Amp output Z

John Woodgate · Apr 12, 2005

I read in sci.electronics.design that Pooh Bear
<rabbitsfriendsandrelations@hotmail.com> wrote (in
<425B9D86.A782F3EA@hotmail.com&gt

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

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

Would you like to explain what you think is wrong with it?

I think the op-amp may be clipping; 14.8 V r.m.s. is awfully big.
--
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

Spehro Pefhany · Apr 12, 2005

On Tue, 12 Apr 2005 10:28:25 +0100, the renowned Tony Williams
<tonyw@ledelec.demon.co.uk> wrote:

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.

Exactly. Harry does some rather cool and important work, and has a
nice lab. Just a glitch.

Probably why I didn't feel like posting multiplication by conjugates
and stuff well after midnight and resorted to a bit of hand-waving to
describe the results... ;-)

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

Jim Thompson · Apr 12, 2005

On Mon, 11 Apr 2005 20:59:38 -0700, Jim Thompson
<thegreatone@example.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?
Harry

Zee math ;-) I'll write it down and scan it in the morning.

...Jim Thompson

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

John Woodgate · Apr 12, 2005

I read in sci.electronics.design that Jim Thompson
<thegreatone@example.com> wrote (in
<40qn51tv3o0838tfbcs2fhi060dhht9jor@4ax.com&gt

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

Vm1/Vm2 = (Ro-jXc+Rm2)/(Ro-jXc+Rm1)*Rm1/Rm2

Since your DMM doesn't give you phase, you can't solve for Ro.

When you rationalise Rm1/(Ro+Rm1+jXc) to Rm1/sqrt{(Ro+Rm1)^2+Xc^2)} you
are automatically including the effect of phase. Likewise for the
similar but inverted expression containing Rm2.
--
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

Jim Thompson · Apr 12, 2005

On Tue, 12 Apr 2005 16:56:34 +0100, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Jim Thompson
thegreatone@example.com> wrote (in
40qn51tv3o0838tfbcs2fhi060dhht9jor@4ax.com&gt about 'Amp output Z', on
Tue, 12 Apr 2005:

Vm1/Vm2 = (Ro-jXc+Rm2)/(Ro-jXc+Rm1)*Rm1/Rm2

Since your DMM doesn't give you phase, you can't solve for Ro.

When you rationalise Rm1/(Ro+Rm1+jXc) to Rm1/sqrt{(Ro+Rm1)^2+Xc^2)} you
are automatically including the effect of phase. Likewise for the
similar but inverted expression containing Rm2.

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

Pooh Bear · Apr 12, 2005

John Woodgate wrote:

I read in sci.electronics.design that Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote (in
425B9D86.A782F3EA@hotmail.com&gt about 'Amp output Z', on Tue, 12 Apr
2005:

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

Would you like to explain what you think is wrong with it?

It gave the wrong answer for Es

I think the op-amp may be clipping; 14.8 V r.m.s. is awfully big.

Sure, but the OP said that the source voltage was *10* volts not 14.8 !

Graham, puzzled

Harry Dellamano · Apr 12, 2005

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

John Woodgate wrote:

I read in sci.electronics.design that Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote (in
425B9D86.A782F3EA@hotmail.com&gt about 'Amp output Z', on Tue, 12 Apr
2005:

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

Would you like to explain what you think is wrong with it?

It gave the wrong answer for Es

I think the op-amp may be clipping; 14.8 V r.m.s. is awfully big.

Sure, but the OP said that the source voltage was *10* volts not 14.8 !

Graham, puzzled

Hey Poop Bear,
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.
Harry

Pooh Bear · Apr 12, 2005

Harry Dellamano wrote:

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

John Woodgate wrote:

I read in sci.electronics.design that Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote (in
425B9D86.A782F3EA@hotmail.com&gt about 'Amp output Z', on Tue, 12 Apr
2005:

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

Would you like to explain what you think is wrong with it?

It gave the wrong answer for Es

I think the op-amp may be clipping; 14.8 V r.m.s. is awfully big.

Sure, but the OP said that the source voltage was *10* volts not 14.8 !

Graham, puzzled

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

John Woodgate · Apr 12, 2005

I read in sci.electronics.design that Jim Thompson
<thegreatone@example.com> wrote (in
<tlvn511djoqu9enbu0skimii42ulthedps@4ax.com&gt

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

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.

