Amp output Z

Harry Dellamano · Apr 12, 2005

"rex" <notat@hotmail.invalid> wrote in message
news:26bm511f2is46vf431udlph0hfifhanp3b@4ax.com...

On Mon, 11 Apr 2005 20:43:14 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 12 Apr 2005 01:24:06 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news:z0F6e.1871$dT4.115@newssvr13.news.prodigy.com...
Hello John,

I believe Harry wants to verify the output impedance of the amp.
Considering that it's in the 10ohm range and the load is a few hundred
the
required accuracy is a bit of a stretch for the AC circuitry of many
DVMs.

Regards, Joerg

http://www.analogconsultants.com

We are trying to measure the output Z across the 600 ohm resistive load
at
400HZ. We measure about 13 ohms using the DVM method and the formula
given
previously. No complex math is necessary because we are looking across a
stepped resistive load. We step the 600 ohm load to 300 ohms to obtain
delta
Eo.
By inspection the output Z is 600 ohms resistance in parallel with a 70
ohm reactance. Is that not about 60 ohms impedance?
What am I missing?

---
Dunno, but here's are your circuits:

+---[10R]---------+------E1A
| |
| [5.6ľF]
| |
[GEN] +------E2A
| |
| [600R]
| |
+-----------------+
| |
GND GND

+---[10R]---------+------E1B
| |
| [5.6ľF]
| |
[GEN] +------E2B
| |
| [300R]
| |
+-----------------+
| |
GND GND

How about replacing E1A, E1B, E2A, and E2B with the voltages you
actually measured at those points?

Harry, If I understand correctly, you are measuring E2 and your ro is
everything back from there including the Xc.

I get this equation:

ro = r1r2(e2-e1)/(e1r2-e2r1)

r1 and r2 are 300 and 600 for your test and e1 and e2 are the voltage
across it. Hope I got the math right.
I don't understand why you have Xc in parallel with something in your
explanation.

Hey Rex,
Love your equation but it yields the dreaded 13 ohms.
E1=14.81Vac, E2=14.50Vac, R1=600 ohms, R2=300 ohms.
Is that the correct answer at 400 Hz?
By inspection I say it's about 70 ohms but cannot prove it. The 71 ohm
reactance is in parallel with the 600 ohm load resistor.
Regards,
Harry

Jim Thompson · Apr 12, 2005

On Tue, 12 Apr 2005 01:24:06 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news:z0F6e.1871$dT4.115@newssvr13.news.prodigy.com...
Hello John,

I believe Harry wants to verify the output impedance of the amp.
Considering that it's in the 10ohm range and the load is a few hundred the
required accuracy is a bit of a stretch for the AC circuitry of many DVMs.

Regards, Joerg

http://www.analogconsultants.com

We are trying to measure the output Z across the 600 ohm resistive load at
400HZ. We measure about 13 ohms using the DVM method and the formula given
previously. No complex math is necessary because we are looking across a
stepped resistive load. We step the 600 ohm load to 300 ohms to obtain delta
Eo.
By inspection the output Z is 600 ohms resistance in parallel with a 70
ohm reactance. Is that not about 60 ohms impedance?
What am I missing?
Harry

480)460-2350 | |
| E-mail Address at Website Fax

Robert · Apr 12, 2005

"Harry Dellamano" <harryd@tdsystems.org> wrote in message
news:cpG6e.8645$jd6.1824@trnddc07...

"qrk" <SpamTrap@reson.com> wrote in message
news:fgam51dbabvmr86kfrnfcpie1rev9bsrqm@4ax.com...
On Mon, 11 Apr 2005 22:00:13 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

[snip]

Mark
Hi Mark,
If I can only measure and vary the 600 ohm resistive load then all
voltages and currents that I can measure are in phase and complex math is
useless. This yields Zo=13 ohms but it sounds wrong. I like your 70 ohms
above but don't know how to go about measuring it.
OT Got to get a new SPICE puter. You were a big help the last time so I
may bug you again.
Thanks,
Harry

Harry, when you change your resistive load you are changing the voltage
divider of it and the series capacitor mentioned. So you have changing phase
angles. Except you claim it's not in series.

It would have been better if you'd have included a diagram of how you are
hooked up and where your measurement point(s) are instead of other people
guessing from your verbal description.

