Amp output Z

[Pooh Bear]

Harry Dellamano wrote:

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in message
news:SwpQ0612yBXCFwav@jmwa.demon.co.uk...
I read in sci.electronics.design that Jim Thompson
I posted the results of using an assumed 10 ohms for Ro earlier. I'm a bit
surprised at the problems people have with rather simple impedance
calculations. However, the calculation of Ro from the measurements is
messy and, under the stated conditions, not very reliable. Rather than
seek an explicit expression (Ro = something huge and horrible), it may be
much easier to put numbers into the equations and work from there. You can
measure the open-circuit output voltage and put that value into the
equations.

Hi John,
I am looking for Zo not Ro. I cannot measure the open circuit output
voltage because the amp is unstable with no load. This is a customers design
that I have to redesign. He believes that the output Z is about 10 ohms
which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but
cannot prove it. The spec requirement is less than 25 ohms.

Have you considered measuring the voltage - not at the load - but at the amp
output - before the coupling cap ?

What do you get ? That result will teel you how much effect is due purely to the
amp's output resistance instead of having to guestimate it.

Wish you'd said all that in the first place.

Graham

 

[John Woodgate]

I read in sci.electronics.design that Harry Dellamano
<harryd@tdsystems.org> wrote (in <op_6e.10644$Zn3.615@trnddc02&gt about
'Amp output Z', on Wed, 13 Apr 2005:

I am looking for Zo not Ro. I cannot measure the open circuit output
voltage because the amp is unstable with no load. This is a customers
design that I have to redesign. He believes that the output Z is about
10 ohms which is easy to measure by Ro=dVo/dRL. I believe it to be 70
ohms but cannot prove it. The spec requirement is less than 25 ohms.

The mere presence of the 5.6 uF capacitor tells you that it is at least
71 ohms at 400 Hz. Of course, it varies with frequency. The *resistive*
part may be only 10 ohms, which give 71.7 ohms for the magnitude of Zo
at 400 Hz.

To get less than 25 ohms, you must increase the capacitance value. There
isn't much point in calculating an exact value, because you can only get
22 uF, 33 uF and 47 uF, and even 33 uF may be difficult. 22 uF is 18
ohms at 400 Hz, so Zo = sqrt(18^2 + 10^2) = 20.6 ohms. I'd go for 47 uF
if 400 Hz is the lowest frequency of interest and you have room for 47
uF.
--
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

 

[Fred Bartoli]

"Harry Dellamano" <harryd@tdsystems.org> a écrit dans le message de
news:NrC6e.8223$jd6.3292@trnddc07...

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

Harry,

I've not seen someone suggest the obvious: why don't you directly measure
the output impedance?

Terminate your amp to 600R to make it stable. Inject some current into the
output through a 1K or more resistor or a current source.

Measure the voltage and phase and then correct for the output load (600R
resistor + current source impedance).

Any measurement error will have much less impact than with your method.


--
Thanks,
Fred.

 

[The Phantom]

On Wed, 13 Apr 2005 01:41:33 GMT, "Harry Dellamano" <harryd@tdsystems.org> wrote:

"Jim Thompson" <thegreatone@example.com> wrote in message
news:tlvn511djoqu9enbu0skimii42ulthedps@4ax.com...
On Tue, 12 Apr 2005 16:56:34 +0100, John Woodgate
Yep, I should have reduced it to what the meter actually shows.
However I'm pretty sure you can't get the answer you want.

...Jim Thompson
Hey Jim, your killing me!
Are you saying that we cannot measure Zo because there is a coupling cap
there? Given that the unit is not stable with an open load but I can measure
the voltage on the either side of the coupling cap and vary the load.
Instead of varying a resistive load, why don't we vary a capacitive load?
That should cause a voltage divider with the coupling cap and yield readings
to prove the Zo is about 70 ohms, not 12 ohms.
Regards,
Harry

You don't have enough measurements to fully determine the output inpedance.

It looks to me as though the locus of possible output impedances is a circle with center
at about 10+j47 and with a radius of about 24.

If there is no relative phase shift between E2A and E2B, the the solution is an output
inpedance consisting of a 28.27 millihenry inductor in series with a 13.1078 ohm resistor.
The reactance of that inductor just cancels the reactance of the 5.6 uF capacitor, and the
13.1078 ohm resistor provides the drop you see when you apply 300 and 600 ohm loads (the
open circuit voltage would be 15.1335 volts).

