In article writes:

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 this is an answer from someone in rec.audio.tech, you aren't very
tech. And if it's an answer from someone in rec.audio.pro, you
obviously didn't read my articles about basic electricity in Recording
Magazine last year (or anythng else about the subject).

We're talking about a capacitor IN SERIES with the unknown resistance
here. And we're also talking about an ohm meter that uses a direct
current and Ohm's Law to display resistance on a screen or scale
(look, Ma, and analog computer!). To a direct current, a capacitor
initially looks like a short circuit, but that only lasts for a very
short time - infintessimally short in fact. Once current starts
flowing through the circuit, the capacitor begins to charge, and the
current drops. Since the fixed resistance in the circuit doesn't
change, and the voltage source (the ohm meter) doesn't change, only
the current can change. Since an ohm meter really measures current,
the ohm meter reading will change.

The ultimate case, at time infinity (though a couple of seconds is
usually close enough for this circuit in audio gear) is that the
capacitor is fully charged to the same voltage as the meter's source,
no current will flow, and the analog computer will try its best to
divide by zero (the current), indicating infinite ohms, an open
circuit, or a very high (hundreds of megohms) resistance, depending on
how the meter treats a value higher than it an measure.

If the capacitor is in parallel with the impedance you're trying to
measure, you're in a slightly better position, because after the
capacitor is charged to the point where it draws much less current
than the impedance you're trying to measure, it will essentially go
away.

However, with practical components and a practical household meter,
you cannot use an ohm meter to measure resistance of a circuit in
which current flows from the meter into a capacitor.



--
I'm really Mike Rivers - )
However, until the spam goes away or Hell freezes over,
lots of IP addresses are blocked from this system. If
you e-mail me and it bounces, use your secret decoder ring
and reach me he double-m-eleven-double-zero at yahoo

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.

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.

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.

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.

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.

--
John Fields

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.

--
John Fields

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.

--
John Fields

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.

--
John Fields

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.