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.

A similar problem occurs when attempting to measure the impedance
looking back into a mains supply (which is important in a number of
situations, including the theoretical bases of IEC/EN 61000-3-2, -3, -11
and -12), although in that case the reactance is inductive. I had a big
argument with someone on a newsgroup a while back about whether a
solution is possible using only resistive loads. There are two equations
and two unknowns, E and Ro, so a solution must be possible unless one
equation is a multiple of the other (so you really only have one
equation) or they are inconsistent (representing parallel lines on a
graph, such as y = x + 3 and y = x + 4.892). I consulted my colleague
who is a professor of electric power engineering in Germany, and he
confirmed that a solution is possible, but messy. One problem is that if
you do the algebra one way, you find that the whole source impedance (Ro
+ jX) cancels from the equations. Talk about babies and bath water!

I just asked Mathcad to solve it symbolically, and the expression for Ro
has eight terms in the numerator.

--
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

Harry Dellamano · Apr 12, 2005

"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.
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.
Harry

John Woodgate · Apr 12, 2005

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.
--
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 · Apr 12, 2005

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?

---
In response to the above, I got the following from Harry via email and
replied to him via email as well as posting it here:
---

From Harry:

On Mon, 11 Apr 2005 19:01:34 -0700, you wrote:

---------+------E1A
| |
| [5.6ľF]
| |
[GEN] +------14.81vac
| |
| [600R]
| |
+-----------------+
| |
GND GND

+---[10R]---------+------E1B
| |
| [5.6ľF]
| |
[GEN] +------14.50vac
| |
| [300R]
| |
+-----------------+
| |
GND GND

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

--
John Fields
Professional Circuit Designer

John,
I added in the voltages in your above circuit with 15.00Vac at 400Hz on the
generator. Node E1A and E1B are not available for measurement. They are
inside a black box that the customer cannot get to. I can make them
available but is it necessary?

My reply:

---
I don't know.

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

1 1
Xc = --------- = ------------------------- ~ 71 ohms
2pi f C 6.28 * 400Hz + 5.6E-6ľF

With the 10 ohm series resistor in there, the impedance of the string
becomes:

Z = sqrt (R˛ + Xc˛) = sqrt (10˛ + 71˛) ~ 71.7 ohms,

the phase angle:

Xc 71
phi = tan^-1 ---- = ---- ~ 82°
R 10

and the power factor:

PF = cos phi ~ 0.14

But now, when we add the 600 ohm resistor:

15V
|
[10R]
|
+--->E1
|
[5.6ľF]
|
+--->E2
|
[600R]
|
GND

and calculate the impedance of the entire string:

Z = sqrt (Rt˛ + Xc˛)

= sqrt ((10R + 600R)˛ + 71˛) ~ 614 ohms,

the phase angle becomes

71
phi = tan^-1 ------------ ~ 6.6°
10R + 600R

and the power factor:

PF = cos phi = 0.9937

So, it seems that the addition of the 600 ohms of resistance has
caused a change in phase great enough to make the imaginary part of
the impedance nearly vanish!

But let's see...

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. Then, since there is 25mA in the string and 0.2V being dropped
across R1 and C1, the impedance developed by R1 and C1 must be:

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.

Just for a quick check I ran a simulation, and I got 14.6V for E2,

E 0.4V
Z = --- = -------- = 16 ohms
I 0.025A

Which looks better to me.

I'm going to post this to the NG, since I'm sure the guys over there
would like to see it as well.

--
John Fields
Professional Circuit Designer

Harry Dellamano · Apr 12, 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.
Regards,
Harry

Harry Dellamano · Apr 13, 2005

"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.
Harry

rex · Apr 13, 2005

On Wed, 13 Apr 2005 00:59:54 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

"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.
Regards,
Harry

I still think that my calculation that arrived at Ro of about 0.5 ohms
is correct for the measurements you made. This is not Zo because of the
largish Xc. Zo is complex, Ro is the real part.

John's point is that an accurate calculation of Ro is pretty much
impossible this way. The slightest error in measurement will change the
answer. There is only a tiny difference in voltage values if Ro is .5 or
10 or 20.

Harry Dellamano · Apr 13, 2005

"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

Jim Thompson · Apr 13, 2005

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

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

Fred Bloggs · Apr 13, 2005

Fred Bloggs 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??