If there is phase shift between E2A and E2B, then the solution lies somewhere else on the
circle diagram.

 

[John Fields]

On Wed, 13 Apr 2005 00:55:21 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:qmco51h5ja548iopr0lpf71vsaj9cskn8a@4ax.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?


My reply:
.

But, at any rate, starting from the beginning and looking at the cap
at 400Hz, its reactance is:


E 0.2V
Z = --- = -------- = 8 ohms
I 0.025A

Considering calculation roundoffs, measurement error, and component
tolerances, that's pretty close to the 10 ohms it should be if C1
wasn't there.

John Fields
Professional Circuit Designer

But John, C1 is there. Are you saying the 71 ohm reactance of C1 has no
effect??

---
No, but first let me repair the snippage above:
---

My reply:

But, at any rate, starting from the beginning and looking at the cap
at 400Hz, its reactance is:

<snip>

E 0.2V
Z = --- = -------- = 8 ohms
I 0.025A

Considering calculation roundoffs, measurement error, and component
tolerances, that's pretty close to the 10 ohms it should be if C1
wasn't there.

<snip>

---
What I said in the first snipped part was that in the case where the
600 ohm resistor wasn't connected, the impedance of the 10 ohm
resistor and the series 5.6ľF cap was about 72 ohms, but when the 600
ohm resistor _was_ connected, the impedance of the entire string
became 614 ohms, of which about 8 ohms was contributed by the 10 ohm
resistor and 5.6ľF cap. In the second snip I changed that to 16 ohms
because of the results of a simulation using ideal components.

So no, I'm _not_ saying that that 71 ohm capacitive reactance has no
effect, merely that its effect becomes very small when the circuit is
loaded because of the change in the phase angle between the unloaded
and loaded circuit. What's happening, in essence, is that the
impedance of the 10 ohm - 5.6ľF string is being _transformed_ from 72
ohms to something around 16 ohms when the 600 ohm resistor is added to
the circuit.

Take a look at my numbers and reasoning again, and if there's an error
there please let me know.

--
John Fields
Professional Circuit Designer

 

[John Woodgate]

I read in sci.electronics.design that John Fields
<jfields@austininstruments.com> wrote (in
<m3jq51li6gru6vu94m548vlci13j9rvirl@4ax.com&gt about 'Amp output Z', on
Wed, 13 Apr 2005:

Take a look at my numbers and reasoning again, and if there's an error
there please let me know.

There IS an error; your result violates Ohm's Law, Thévenin's Theorem
and the Superposition Theorem. You can't change 10 - j71 into 8 + j0 be
adding 600 ohms externally. But I can't spare time at present to delve
into what you did, in order to locate the error precisely.
--
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]

I read in sci.electronics.design that The Phantom <phantom@aol.com>
wrote (in <6sup5193h1jtbq6a33fms79b7t4de481km@4ax.com&gt about 'Amp
output Z', on Wed, 13 Apr 2005:

It looks to me as though the locus of possible output impedances is a
circle with center at about 10+j47 and with a radius of about 24.

If there is no relative phase shift between E2A and E2B, the the
solution is an output inpedance consisting of a 28.27 millihenry
inductor in series with a 13.1078 ohm resistor. The reactance of that
inductor just cancels the reactance of the 5.6 uF capacitor, and the
13.1078 ohm resistor provides the drop you see when you apply 300 and
600 ohm loads (the open circuit voltage would be 15.1335 volts).

The mind boggles! How on earth do you imagine you can generate an
inductor from a circuit that has two resistors and a capacitor in
series?
--
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 Fields]

On Wed, 13 Apr 2005 19:23:55 +0100, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields
jfields@austininstruments.com> wrote (in
m3jq51li6gru6vu94m548vlci13j9rvirl@4ax.com&gt about 'Amp output Z', on
Wed, 13 Apr 2005:

Take a look at my numbers and reasoning again, and if there's an error
there please let me know.

There IS an error; your result violates Ohm's Law, Thévenin's Theorem
and the Superposition Theorem. You can't change 10 - j71 into 8 + j0 be
adding 600 ohms externally. But I can't spare time at present to delve
into what you did, in order to locate the error precisely.

---
I'll wait...