Robert

Pooh Bear · Apr 12, 2005

Harry Dellamano wrote:

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
news:1lam5154fnuhm080jnc1690dk88prbjaah@4ax.com...
On Tue, 12 Apr 2005 01:24:06 GMT, the renowned "Harry Dellamano"
harryd@tdsystems.org> wrote:

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news:z0F6e.1871$dT4.115@newssvr13.news.prodigy.com...
Hello John,

I believe Harry wants to verify the output impedance of the amp.
Considering that it's in the 10ohm range and the load is a few hundred
the
required accuracy is a bit of a stretch for the AC circuitry of many
DVMs.

Regards, Joerg

http://www.analogconsultants.com

We are trying to measure the output Z across the 600 ohm resistive load
at
400HZ. We measure about 13 ohms using the DVM method and the formula given
previously. No complex math is necessary because we are looking across a
stepped resistive load. We step the 600 ohm load to 300 ohms to obtain
delta
Eo.
By inspection the output Z is 600 ohms resistance in parallel with a 70
ohm reactance. Is that not about 60 ohms impedance?

Mmm.. I think more like 69.5 ohms..

Best regards,
Spehro Pefhany
--

Hey Spehro,
If it is 69.5 ohms how do I measure/calculate this given only the load to
vary and measure with a DMM. I get 13 ohms when I measure/calculate. Do I
need other equipment and/or the node on the other side of the coupling cap?
I do like your 69.5 ohms but can't measure it.
How is your cooking show doing?
Harry

Why don't you have any real test gear ? Life's so much simpler.......

Graham

Pooh Bear · Apr 12, 2005

Joerg wrote:

Hello Spehro,

Mmm.. I think more like 69.5 ohms..

Once I pulled a little prank when the financials were discussed. Asked
about what my confidence level would be for the margin I replied "plus
minus 3dB". Some jaws dropped so I quickly explained that this was a
joke and promised to myself never to do that again.

Humourless shits !

Graham

Pooh Bear · Apr 12, 2005

Joerg wrote:

Hello Spehro,

Mmm.. I think more like 69.5 ohms..

Once I pulled a little prank when the financials were discussed. Asked
about what my confidence level would be for the margin I replied "plus
minus 3dB". Some jaws dropped so I quickly explained that this was a
joke and promised to myself never to do that again.

Ummmm - as a follow-up. I prefer +0 -0.5 dB. That's what I design to
anyway. On a good day, -0.3dB even !

Graham

Spehro Pefhany · Apr 12, 2005

On Tue, 12 Apr 2005 02:08:44 GMT, the renowned "Harry Dellamano"
<harryd@tdsystems.org> wrote:

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
news:1lam5154fnuhm080jnc1690dk88prbjaah@4ax.com...
On Tue, 12 Apr 2005 01:24:06 GMT, the renowned "Harry Dellamano"
harryd@tdsystems.org> wrote:

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news:z0F6e.1871$dT4.115@newssvr13.news.prodigy.com...
Hello John,

I believe Harry wants to verify the output impedance of the amp.
Considering that it's in the 10ohm range and the load is a few hundred
the
required accuracy is a bit of a stretch for the AC circuitry of many
DVMs.

Regards, Joerg

http://www.analogconsultants.com

We are trying to measure the output Z across the 600 ohm resistive load
at
400HZ. We measure about 13 ohms using the DVM method and the formula given
previously. No complex math is necessary because we are looking across a
stepped resistive load. We step the 600 ohm load to 300 ohms to obtain
delta
Eo.
By inspection the output Z is 600 ohms resistance in parallel with a 70
ohm reactance. Is that not about 60 ohms impedance?

Mmm.. I think more like 69.5 ohms..

Best regards,
Spehro Pefhany
--

Hey Spehro,
If it is 69.5 ohms how do I measure/calculate this given only the load to
vary and measure with a DMM. I get 13 ohms when I measure/calculate. Do I
need other equipment and/or the node on the other side of the coupling cap?
I do like your 69.5 ohms but can't measure it.

Assuming my guess is right on your setup, I think you'll have problems
with this way of measuring Zo. Xc will dominate (95% of the output
change) the *magnitude* of output change, and you're not changing the
loading a huge amount, so there's not a lot left to get any kind of
usable result, even with fairly bold assumptions. But I'll wait for
the equation from Jim to plug the numbers in, it's late. ;-)

Maybe you can get rid of the cap or increase it by several orders of
magnitude for this test?

How is your cooking show doing?
Harry

I'll refrain from mentioning "boil"..

