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#161
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How Do I Determine A Power-Amp's Input Impedance?
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#162
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. |
#163
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. |
#164
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. |
#165
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. |
#166
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How Do I Determine A Power-Amp's Input Impedance?
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#167
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How Do I Determine A Power-Amp's Input Impedance?
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#168
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How Do I Determine A Power-Amp's Input Impedance?
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#169
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How Do I Determine A Power-Amp's Input Impedance?
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#171
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#172
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#173
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How Do I Determine A Power-Amp's Input Impedance?
John Fields wrote in message . ..
On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#174
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How Do I Determine A Power-Amp's Input Impedance?
I've only been following this thread loosely, so forgive me if I've missed
something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. Mark Z. -- Please reply only to Group. I regret this is necessary. Viruses and spam have rendered my regular e-mail address useless. "Svante" wrote in message om... John Fields wrote in message . .. On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#175
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How Do I Determine A Power-Amp's Input Impedance?
I've only been following this thread loosely, so forgive me if I've missed
something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. Mark Z. -- Please reply only to Group. I regret this is necessary. Viruses and spam have rendered my regular e-mail address useless. "Svante" wrote in message om... John Fields wrote in message . .. On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#176
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How Do I Determine A Power-Amp's Input Impedance?
I've only been following this thread loosely, so forgive me if I've missed
something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. Mark Z. -- Please reply only to Group. I regret this is necessary. Viruses and spam have rendered my regular e-mail address useless. "Svante" wrote in message om... John Fields wrote in message . .. On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#177
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How Do I Determine A Power-Amp's Input Impedance?
I've only been following this thread loosely, so forgive me if I've missed
something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. Mark Z. -- Please reply only to Group. I regret this is necessary. Viruses and spam have rendered my regular e-mail address useless. "Svante" wrote in message om... John Fields wrote in message . .. On 17 Jan 2004 12:14:29 -0800, (Svante) wrote: John Fields wrote in message . .. On Fri, 16 Jan 2004 22:50:00 GMT, Jeff Wiseman wrote: John Fields wrote: stuff deleted With a resistor in parallel with a capacitor, once the capacitor charges to the test voltage, current will cease to flow into it and the total resistance will look like a single resitor in parallel with an open circuit. The problem is that sometimes the capacitor is in SERIES with the rest of the circuit (and is relatively large at that). --- I don't see that as a problem in that as long as the capacitor keeps charging, the meter will display increasing resistance until the meter can no longer supply current into the cap, at which time the meter will read OL (OverLoad) or something like that. If you've got the meter leads across the _resistor_ and something like that happens, it means that either the resistor's blown open or that the OHMS range was set too low. On top of that, if you've got the meter leads across the resistor and you see resistance increasing, it means that, somehow, there's a cap in there being charged up. If it _seems_ like it's in series and not connected to the other side of the resistor, then there's something wrong because there's no way it could charge up with out a voltage difference across it, and that implies that there is a path to both sides of it which is letting current flow into it from the meter. Maybe we have to agree on the cicuit diagram of the input of the amplifier. My assumption was something like this: A --------- In+ *---Cap-----*-------| |---- | | Amp | Res | | | --------- B | | In- *-----------*------------*--------- Measuring between In+ and In- would not work IMO, due to the typically short time constant formed by the Res and Cap. If you succed to find terminals A and B it would work OK. --- I don't know what you mean by "it would not work", but if you're referring to measuring the input impedance of the amp with only a multimeter or measuring the resistance of the resistor with only a multimeter, of course it wouldn't work. I meant that connecting a standard digital multimeter, set to measure resistance, would NOT show an accurate measure of the amplifiers input impedance, not even during the first "flash" of numbers in the display. |
#178
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How Do I Determine A Power-Amp's Input Impedance?
Yes. It was one of the first things suggested, about 50 postings back.
The original poster didn't have the necessary test equipment, and as the correct setting of the crossover might interact with room acoustics, I suggested she simply assume the nominal input impedance of the unbalanced version of the amp and go from there. No doubt she has already done that and is enjoying her Infinity speakers. Has she? Inquiring minds want to know! Mark D. Zacharias wrote... I've only been following this thread loosely, so forgive me if I've missed something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. |
#179
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How Do I Determine A Power-Amp's Input Impedance?
Yes. It was one of the first things suggested, about 50 postings back.
The original poster didn't have the necessary test equipment, and as the correct setting of the crossover might interact with room acoustics, I suggested she simply assume the nominal input impedance of the unbalanced version of the amp and go from there. No doubt she has already done that and is enjoying her Infinity speakers. Has she? Inquiring minds want to know! Mark D. Zacharias wrote... I've only been following this thread loosely, so forgive me if I've missed something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. |
#180
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How Do I Determine A Power-Amp's Input Impedance?
Yes. It was one of the first things suggested, about 50 postings back.
The original poster didn't have the necessary test equipment, and as the correct setting of the crossover might interact with room acoustics, I suggested she simply assume the nominal input impedance of the unbalanced version of the amp and go from there. No doubt she has already done that and is enjoying her Infinity speakers. Has she? Inquiring minds want to know! Mark D. Zacharias wrote... I've only been following this thread loosely, so forgive me if I've missed something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. |
#181
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How Do I Determine A Power-Amp's Input Impedance?
Yes. It was one of the first things suggested, about 50 postings back.
The original poster didn't have the necessary test equipment, and as the correct setting of the crossover might interact with room acoustics, I suggested she simply assume the nominal input impedance of the unbalanced version of the amp and go from there. No doubt she has already done that and is enjoying her Infinity speakers. Has she? Inquiring minds want to know! Mark D. Zacharias wrote... I've only been following this thread loosely, so forgive me if I've missed something, but has anyone AT ALL suggested using series resistors and applying an AC input signal, then using an RMS meter to measure across the resistor and amp input? The ratio will tell the impedance. For a balanced input, use a resistor in each of the signal wires. For unbalanced, only one resistor is needed. |
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