-and-

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.

Your equation is based upon the amplifier open circuit voltage, Voc, and
the effective output resistance, Ro, taking the load current as the
circuit phasor reference angle so that Voc/_phi1,2=I1,2*(Ro-jXc)+E1,2
for the two different load conditions, R1,2. Note that phi1 is not equal
to phi2, only Voc amplitude stays constant with loading. The so-called
output impedance is *always* the coefficient of the I term which is
Ro-jXc and has nothing to do with R1 or R2 assuming the circuit stays
linear. Since the Voc phase angle changes, your only option is to work
with scalar magnitude or magnitude squared in I1 and I2, the reference,
so this gives you I1^2*((Ro+R1)^2 + Xc^2)=I2^2*( (Ro+R2)^2 + Xc^2) as
the only recoverable equation given your measurements. This develops
into Xc=sqrt{(I1^2*(Ro+R1)^2-I2^2*(Ro+R2)^2)/(I2^2-I1^2)}. Plugging in
your measurements R1=600, R2=300, Ro=10, I1=14.81/600, I2=14.50/300,
yields
Xc=600*sqrt{(14.81^2*(1+10/600)^2-14.50^2*(1+10/300)^2)/(4*14.50^2-14.81^2)}
or Xc=600*sqrt{2.2079523/(621.6639)}=600*0.059596=36 ohms so that
C=1/(36*6.28*400)=11uF and not 5.6u- this is suspiciously a factor of
x2. Also, the output impedance is 10-j36 or 37 ohms reactive.

Well- that is all well and good but what does it accomplish? Starting
from the basic scalar equation:I1^2*((Ro+R1)^2 + Xc^2)=I2^2*( (Ro+R2)^2
+ Xc^2), this rearranges to:
|Zout|^2=Ro^2+Xc^2=(I2^2*(2*Ro*R2+R2^2)-I1^2*(2*Ro*R1+R1^2))/(I1^2-I2^2)
so that if a third measurement was taken at load R3 with resulting I3,
then you would have a second equation of form:
(I2^2*(2*Ro*R2+R2^2)-I1^2*(2*Ro*R1+R1^2))/(I1^2-I2^2)=
(I3^2*(2*Ro*R3+R3^2)-I1^2*(2*Ro*R1+R1^2))/(I1^2-I3^2), a messy but
linear equation in Ro which is then solved and result substituted back
into the |Zout|^2 equation.

John Popelish · Apr 13, 2005

Fred Bloggs 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??

-and-

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.

Your equation is based upon the amplifier open circuit voltage, Voc, and
the effective output resistance, Ro, taking the load current as the
circuit phasor reference angle so that Voc/_phi1,2=I1,2*(Ro-jXc)+E1,2
for the two different load conditions, R1,2. Note that phi1 is not equal
to phi2, only Voc amplitude stays constant with loading. The so-called
output impedance is *always* the coefficient of the I term which is
Ro-jXc and has nothing to do with R1 or R2 assuming the circuit stays
linear. Since the Voc phase angle changes, your only option is to work
with scalar magnitude or magnitude squared in I1 and I2, the reference,
so this gives you I1^2*((Ro+R1)^2 + Xc^2)=I2^2*( (Ro+R2)^2 + Xc^2) as
the only recoverable equation given your measurements. This develops
into Xc=sqrt{(I1^2*(Ro+R1)^2-I2^2*(Ro+R2)^2)/(I2^2-I1^2)}. Plugging in
your measurements R1=600, R2=300, Ro=10, I1=14.81/600, I2=14.50/300,
yields
Xc=600*sqrt{(14.81^2*(1+10/600)^2-14.50^2*(1+10/300)^2)/(4*14.50^2-14.81^2)}
or Xc=600*sqrt{2.2079523/(621.6639)}=600*0.059596=36 ohms so that
C=1/(36*6.28*400)=11uF and not 5.6u- this is suspiciously a factor of
x2. Also, the output impedance is 10-j36 or 37 ohms reactive.

Nice work, but I think the capacitance is a component and therefore
assumed to be known. It is the 10 ohms resistive component that is in
question (the effective impedance of the amplifier) and is to be
solved for. The trouble is that a value so much lower than the lowest
load (300 ohms), if it really is that low, it produces little real,
measurable voltage change.

John Woodgate · Apr 13, 2005

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.
--
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

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