Check your amp's manual. A lot of times there's a value for input impedance. It's generally in the 10-30k range. Then just treat it as a voltage divider. PS Are you saying that this output isn't controlled by the volume knob? Is this plan valid??? Another idea, better, would be to able to sum the mids & highs for the back speakers *without having them be summed to the front speakers. That way, fronts and rear would be controlled by the head unit's volume control. I don't know how to build this circuit, other than buying a little mixer. It's a pain. Leave it alone. Why not run just the low output to the rears? Where's the cutoff? |
#3
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"Fixed" 8 volt output o-loads eq
Yes the "fixed" output I am trying to tone down is not controlled by any voulme control on the head unit, other than the mute control. I need a full range signal and not just mid-bass at the request of my daughter passengers in the back. The xover is set for 1.6k hz, nice and low The spec in the manual says 55 ohm output impedance. (not 55k) So does that mean if I build an interconnect with a 55 ohm resistor in the tip, the signal will be reduced 3 db?? On Wed, 18 Feb 2004 08:11:31 -0500, "MZ" wrote: My plan is to try a 10k resistor (guessing on the value here) inline with the tip of the interconnect rca before the EQ. Check your amp's manual. A lot of times there's a value for input impedance. It's generally in the 10-30k range. Then just treat it as a voltage divider. PS Are you saying that this output isn't controlled by the volume knob? Is this plan valid??? Another idea, better, would be to able to sum the mids & highs for the back speakers *without having them be summed to the front speakers. That way, fronts and rear would be controlled by the head unit's volume control. I don't know how to build this circuit, other than buying a little mixer. It's a pain. Leave it alone. Why not run just the low output to the rears? Where's the cutoff? |
#4
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"Fixed" 8 volt output o-loads eq
The spec in the manual says 55 ohm output impedance. (not 55k)
So does that mean if I build an interconnect with a 55 ohm resistor in the tip, the signal will be reduced 3 db?? No, that's the output impedance of the head unit. What you need is the input impedance of the amplifier. Oh yeah, and you may want to reduce by more than 3dB. |
#5
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"Fixed" 8 volt output o-loads eq
The spec in the manual says 55 ohm output impedance. (not 55k)
So does that mean if I build an interconnect with a 55 ohm resistor in the tip, the signal will be reduced 3 db?? No, that's the output impedance of the head unit. What you need is the input impedance of the amplifier. Oh yeah, and you may want to reduce by more than 3dB. Oh yeah, the other way to do it is to use 2 resistors and create a voltage divider. Really, you don't need to know the input impedance. All it does for you is give you a good starting point. If you can't find it out, just assume 20k ohms. |
#6
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"Fixed" 8 volt output o-loads eq
PS-
If you use a potentiometer (say, 200k ohms or more range), you can dial in the ideal value of resistance, then use a resistor of that value as the permanent solution. -- Mark remove "remove" and "spam" to reply "MZ" wrote in message ... The spec in the manual says 55 ohm output impedance. (not 55k) So does that mean if I build an interconnect with a 55 ohm resistor in the tip, the signal will be reduced 3 db?? No, that's the output impedance of the head unit. What you need is the input impedance of the amplifier. Oh yeah, and you may want to reduce by more than 3dB. Oh yeah, the other way to do it is to use 2 resistors and create a voltage divider. Really, you don't need to know the input impedance. All it does for you is give you a good starting point. If you can't find it out, just assume 20k ohms. |
#7
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"Fixed" 8 volt output o-loads eq
200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it. In article , "MZ" wrote: PS- If you use a potentiometer (say, 200k ohms or more range), you can dial in the ideal value of resistance, then use a resistor of that value as the permanent solution. -- Mark remove "remove" and "spam" to reply "MZ" wrote in message ... The spec in the manual says 55 ohm output impedance. (not 55k) So does that mean if I build an interconnect with a 55 ohm resistor in the tip, the signal will be reduced 3 db?? No, that's the output impedance of the head unit. What you need is the input impedance of the amplifier. Oh yeah, and you may want to reduce by more than 3dB. Oh yeah, the other way to do it is to use 2 resistors and create a voltage divider. Really, you don't need to know the input impedance. All it does for you is give you a good starting point. If you can't find it out, just assume 20k ohms. |
#8
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"Fixed" 8 volt output o-loads eq
200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it. No, 1kohm would be worthless. It would provide somewhere on the order of 0.2dB attenuation, which would be inaudible. I suggested 200k ohm as a POT value, so that he has sufficient range to make the determination. I suspect the resistance value that will provide the results he's after won't exceed ~50kohm. |
#9
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"Fixed" 8 volt output o-loads eq
I'll try the pot. How do I wire it? Like this: _____~~~~~~~~_ to head __ ^ | |________________ |_______________________to eq Or do I want the wiper to sweep to ground? On Thu, 19 Feb 2004 14:18:40 -0500, "MZ" wrote: 200K is way too high. The impedance will be too high to overcome cable capacitance and electric noise. 1K would be more like it. No, 1kohm would be worthless. It would provide somewhere on the order of 0.2dB attenuation, which would be inaudible. I suggested 200k ohm as a POT value, so that he has sufficient range to make the determination. I suspect the resistance value that will provide the results he's after won't exceed ~50kohm. |
#10
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"Fixed" 8 volt output o-loads eq
No, don't run it to ground. You don't want to short things. Just run the
wiper out so that it forms a series circuit with the amplifier input. Essentially, you're just treating it as a variable series resistor. -- Mark remove "remove" and "spam" to reply "Jim Hoff" wrote in message ... I'll try the pot. How do I wire it? Like this: _____~~~~~~~~_ to head __ ^ | |________________ |_______________________to eq Or do I want the wiper to sweep to ground? On Thu, 19 Feb 2004 14:18:40 -0500, "MZ" wrote: 200K is way too high. The impedance will be too high to overcome cable capacitance and electric noise. 1K would be more like it. No, 1kohm would be worthless. It would provide somewhere on the order of 0.2dB attenuation, which would be inaudible. I suggested 200k ohm as a POT value, so that he has sufficient range to make the determination. I suspect the resistance value that will provide the results he's after won't exceed ~50kohm. |
#11
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"Fixed" 8 volt output o-loads eq
Ok I'll try it like my diagram above and post if it works.