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Ross Herbert · Apr 12, 2005

On Mon, 11 Apr 2005 22:00:13 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

Ok, so I got this amplifier with an internal output resistance of 10 ohms
and we are outside of all feedback loops. It is coupled to the 600 ohm
resistive load thru a 5.6uF capacitor. The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected. If I measure the Zo with SPICE
using a swept 1 amp current source at the output, I get about 71 ohms which
is the Xc of the coupling cap. If I use a DVM on the real circuit and
measure the output voltage at two different loads, 600 ohms and 300 ohms, I
get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure
the Zo using a DVM and not Ro. By observation it must be 71 ohms. What must
I do??
Harry

See
http://www.audioholics.com/techtips/audioprinciples/amplifiers/basicamplifiermeasurements.php

Tony Williams · Apr 12, 2005

In article <NrC6e.8223$jd6.3292@trnddc07>,
Harry Dellamano <harryd@tdsystems.org> wrote:

Ok, so I got this amplifier with an internal output resistance
of 10 ohms and we are outside of all feedback loops. It is
coupled to the 600 ohm resistive load thru a 5.6uF capacitor.
The output voltage of the amp is 10.0Vrms at 400 Hz with no
loads connected. If I measure the Zo with SPICE using a swept 1
amp current source at the output, I get about 71 ohms which is
the Xc of the coupling cap. If I use a DVM on the real circuit
and measure the output voltage at two different loads, 600 ohms
and 300 ohms, I get about 13.0 ohms using
Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure the Zo using a
DVM and not Ro. By observation it must be 71 ohms. What must I
do??

Hello Harry. I think you are missing something.

I Ro X
+->-----/\/\----||----+ <---E = I*RL
| |
| \
[Vgen] /RL = R1 or R2
| \
| |
+---------------------+---0v

2 2 2
I = Vgen/sq-rt(Ro + X + RL)

2 2 2
E1 = R1*Vgen/sq-rt(Ro + X + R1)

E2 is a similar sum. Square+Divide them to give:-

2 2 2 2 2
Ro + X + R1 R1 E2
----------- = ---- * ----
2 2 2 2 2
Ro + X + R2 E1 R2

So if you know Ro/R1/R2 and measure E1/E2 you can
(in theory) calculate X.

Trouble is, when X is small compared with delta-R
an almost impossible precision is required.
For example, your 13-ohm calc suggests an E1/E2 ratio
of 0.97785, whereas a 71ohm Xc via the sums above would
need an E1/E2 ratio of 0.97847.... and that's assuming
that you know the 300/600 ohms to the same precision.

You *might* improve things by keeping the DVM across a
300 ohm, and switching the second 300 ohm in/out.
Measure the change in current to get X. Still marginal.

--
Tony Williams.

rex · Apr 12, 2005

On Tue, 12 Apr 2005 03:13:05 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

Hey Rex,
Love your equation but it yields the dreaded 13 ohms.
E1=14.81Vac, E2=14.50Vac, R1=600 ohms, R2=300 ohms.
Is that the correct answer at 400 Hz?
By inspection I say it's about 70 ohms but cannot prove it. The 71 ohm
reactance is in parallel with the 600 ohm load resistor.
Regards,
Harry

Ok, forget all that. As I mentioned, my first approach was wrong.

Here's my attempt at one approach to a proper attempt to find what I
think you are trying to get...

Here is your circuit as I see it.

----Ri-------Xc-------
| |
Es Rm Vm
| |
----------------------

Es is the source voltage and Rm is the load you are applying.

I think you are changing the value of Rm and measuring the ac voltage Vm
across that resistance.

As I understand it, you are trying to calculate Ri from the measured
values. You said you measured (Vm, Rm) as (14.81, 600) and (14.50, 300)

The total impedance that Es sees is Z = Rm + Ri -jXc

Xc = 1/(2 pi f C) = 1/(6.283 * 400 * 5.6E-6) = 71.05 ; lets call it 71

The current flowing:
i = Es / |Z|

or from the measurement:
i = Vm / Rm

Equating the i equations and solving for Es:

Es = Vm/Rm * |Rm + Ri - jXc| = Vm/Rm (sqrt((Ri + Rm)^2 + 71^2))

Es is constant for both measurements so, substituting your measurements:

14.81/600 (sqrt((Ri + 600)^2 + 71^2)) = 14.50/300 (sqrt((Ri + 300)^2 +
71^2))

when I solve that for Ri (with the help of math software), I get

Ri = 0.567 ohm

Just for fun, I calculated Es at about 14.9 V

Hopefully, I didn't completely screw this up.

