I need to physically measure the output impedance of an operational
amplifier (LM741), working as an inverter and with a capacitor as
load. Any ideas on how to do this? Impedance analyzer is out of the
question.
Thanx in advance.

"Martin Sigwald"

>
>I need to physically measure the output impedance of an operational
> amplifier (LM741), working as an inverter and with a capacitor as
> load. Any ideas on how to do this? Impedance analyzer is out of the
> question.
>


** Drive the output from a sine wave generator ( 600 ohms source Z or
thereabouts) and measure the resulting voltage imposed across the output
with a scope. Calculate Z from the current and voltage.

You may be able to measure phase angle as well, if you synch the scope to
the generator .

The signal voltage at the 741's output may be very small if the test
frequency and closed loop gain are low.

Maybe use another op-amp ( TL071 etc) as a fixed ( 100 times) gain pre-amp
to up the level to the scope.



........ Phil

On Wed, 15 Aug 2007 20:41:46 -0700, Martin Sigwald
<msigwald@gmail.com> wrote:

>I need to physically measure the output impedance of an operational
>amplifier (LM741), working as an inverter and with a capacitor as
>load. Any ideas on how to do this? Impedance analyzer is out of the
>question.
>Thanx in advance.

Phil has given a short description of the method but it won't hurt to
expand it a little.

The output impedance can be measured by connecting the input to ground
with a 1K resistor across pins 2 and 3, and then injecting a sine-wave
(eg. 1KHz) into the output through a resistor R (eg.1K). Measure with
an oscilloscope the sine-wave voltage Vo between the output and ground
(will be quite small). The arrangement makes a voltage divider out of
R and Zout, so Vo/Vgen = Zout/(R + Zout), and since Zout is quite
small Zout ~ R(Vo/Vgen).

On Aug 16, 5:13 am, Ross Herbert <rherb...@bigpond.net.au> wrote:
> On Wed, 15 Aug 2007 20:41:46 -0700, Martin Sigwald
>
> <msigw...@gmail.com> wrote:
> >I need to physically measure the output impedance of an operational
> >amplifier (LM741), working as an inverter and with a capacitor as
> >load. Any ideas on how to do this? Impedance analyzer is out of the
> >question.
> >Thanx in advance.
>
> Phil has given a short description of the method but it won't hurt to
> expand it a little.
>
> The output impedance can be measured by connecting the input to ground
> with a 1K resistor across pins 2 and 3, and then injecting a sine-wave
> (eg. 1KHz) into the output through a resistor R (eg.1K). Measure with
> an oscilloscope the sine-wave voltage Vo between the output and ground
> (will be quite small). The arrangement makes a voltage divider out of
> R and Zout, so Vo/Vgen = Zout/(R + Zout), and since Zout is quite
> small Zout ~ R(Vo/Vgen).

It is my understanding that you can`t inject a signal through the
output, since this is an active component. Furthermore, I already
tried and since my input was 0, my output was also 0 so I couldn`t
measure anything. Have you tried doing this yourself? Because if you
did please explain further, so I can see what I'm doing wrong.

Phil: Where you referring to the same method in your post? Cause I
didn`t understand what you meant by "drive the output" (sorry, english
is not my native language).

Thanx for your answers.

"Martin Sigwald"
>
>
> It is my understanding that you can`t inject a signal through the
> output, since this is an active component.


** Then you simply have no damn " understanding ".


> Furthermore, I already
> tried and since my input was 0, my output was also 0 so I couldn`t
> measure anything.


** Only demonstrates your inability - at everything.


> Have you tried doing this yourself? Because if you
> did please explain further, so I can see what I'm doing wrong.


** Has been fully explained to you already.


> Phil: Where you referring to the same method in your post? Cause I
> didn`t understand what you meant by "drive the output" (sorry, english
> is not my native language).


** Wot - so you speak good pigeon boss ?

Or just another dodgy code scribbling, pig ignorant useless turd ?




......... Phil

On 16 Aug, 04:41, Martin Sigwald <msigw...@gmail.com> wrote:
> I need to physically measure the output impedance of an operational
> amplifier (LM741), working as an inverter and with a capacitor as
> load. Any ideas on how to do this? Impedance analyzer is out of the
> question.
> Thanx in advance.

1) With no load connected measure the output level with scope.
2) Connect either a pot or a (decade resistance box) and adjust until
the level is half the original level.
3) Measure the pot's resistance.