--
John Fields
Professional Circuit Designer

 

[Pooh Bear]

Harry Dellamano 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?

The 70 ohm reactance isn't in *parallel* with anything ! It's in series.

Vacross600R = 10V * ( 600 / modulus ( 610-71j )) I think. Sorry, don't fancy
doing the complex math right now, it's late


Graham

 

[Rich Grise]

On Wed, 13 Apr 2005 19:25:45 +0100, John Woodgate wrote:

I read in sci.electronics.design that The Phantom <phantom@aol.com> wrote
(in <6sup5193h1jtbq6a33fms79b7t4de481km@4ax.com&gt about 'Amp output Z', on
Wed, 13 Apr 2005:

It looks to me as though the locus of possible output impedances is a
circle with center at about 10+j47 and with a radius of about 24.

If there is no relative phase shift between E2A and E2B, the the solution
is an output inpedance consisting of a 28.27 millihenry inductor in
series with a 13.1078 ohm resistor. The reactance of that inductor just
cancels the reactance of the 5.6 uF capacitor, and the 13.1078 ohm
resistor provides the drop you see when you apply 300 and 600 ohm loads
(the open circuit voltage would be 15.1335 volts).

The mind boggles! How on earth do you imagine you can generate an inductor
from a circuit that has two resistors and a capacitor in series?

Nothing to it!

Just measure it at 13 GHz. ;-)

Cheers!
Rich

 

[Pooh Bear]

Harry Dellamano wrote:

"Pooh Bear" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:425C1405.7EFD866E@hotmail.com...


Harry Dellamano wrote:

"Pooh Bear" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:425C071F.83BBAA67@hotmail.com...


Hey Poop Bear,

Oi ! wash your mouth out !

I originally stated that the voltage was 10 volts to keep everything in
simple terms but when someone wanted real data, it was taken with a
15.0Vrms
voltage source. The amplifier is not clipping, has +/-50Vpk compliance
and
as stated all pure sine waves with <1% THD. The 14.8Vrms is across a 600
ohm
load and 14.50Vrms across a 300 ohm load.

OK I just did a quick simulation with 15V open circuit volts @ 400Hz,10
ohms
amplifier output resistance, 5.6 uF coupling cap and a choice of 600 or
300 ohm
loads.

I get 14.66 V across the 600R load

I get 14.16 V across the 300R load.

Did you measure the voltage in the no load situation with the DVM ?


Graham

Did I measure the output voltage at the open 600 ohm load? No, this unit
is not stable with open loads so that measurement is not possible. I do have
the node on the other side of the 5U6 cap if that is helpful but no readings
at present.

Those readings would pretty much give an instant first order approximation to
Rout.

Your twin sure did give me a lot of POOP about my technical abilities, are
you the same bear? You don't seem smarter than the average bear.

Well I know how to measure and calculate the output imdeance of amplifiers for
sure which is more than you seem capable of ( never mind even explaining in your
original post what you were trying to do ).

I'm still not sure if you want to know Rout or Zout of your amplifier.

Since 5.6uF measures -70j @ 400 Hz you clearly have a problem if it's included.

Graham

 

[Pooh Bear]

John Woodgate wrote:

I read in sci.electronics.design that Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote (in
425C1405.7EFD866E@hotmail.com&gt about 'Amp output Z', on Tue, 12 Apr
2005:
OK I just did a quick simulation with 15V open circuit volts @ 400Hz,10
ohms amplifier output resistance, 5.6 uF coupling cap and a choice of
600 or 300 ohm loads.

I get 14.66 V across the 600R load

I get 14.16 V across the 300R load.

I calculated the ratios, and based on 15 V open-circuit, and the results
are:

V600/Voc Mine: 0.9770 Pooh Bear: 0.9773

V300/Voc Mine: 0.9433 Pooh Bear: 0.9440

V300/V600 Mine: 0.9655 Pooh Bear 0.9658

I guess we agree a bit.

I have no difficulty understanding that John. ;-)


Graham

 

[rex]

On Wed, 13 Apr 2005 13:51:23 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 13 Apr 2005 19:23:55 +0100, John Woodgate
jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields
jfields@austininstruments.com> wrote (in
m3jq51li6gru6vu94m548vlci13j9rvirl@4ax.com&gt about 'Amp output Z', on
Wed, 13 Apr 2005:

Take a look at my numbers and reasoning again, and if there's an error
there please let me know.