So Spehro is right. There is not much there to measure with your setup
and Xc dominates.

Just as a sanity check, I removed Ri assuming the circuit is only Xc and
Rm. I let Es be 14.9 V and calculated Vm for Rm=600 and Rm=300. I got
14.80 and 14.50, which is almost exactly what you measured. So we can
assume Ri is low enough in your circuit to be in the noise with your
measurement.

John Woodgate · Apr 12, 2005

I read in sci.electronics.design that Harry Dellamano
<harryd@tdsystems.org> wrote (in <NrC6e.8223$jd6.3292@trnddc07&gt

about
'Amp output Z', on Mon, 11 Apr 2005:

Ok, so I got this amplifier with an internal output resistance of 10 ohms
and we are outside of all feedback loops. It is coupled to the 600 ohm
resistive load thru a 5.6uF capacitor. The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected. If I measure the Zo with SPICE
using a swept 1 amp current source at the output, I get about 71 ohms which
is the Xc of the coupling cap. If I use a DVM on the real circuit and
measure the output voltage at two different loads, 600 ohms and 300 ohms, I
get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure
the Zo using a DVM and not Ro. By observation it must be 71 ohms. What must
I do??
Harry

I am astonished at all the discussion on this. It actually isn't helped

by your loose terminology. You've been using 'output impedance' to mean
both the output source impedance and the load impedance. You also seem
confused between series and parallel (or perhaps you are mixed up
between Thévenin and Norton equivalents).

You DON'T need complex numbers as such, but you DO need to take the
reactance of the capacitor into account.

Assuming that the output source resistance is 10 ohms, as shown in the
various circuits that have been posted, The voltages you should measure
with your DMM are related by:

600/{sqrt(610^2 + 71^2)} and 300/(sqrt{310^2 + 71^2)}

Their ratio is 0.9655. How does that compare with the ratio of the
voltages you measure?
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

John Woodgate · Apr 12, 2005

I read in sci.electronics.design that Harry Dellamano
<harryd@tdsystems.org> wrote (in <51H6e.8649$jd6.6481@trnddc07&gt

about
'Amp output Z', on Tue, 12 Apr 2005:

Love your equation but it yields the dreaded 13 ohms.
E1=14.81Vac, E2=14.50Vac, R1=600 ohms, R2=300 ohms.

To get 14.81 V out, you need supplies exceeding +/- 21 V, otherwise your
output waveform is clipped.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Pooh Bear · Apr 12, 2005

Harry Dellamano wrote:

"qrk" <SpamTrap@reson.com> wrote in message
news:fgam51dbabvmr86kfrnfcpie1rev9bsrqm@4ax.com...
On Mon, 11 Apr 2005 22:00:13 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

Ok, so I got this amplifier with an internal output resistance of 10
ohms
and we are outside of all feedback loops. It is coupled to the 600 ohm
resistive load thru a 5.6uF capacitor. The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected. If I measure the Zo with SPICE
using a swept 1 amp current source at the output, I get about 71 ohms
which
is the Xc of the coupling cap. If I use a DVM on the real circuit and
measure the output voltage at two different loads, 600 ohms and 300 ohms,
I
get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300)). I need to
measure
the Zo using a DVM and not Ro. By observation it must be 71 ohms. What
must
I do??

Harry,
Few things are suspect.
Your equation has some sign and parentheses problems.
It might be Ro=(E1-E2)/((E2/300)-(E1/600))

Your equation is for pure resistance. This is a complex number
problem. Use Spice to measure the output amplitude like you would with
a DMM. I'll bet you end up with the same wrong answer.

You need to deal with complex numbers since your series capacitor
introduces a significant phase angle at 400 Hz. At 400 Hz your output
impedance is 10-j71 Ohms.

Mark
Hi Mark,
If I can only measure and vary the 600 ohm resistive load then all voltages
and currents that I can measure are in phase and complex math is useless.
This yields Zo=13 ohms but it sounds wrong. I like your 70 ohms above but
don't know how to go about measuring it.

You could try measuring the voltage across the cap for starters. Won't be in
phase with the load - but it'll tell you the load current !

OT Got to get a new SPICE puter.