Robin

On Thu, 16 Aug 2007 04:30:47 -0700, Martin Sigwald
<msigwald@gmail.com> wrote:

>On Aug 16, 5:13 am, Ross Herbert <rherb...@bigpond.net.au> wrote:
>> On Wed, 15 Aug 2007 20:41:46 -0700, Martin Sigwald
>>
>> <msigw...@gmail.com> wrote:
>> >I need to physically measure the output impedance of an operational
>> >amplifier (LM741), working as an inverter and with a capacitor as
>> >load. Any ideas on how to do this? Impedance analyzer is out of the
>> >question.
>> >Thanx in advance.
>>
>> Phil has given a short description of the method but it won't hurt to
>> expand it a little.
>>
>> The output impedance can be measured by connecting the input to ground
>> with a 1K resistor across pins 2 and 3, and then injecting a sine-wave
>> (eg. 1KHz) into the output through a resistor R (eg.1K). Measure with
>> an oscilloscope the sine-wave voltage Vo between the output and ground
>> (will be quite small). The arrangement makes a voltage divider out of
>> R and Zout, so Vo/Vgen = Zout/(R + Zout), and since Zout is quite
>> small Zout ~ R(Vo/Vgen).
>
>It is my understanding that you can`t inject a signal through the
>output, since this is an active component.

Can't? Who says so? Just do it!


Furthermore, I already
>tried and since my input was 0, my output was also 0 so I couldn`t
>measure anything.

It wasn't zero, it was just very, very small. That's because the
output impedance is very low.

Note that closed-loop Zout will increase with frequency as the opamp
gain drops. So try it again at a higher frequency. Zout of a 741
follower will approach the open-loop impedance as frequency approaches
1 MHz, in the rough turf of 100 ohms. Zout will be about inverse on
frequency below that, down to 10 Hz maybe.

John

On Thu, 16 Aug 2007 08:27:39 -0700, robin.pain@googlemail.com wrote:

>On 16 Aug, 04:41, Martin Sigwald <msigw...@gmail.com> wrote:
>> I need to physically measure the output impedance of an operational
>> amplifier (LM741), working as an inverter and with a capacitor as
>> load. Any ideas on how to do this? Impedance analyzer is out of the
>> question.
>> Thanx in advance.
>
>1) With no load connected measure the output level with scope.
>2) Connect either a pot or a (decade resistance box) and adjust until
>the level is half the original level.
>3) Measure the pot's resistance.
>

That will be driving the opamp into current limit, not the same thing
at all.

John

On Aug 16, 8:42 am, "Phil Allison" <philalli...@tpg.com.au> wrote:
> "Martin Sigwald"
>
>
>
> > It is my understanding that you can`t inject a signal through the
> > output, since this is an active component.
>
> ** Then you simply have no damn " understanding ".
>
> > Furthermore, I already
> > tried and since my input was 0, my output was also 0 so I couldn`t
> > measure anything.
>
> ** Only demonstrates your inability - at everything.
>
> > Have you tried doing this yourself? Because if you
> > did please explain further, so I can see what I'm doing wrong.
>
> ** Has been fully explained to you already.
>
> > Phil: Where you referring to the same method in your post? Cause I
> > didn`t understand what you meant by "drive the output" (sorry, english
> > is not my native language).
>
> ** Wot - so you speak good pigeon boss ?
>
> Or just another dodgy code scribbling, pig ignorant useless turd ?
>
> ........ Phil

I think I was more than polite when asking my questions, so I really
think the tone of your answer was uncalled for. Never the less, you
are free to answer questions the way you want, so so be it. Lets get
back to the topic in discussion.
First of all, my professor (which is a PhD in EE and chairman of IEEE
in Argentina, so I reckon she knows what she is talking about) told me
you can`t use the common aproach of impedance measurements used in
passive components to measure an active one. When I tried injecting a
signal into the output with a grounded input, a friend of mine
suggested that since the impedance was to small the voltage was
equally small, far beyond the oscilloscope's range and suggested
amplifying it. Before doing that I asked my proffesor who, as I told
you already, said that wasn't a valid approach.
I haven`t found anything either on the internet or in books, since
they always refer to the ideal opamp, so I am at a loss here. Any help
will be welcomed.

"Martin Sigwald = PITA IDIOT "


> I think I was more than polite when asking my questions, so I really
> think the tone of your answer was uncalled for. Never the less, you
> are free to answer questions the way you want, so so be it. Lets get
> back to the topic in discussion.