There IS an error; your result violates Ohm's Law, Thévenin's Theorem
and the Superposition Theorem. You can't change 10 - j71 into 8 + j0 be
adding 600 ohms externally. But I can't spare time at present to delve
into what you did, in order to locate the error precisely.

---
I'll wait...

I think I see the problem. Been a while since I thought about math with
complex numbers, so correct me if I'm wrong.

You said:

If we look at the total circuit once again, and insert the value of
the voltage you measured at E2:

15V
|
R1 [10R]
|
+--->E1
|
C1 [5.6ľF]
|
+--->14.8V
|
R2 [600R]
|
GND

then calculate the current through R2, we have:

E 14.8V
I = --- = ------- ~ 0.025A
R 600R


Now, since we have 25mA flowing through the string, a 15V supply, and
a 14.8V drop across R2, the drop across the impedance represented by
R1 and C1 must be the difference in voltage between 15V and 14.8V;
0.2V.

So you subtracted 14.8V from 15V and got .2V. That is treating the
voltages like they are in phase, but they aren't. You have to subtract
them as phasors which will give you a different scalar answer than 0.2V.

 

[The Phantom]

On Wed, 13 Apr 2005 19:25:45 +0100, John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that The Phantom <phantom@aol.com
wrote (in <6sup5193h1jtbq6a33fms79b7t4de481km@4ax.com&gt about 'Amp
output Z', on Wed, 13 Apr 2005:

It looks to me as though the locus of possible output impedances is a
circle with center at about 10+j47 and with a radius of about 24.

If there is no relative phase shift between E2A and E2B, the the
solution is an output inpedance consisting of a 28.27 millihenry
inductor in series with a 13.1078 ohm resistor. The reactance of that
inductor just cancels the reactance of the 5.6 uF capacitor, and the
13.1078 ohm resistor provides the drop you see when you apply 300 and
600 ohm loads (the open circuit voltage would be 15.1335 volts).

The mind boggles! How on earth do you imagine you can generate an
inductor from a circuit that has two resistors and a capacitor in
series?

The OP said:
" 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))."

There is obviously a discrepancy between the real circuit and his model.

Then somebody else said:

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?

Then Harry responded:

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

What I have done is to determine what the output impedance of the *real* circuit
*must be* to account for the *measured* E2A and E2B. (replace the 10 ohm resistor with
the impedance I have described.)

Harry said: "..so I got this amplifier with an internal output resistance of 10 ohms"

In the real amp it must be something other a pure resistance of 10 ohms to account for the
voltages he measures.

The E2A and E2B he gives can only happen if the effective output impedance is
substantially less than the 71 ohms attributable to the 5.6 uF capacitor alone. Harry
himself calculated it to be about 13 ohms. So there must effectively be some inductive
reactance in there somewhere. If he could tell us the relative phases of the E2A and E2B
voltages, there would be only one solution instead a circle diagram of possible
impedances.

 

[John Woodgate]

I read in sci.electronics.design that The Phantom <phantom@aol.com>
wrote (in <0q4r51pg9l8v7sin5c9jenvd4v6opa9k90@4ax.com&gt about 'Amp
output Z', on Wed, 13 Apr 2005:

In the real amp it must be something other a pure resistance of 10 ohms
to account for the voltages he measures.

I see you have assumed that they are accurate. But do you not think that
your inference from them is, on reflection, improbable? How could such
an inductance arise?
--
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]

Harry Dellamano wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:qmco51h5ja548iopr0lpf71vsaj9cskn8a@4ax.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?


My reply:
.

But, at any rate, starting from the beginning and looking at the cap
at 400Hz, its reactance is:


E 0.2V
Z = --- = -------- = 8 ohms
I 0.025A

Considering calculation roundoffs, measurement error, and component
tolerances, that's pretty close to the 10 ohms it should be if C1
wasn't there.

John Fields
Professional Circuit Designer

But John, C1 is there. Are you saying the 71 ohm reactance of C1 has no
effect??
Harry

Compared to the total *resistance* in the chain of 610 ohms, -71j makes a minor
contribution to the total load impedance.