I'm sorry but you have to be the biggest example I've seen to date of believing
simulations over real practice. If you'd learnt Electronics 101 you wouldn't
have to ask this kind of stuff.

Graham

Pooh Bear · Apr 12, 2005

Tony Williams wrote:

In article <NrC6e.8223$jd6.3292@trnddc07>,
Harry Dellamano <harryd@tdsystems.org> wrote:
Ok, so I got this amplifier with an internal output resistance
of 10 ohms and we are outside of all feedback loops. It is
coupled to the 600 ohm resistive load thru a 5.6uF capacitor.
The output voltage of the amp is 10.0Vrms at 400 Hz with no
loads connected. If I measure the Zo with SPICE using a swept 1
amp current source at the output, I get about 71 ohms which is
the Xc of the coupling cap. If I use a DVM on the real circuit
and measure the output voltage at two different loads, 600 ohms
and 300 ohms, I get about 13.0 ohms using
Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure the Zo using a
DVM and not Ro. By observation it must be 71 ohms. What must I
do??

Hello Harry. I think you are missing something.

I Ro X
+->-----/\/\----||----+ <---E = I*RL
| |
| \
[Vgen] /RL = R1 or R2
| \
| |
+---------------------+---0v

2 2 2
I = Vgen/sq-rt(Ro + X + RL)

2 2 2
E1 = R1*Vgen/sq-rt(Ro + X + R1)

E2 is a similar sum. Square+Divide them to give:-

2 2 2 2 2
Ro + X + R1 R1 E2
----------- = ---- * ----
2 2 2 2 2
Ro + X + R2 E1 R2

So if you know Ro/R1/R2 and measure E1/E2 you can
(in theory) calculate X.

Trouble is, when X is small compared with delta-R
an almost impossible precision is required.
For example, your 13-ohm calc suggests an E1/E2 ratio
of 0.97785, whereas a 71ohm Xc via the sums above would
need an E1/E2 ratio of 0.97847.... and that's assuming
that you know the 300/600 ohms to the same precision.

You *might* improve things by keeping the DVM across a
300 ohm, and switching the second 300 ohm in/out.
Measure the change in current to get X. Still marginal.

OMG - you've had to resort to teaching him basic *electricity* !

Why, Why, Why do ppl post questions here illustrating their total
absence of knowledge of the art / science of electronics ?

Is it that the education sytem is fucked ? ( most likely IMHO actually )
or do ppl just get degrees by attending class ? ( also quite probable
IMHO )

Graham

Tony Williams · Apr 12, 2005

On 12 Apr, tonyw@ledelec.demon.co.uk wrote:

I Ro X
+->-----/\/\----||----+ <---E = I*RL
| |
| \
[Vgen] /RL = R1 or R2
| \
| |
+---------------------+---0v
[snip]

2 2 2 2 2
Ro + X + R1 R1 E2
----------- = ---- * ----
2 2 2 2 2
Ro + X + R2 E1 R2

Harry. I see now that you gave in another post,
E1= 14.81Vac, E2= 14.5Vac for R1= 600 and R2= 300.

Plugging those numbers into the above sum gives
X= 35.8 ohms.

That's with an assumed Ro of 10 ohms.

If Ro= 0 the sum gives X= 72.5 ohms.

--
Tony Williams.

Tony Williams · Apr 12, 2005

In article <425B8517.6C6454FE@hotmail.com>,
Pooh Bear <rabbitsfriendsandrelations@hotmail.com> wrote:

OMG - you've had to resort to teaching him basic *electricity* !

I suspect that Harry has just had a temporary brain
phart, that's all..... I recognise the symptoms.

--
Tony Williams.

John Woodgate · Apr 12, 2005

I read in sci.electronics.design that Tony Williams
<tonyw@ledelec.demon.co.uk> wrote (in
<4d5a52d515tonyw@ledelec.demon.co.uk&gt

about 'Amp output Z', on Tue, 12
Apr 2005:

Hello Harry. I think you are missing something.