> First of all, my professor (which is a PhD in EE and chairman of IEEE
> in Argentina, so I reckon she knows what she is talking about) told me
> you can`t use the common aproach of impedance measurements used in
> passive components to measure an active one. When I tried injecting a
> signal into the output with a grounded input, a friend of mine
> suggested that since the impedance was to small the voltage was
> equally small, far beyond the oscilloscope's range and suggested
> amplifying it. Before doing that I asked my proffesor who, as I told
> you already, said that wasn't a valid approach.
> I haven`t found anything either on the internet or in books, since
> they always refer to the ideal opamp, so I am at a loss here. Any help
> will be welcomed.


** You are an utter ass.

You " friend " is a fool.

Your prof is a dumb bitch ASS !!

The topic has NOTHING to do with anyone's opinion.

Do the test, just as described and find out just how WRONG you all are.





........ Phil

On 16 Aug, 17:00, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
> On Thu, 16 Aug 2007 08:27:39 -0700, robin.p...@googlemail.com wrote:
> >On 16 Aug, 04:41, Martin Sigwald <msigw...@gmail.com> wrote:
> >> I need to physically measure the output impedance of an operational
> >> amplifier (LM741), working as an inverter and with a capacitor as
> >> load. Any ideas on how to do this? Impedance analyzer is out of the
> >> question.
> >> Thanx in advance.
>
> >1) With no load connected measure the output level with scope.
> >2) Connect either a pot or a (decade resistance box) and adjust until
> >the level is half the original level.
> >3) Measure the pot's resistance.
>
> That will be driving the opamp into current limit, not the same thing
> at all.
>
> John

Yes that's true.

Robin

On Aug 16, 12:40 pm, Martin Sigwald <msigw...@gmail.com> wrote:
> On Aug 16, 8:42 am, "Phil Allison" <philalli...@tpg.com.au> wrote:
>
>
>
>
>
> > "Martin Sigwald"
>
> > > It is my understanding that you can`t inject a signal through the
> > > output, since this is an active component.
>
> > ** Then you simply have no damn " understanding ".
>
> > > Furthermore, I already
> > > tried and since my input was 0, my output was also 0 so I couldn`t
> > > measure anything.
>
> > ** Only demonstrates your inability - at everything.
>
> > > Have you tried doing this yourself? Because if you
> > > did please explain further, so I can see what I'm doing wrong.
>
> > ** Has been fully explained to you already.
>
> > > Phil: Where you referring to the same method in your post? Cause I
> > > didn`t understand what you meant by "drive the output" (sorry, english
> > > is not my native language).
>
> > ** Wot - so you speak good pigeon boss ?
>
> > Or just another dodgy code scribbling, pig ignorant useless turd ?
>
> > ........ Phil
>
> I think I was more than polite when asking my questions, so I really
> think the tone of your answer was uncalled for. Never the less, you
> are free to answer questions the way you want, so so be it. Lets get
> back to the topic in discussion.
> First of all, my professor (which is a PhD in EE and chairman of IEEE
> in Argentina, so I reckon she knows what she is talking about) told me
> you can`t use the common aproach of impedance measurements used in
> passive components to measure an active one. When I tried injecting a
> signal into the output with a grounded input, a friend of mine
> suggested that since the impedance was to small the voltage was
> equally small, far beyond the oscilloscope's range and suggested
> amplifying it. Before doing that I asked my proffesor who, as I told
> you already, said that wasn't a valid approach.
> I haven`t found anything either on the internet or in books, since
> they always refer to the ideal opamp, so I am at a loss here. Any help
> will be welcomed.- Hide quoted text -
>
> - Show quoted text -

Hi, Martin. In fact, Mr. Allison gave you the correct answer -- your
teacher is wrong. Look at the bottom of page one of National
Semiconductor's Applications Brief 108 -- that's exactly the way
output impedance in op amps is measured.

http://www.national.com/appbriefs/files ... pdf#page=1

I'd assume you'll regard the manufacturer of opamps as a sufficient
authority.

By the way, it only took a minute or two of poking around National's
website to find this. Try using the manufacturers' websites as a
source of practical information -- they'll tell you things that'd make
your teachers blush.

Good luck in your studies
Chris

"Chris"
>
> Hi, Martin. In fact, Mr. Allison gave you the correct answer -- your
> teacher is wrong. Look at the bottom of page one of National
> Semiconductor's Applications Brief 108 -- that's exactly the way
> output impedance in op amps is measured.
>
> http://www.national.com/appbriefs/files ... pdf#page=1


** Figure 1 (page 1 ) shows a HP network analyser being used, a HP 4195A .

Though I much appreciate your backing my comments up - the OP clearly
specified no high tech equipment and obediently complied with his request.

Till the Argentinean shithead abused me, that is.




