Reply by Martin Sigwald August 24, 20072007-08-24
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.
Reply by August 20, 20072007-08-20
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.
Reply by John Larkin August 18, 20072007-08-18
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
Reply by Phil Allison August 18, 20072007-08-18
"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/AppBrief108.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
Reply by Chris August 18, 20072007-08-18
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/AppBrief108.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
Reply by August 17, 20072007-08-17
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
Reply by Phil Allison August 16, 20072007-08-16
"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
Reply by Martin Sigwald August 16, 20072007-08-16
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.
Reply by John Larkin August 16, 20072007-08-16
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
Reply by John Larkin August 16, 20072007-08-16
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