I Ro X
+->-----/\/\----||----+ <---E = I*RL
| |
| \
[Vgen] /RL = R1 or R2
| \
| |
+---------------------+---0v

2 2 2
I = Vgen/sq-rt(Ro + X + RL)

The current I = Vgen/sqrt(R^2 + X^2), where R = Ro + RL. Series
resistances do not add in quadrature.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Pooh Bear · Apr 12, 2005

rex wrote:

On Tue, 12 Apr 2005 03:13:05 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

Hey Rex,
Love your equation but it yields the dreaded 13 ohms.
E1=14.81Vac, E2=14.50Vac, R1=600 ohms, R2=300 ohms.
Is that the correct answer at 400 Hz?
By inspection I say it's about 70 ohms but cannot prove it. The 71 ohm
reactance is in parallel with the 600 ohm load resistor.
Regards,
Harry

Ok, forget all that. As I mentioned, my first approach was wrong.

Here's my attempt at one approach to a proper attempt to find what I
think you are trying to get...

Here is your circuit as I see it.

----Ri-------Xc-------
| |
Es Rm Vm
| |
----------------------

Es is the source voltage and Rm is the load you are applying.

I think you are changing the value of Rm and measuring the ac voltage Vm
across that resistance.

As I understand it, you are trying to calculate Ri from the measured
values. You said you measured (Vm, Rm) as (14.81, 600) and (14.50, 300)

The total impedance that Es sees is Z = Rm + Ri -jXc

Xc = 1/(2 pi f C) = 1/(6.283 * 400 * 5.6E-6) = 71.05 ; lets call it 71

The current flowing:
i = Es / |Z|

or from the measurement:
i = Vm / Rm

Equating the i equations and solving for Es:

Es = Vm/Rm * |Rm + Ri - jXc| = Vm/Rm (sqrt((Ri + Rm)^2 + 71^2))

Es is constant for both measurements so, substituting your measurements:

14.81/600 (sqrt((Ri + 600)^2 + 71^2)) = 14.50/300 (sqrt((Ri + 300)^2 +
71^2))

when I solve that for Ri (with the help of math software), I get

Ri = 0.567 ohm

Just for fun, I calculated Es at about 14.9 V

Hopefully, I didn't completely screw this up.

So Spehro is right. There is not much there to measure with your setup
and Xc dominates.

Just as a sanity check, I removed Ri assuming the circuit is only Xc and
Rm. I let Es be 14.9 V and calculated Vm for Rm=600 and Rm=300. I got
14.80 and 14.50, which is almost exactly what you measured. So we can
assume Ri is low enough in your circuit to be in the noise with your
measurement.

Can you explain that again using proper science and analytical methods ?

Graham

Pooh Bear · Apr 12, 2005

John Woodgate wrote:

I read in sci.electronics.design that Harry Dellamano
harryd@tdsystems.org> wrote (in <NrC6e.8223$jd6.3292@trnddc07&gt about
'Amp output Z', on Mon, 11 Apr 2005:
Ok, so I got this amplifier with an internal output resistance of 10 ohms
and we are outside of all feedback loops. It is coupled to the 600 ohm
resistive load thru a 5.6uF capacitor. The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected. If I measure the Zo with SPICE
using a swept 1 amp current source at the output, I get about 71 ohms which
is the Xc of the coupling cap. If I use a DVM on the real circuit and
measure the output voltage at two different loads, 600 ohms and 300 ohms, I
get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure
the Zo using a DVM and not Ro. By observation it must be 71 ohms. What must
I do??
Harry

I am astonished at all the discussion on this. It actually isn't helped
by your loose terminology. You've been using 'output impedance' to mean
both the output source impedance and the load impedance. You also seem
confused between series and parallel (or perhaps you are mixed up
between Thévenin and Norton equivalents).

You DON'T need complex numbers as such, but you DO need to take the
reactance of the capacitor into account.

Assuming that the output source resistance is 10 ohms, as shown in the
various circuits that have been posted, The voltages you should measure
with your DMM are related by:

600/{sqrt(610^2 + 71^2)} and 300/(sqrt{310^2 + 71^2)}

WooHoo ! John came to the same conclusions as me ! OK I used 'mod' as shorthand.

Their ratio is 0.9655. How does that compare with the ratio of the
voltages you measure?

I bet it's spot on !

Graham

Tony Williams · Apr 12, 2005

In article <tBWPIhQOb5WCFw3b@jmwa.demon.co.uk>,
John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Tony Williams
tonyw@ledelec.demon.co.uk> wrote
2 2 2
I = Vgen/sq-rt(Ro + X + RL)

The current I = Vgen/sqrt(R^2 + X^2), where R = Ro + RL. Series
resistances do not add in quadrature.

"Brain Phart. I recognise the symptoms."

--
Tony Williams.

Amp output Z

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