Graham

 

[Pooh Bear]

John Woodgate wrote:

I read in sci.electronics.design that Harry Dellamano
harryd@tdsystems.org> wrote (in <ea_6e.10637$Zn3.4787@trnddc02&gt about
'Amp output Z', on Wed, 13 Apr 2005:

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in message
news:swFf0l2G7BXCFw8R@jmwa.demon.co.uk...
I read in sci.electronics.design that Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote (in
425C1405.7EFD866E@hotmail.com&gt about 'Amp output Z', on Tue, 12 Apr 2005:
OK I just did a quick simulation with 15V open circuit volts @ 400Hz,10
ohms amplifier output resistance, 5.6 uF coupling cap and a choice of 600
or 300 ohm loads.

I get 14.66 V across the 600R load

I get 14.16 V across the 300R load.

I calculated the ratios, and based on 15 V open-circuit, and the results
are:

V600/Voc Mine: 0.9770 Pooh Bear: 0.9773

V300/Voc Mine: 0.9433 Pooh Bear: 0.9440

V300/V600 Mine: 0.9655 Pooh Bear 0.9658

I guess we agree a bit.
--
John,
What do your numbers mean? Is Zo around 10 ohms or 70 ohms, and if 70 how
do I measure it.

My results are the calculated voltages which your 600/300 ohm test would
produce if the resistive component of the output source impedance is 10
ohms and the (capacitive) reactive component is 71 ohms. The magnitude
is sqrt(71^2 + 10^2) = 71.7 ohms

If you want to determine the resistive and reactive components
separately, and you can't measure phase difference, don't use two
different resistive loads, use two different frequencies, and measure
the open-circuit output voltage and the output voltage with a single
resistive load at both frequencies. Of course, the open-circuit voltage
should be (nearly) the same at the two frequencies.

In this case, too, the explicit equations are Ro = huge mess and Co =
another huge mess, I think. So write the equations you get from Ohm's
Law, which have Ro and Co implicit, plug the known numbers into the
equations and go from there.

Note 'Co' in the above, because with two different frequencies you have
two Xo values, giving you three unknowns, Ro, Xo1 and Xo2. But Xo1 and
Xo2 are related by f2/f1, because Co is constant. That is the third
equation you need to reduce back to two variables, either Ro and Xo1 or
Ro and Co.

He could also measure 'the other side' of the 5.6uF and back-calculate Rout to a
high degree of accuracy essentially ignoring the reactive component for practical
purposes.


Graham

 

[The Phantom]

On Thu, 14 Apr 2005 00:06:16 +0100, John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that The Phantom <phantom@aol.com
wrote (in <0q4r51pg9l8v7sin5c9jenvd4v6opa9k90@4ax.com&gt about 'Amp
output Z', on Wed, 13 Apr 2005:

In the real amp it must be something other a pure resistance of 10 ohms
to account for the voltages he measures.

I see you have assumed that they are accurate. But do you not think that
your inference from them is, on reflection, improbable? How could such
an inductance arise?

This problem is not well described or measured. I can only do calculations on what Harry
has provided. I have determined the locus of output impedances that must exist to account
for the voltages Harry said he measured. If that leads to an impossible situation, one
might take that as an indicator that better measurements are called for, or a better Spice
model. I'm not sure we can figure out the source of the discrepancy based on what Harry
has told us.

 

[qrk]

On Tue, 12 Apr 2005 02:30:32 GMT, "Harry Dellamano"
<harryd@tdsystems.org> 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.
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,
If you can change the output capacitor to a large value, like
10,000uF, then you can treat this like a pure resistance voltage
divider. Or, connect up a network analyzer and measure directly.

I'm in process of building up a dual AMD Opteron computer. I'll pass
on details if you want. It will be a week or so until I get my
benchmarks tested out. I'm interested to see what happens with the
better memory access speed.

Mark

 

[Jim Thompson]

On Wed, 13 Apr 2005 18:26:23 -0700, qrk <SpamTrap@reson.com> wrote:

On Tue, 12 Apr 2005 02:30:32 GMT, "Harry Dellamano"
harryd@tdsystems.org> 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.
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,
If you can change the output capacitor to a large value, like
10,000uF, then you can treat this like a pure resistance voltage
divider. Or, connect up a network analyzer and measure directly.

I'm in process of building up a dual AMD Opteron computer. I'll pass
on details if you want. It will be a week or so until I get my
benchmarks tested out. I'm interested to see what happens with the
better memory access speed.

Mark

Run our "benchmark" and see how it compares ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.