........ Phil

On Thu, 16 Aug 2007 10:40:50 -0700, Martin Sigwald
<msigwald@gmail.com> wrote:

>On Aug 16, 8:42 am, "Phil Allison" <philalli...@tpg.com.au> wrote:
>> "Martin Sigwald"
>>
>>
>>
>> > It is my understanding that you can`t inject a signal through the
>> > output, since this is an active component.
>>
>> ** Then you simply have no damn " understanding ".
>>
>> > Furthermore, I already
>> > tried and since my input was 0, my output was also 0 so I couldn`t
>> > measure anything.
>>
>> ** Only demonstrates your inability - at everything.
>>
>> > Have you tried doing this yourself? Because if you
>> > did please explain further, so I can see what I'm doing wrong.
>>
>> ** Has been fully explained to you already.
>>
>> > Phil: Where you referring to the same method in your post? Cause I
>> > didn`t understand what you meant by "drive the output" (sorry, english
>> > is not my native language).
>>
>> ** Wot - so you speak good pigeon boss ?
>>
>> Or just another dodgy code scribbling, pig ignorant useless turd ?
>>
>> ........ Phil
>
>I think I was more than polite when asking my questions, so I really
>think the tone of your answer was uncalled for. Never the less, you
>are free to answer questions the way you want, so so be it. Lets get
>back to the topic in discussion.
>First of all, my professor (which is a PhD in EE and chairman of IEEE
>in Argentina, so I reckon she knows what she is talking about) told me
>you can`t use the common aproach of impedance measurements used in
>passive components to measure an active one. When I tried injecting a
>signal into the output with a grounded input, a friend of mine
>suggested that since the impedance was to small the voltage was
>equally small, far beyond the oscilloscope's range and suggested
>amplifying it. Before doing that I asked my proffesor who, as I told
>you already, said that wasn't a valid approach.
>I haven`t found anything either on the internet or in books, since
>they always refer to the ideal opamp, so I am at a loss here. Any help
>will be welcomed.

Phil lives in a constant state of profane rage, sort of a textual
Tourette's. He's an audio tech or something. Ignore him.

John

On Aug 17, 1:27 am, robin.p...@googlemail.com wrote:
> On 16 Aug, 04:41, Martin Sigwald <msigw...@gmail.com> wrote:
>
> > I need to physically measure the output impedance of an operational
> > amplifier (LM741), working as an inverter and with a capacitor as
> > load. Any ideas on how to do this? Impedance analyzer is out of the
> > question.
> > Thanx in advance.
>
> 1) With no load connected measure the output level with scope.
> 2) Connect either a pot or a (decade resistance box) and adjust until
> the level is half the original level.
> 3) Measure the pot's resistance.
>
> Robin

Only one comment.... the measurement must be done at a level
that does not overload the amp and at a frequency within the
opamp's range.

Feed a signal into the opamp that is well below the clipping level
at the output measured with the scope.
With no load connected measure the output level with the scope.
Connect either a pot or a (decade resistance box) and adjust until
the level is half the original level (The resistance will be small
as
the source impedance is low).
Measure the pot's resistance.This value is the output source
impedance.

On Aug 20, 5:34 am, 100246.2...@compuserve.com wrote:
> On Aug 17, 1:27 am, robin.p...@googlemail.com wrote:
>
> > On 16 Aug, 04:41, Martin Sigwald <msigw...@gmail.com> wrote:
>
> > > I need to physically measure the output impedance of an operational
> > > amplifier (LM741), working as an inverter and with a capacitor as
> > > load. Any ideas on how to do this? Impedance analyzer is out of the
> > > question.
> > > Thanx in advance.
>
> > 1) With no load connected measure the output level with scope.
> > 2) Connect either a pot or a (decade resistance box) and adjust until
> > the level is half the original level.
> > 3) Measure the pot's resistance.
>
> > Robin
>
> Only one comment.... the measurement must be done at a level
> that does not overload the amp and at a frequency within the
> opamp's range.
>
> Feed a signal into the opamp that is well below the clipping level
> at the output measured with the scope.
> With no load connected measure the output level with the scope.
> Connect either a pot or a (decade resistance box) and adjust until
> the level is half the original level (The resistance will be small
> as
> the source impedance is low).
> Measure the pot's resistance.This value is the output source
> impedance.

The thin is I need to measure the output impedance of the whole
circuit, opamp+capacitor, in terms of frequency. Doing what you
propose would take an enourmous amount of time.
What I did and think it works, since it produced something very
similar to my analitical calculations, is measure the output voltage
and phase of the opamp+capacitor, then add a parallel resistor and
measure again. Using Thevenin the equivalent output resistance of the
circuit will be Zth= (V1/V2-1) R. As I said, it seems to work.