The only stereo potentiometer in town (Juneau, AK) was a 100k ohm. Thanks. "MZ" wrote in message ... No, don't run it to ground. You don't want to short things. Just run the wiper out so that it forms a series circuit with the amplifier input. Essentially, you're just treating it as a variable series resistor. -- Mark remove "remove" and "spam" to reply "Jim Hoff" wrote in message ... I'll try the pot. How do I wire it? Like this: _____~~~~~~~~_ to head __ ^ | |________________ |_______________________to eq Or do I want the wiper to sweep to ground? On Thu, 19 Feb 2004 14:18:40 -0500, "MZ" wrote: 200K is way too high. The impedance will be too high to overcome cable capacitance and electric noise. 1K would be more like it. No, 1kohm would be worthless. It would provide somewhere on the order of 0.2dB attenuation, which would be inaudible. I suggested 200k ohm as a POT value, so that he has sufficient range to make the determination. I suspect the resistance value that will provide the results he's after won't exceed ~50kohm. |
#12
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"Fixed" 8 volt output o-loads eq
One side to the 8V audio output, one side to the signal ground, and the
wiper to the low voltage amp input. Use a pot in the area of 1K to 4K. The 200K series-only pot idea doesn't work because the impedance will become too high. The impedance to the amp should be no higher than a few K Ohms to avoid loss of trebble, electrical noise pickup, and distortion due to non-linearities of the amp input's impedance. In article , Jim Hoff wrote: I'll try the pot. How do I wire it? Like this: _____~~~~~~~~_ to head __ ^ | |________________ |_______________________to eq Or do I want the wiper to sweep to ground? On Thu, 19 Feb 2004 14:18:40 -0500, "MZ" wrote: 200K is way too high. The impedance will be too high to overcome cable capacitance and electric noise. 1K would be more like it. No, 1kohm would be worthless. It would provide somewhere on the order of 0.2dB attenuation, which would be inaudible. I suggested 200k ohm as a POT value, so that he has sufficient range to make the determination. I suspect the resistance value that will provide the results he's after won't exceed ~50kohm. |
#13
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"Fixed" 8 volt output o-loads eq
One side to the 8V audio output, one side to the signal ground, and the
wiper to the low voltage amp input. Use a pot in the area of 1K to 4K. dB = 10ln(A/B) = 10ln(20/21) = -0.4dB Sorry. Won't be effective. The 200K series-only pot idea doesn't work because the impedance will become too high. The impedance to the amp should be no higher than a few K Ohms to avoid loss of trebble, electrical noise pickup, and distortion due to non-linearities of the amp input's impedance. What nonlinearities? Um...and he can't set the amp's input impedance. It's predetermined by the 10kohm or greater resistor inside the chassis. He'll modify the input impedance some by providing a series resistance, but it still shouldn't be too high unless he has to introduce some major attenuation. |
#14
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"Fixed" 8 volt output o-loads eq
In article ,
"MZ" wrote: One side to the 8V audio output, one side to the signal ground, and the wiper to the low voltage amp input. Use a pot in the area of 1K to 4K. dB = 10ln(A/B) = 10ln(20/21) = -0.4dB Sorry. Won't be effective. It will be perfectly effective when wired correctly. This isn't a series resistor. It's a voltage divider where the wiper varies between the signal and the signal's ground. The 200K series-only pot idea doesn't work because the impedance will become too high. The impedance to the amp should be no higher than a few K Ohms to avoid loss of trebble, electrical noise pickup, and distortion due to non-linearities of the amp input's impedance. What nonlinearities? Amplifiers with the classic differential bipolar transistor inputs can present a varying impedance if the circuit is not designed well. Since car amps are often cheaply designed and a low impedance input is expected, there could be significant distortion if a high value series resistor is used. Um...and he can't set the amp's input impedance. It's predetermined by the 10kohm or greater resistor inside the chassis. He'll modify the input impedance some by providing a series resistance, but it still shouldn't be too high unless he has to introduce some major attenuation. |
#15
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"Fixed" 8 volt output o-loads eq
One side to the 8V audio output, one side to the signal ground, and
the wiper to the low voltage amp input. Use a pot in the area of 1K to 4K. dB = 10ln(A/B) = 10ln(20/21) = -0.4dB Sorry. Won't be effective. It will be perfectly effective when wired correctly. This isn't a series resistor. It's a voltage divider where the wiper varies between the signal and the signal's ground. He'd be better off using a resistor for the final solution rather than a pot, considering his application. Again, I only recommended the pot so that he could pinpoint the correct resistance value. As I mentioned in a previous post, a v divider would work fine. But it's not necessary. The 200K series-only pot idea doesn't work because the impedance will become too high. The impedance to the amp should be no higher than a few K Ohms to avoid loss of trebble, electrical noise pickup, and distortion due to non-linearities of the amp input's impedance. What nonlinearities? Amplifiers with the classic differential bipolar transistor inputs can present a varying impedance if the circuit is not designed well. Since car amps are often cheaply designed and a low impedance input is expected, there could be significant distortion if a high value series resistor is used. Low impedance input isn't expected. The vast majority of amplifiers have input impedances greater than 10kohms. Most of the time, the inputs look like a load resistor and an op amp. |
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