Measuring output impedance

abbey road d enfer said:

Important caveat: 1000 ohms must be the sum of additional resistor + output impedance of the generator. For example, if the output Z of the generator, teh additional resistor must be 400r.

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This already accounted for by measuring the the voltage on either side of the the 1K resistor, The current is V2/Rout. The voltage across the 1K resistor is this current multiplied by 1K. Some simple manipulation yields the formula I gave. The generator impedance will alter the values of V1 and V2 but the formula still holds.

Cheers

Ian

moamps said:

You should replace the number "1000" in numerator of  formula in G9 with "G5".

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Thanks for picking that up. I have corrected the spreadsheet and re-attached it to the original post.

Cheers

Ian

johnheath said:

Some data based on how I understood how to do it… It is probably just me  :

A = 0,046V p-p
B = 0,36V p-p

X factor would be 0.1278

Rout=10(X-1)/(10-X) = 10(0.1278-1)/(10-0.1278) = (10 x -0.8722)/9.8722 = 0.883

Which would mean that the output impedance is roughly 883R




According to this the output impedance would be roughly 883R

But if I use the method of measuring according to "radardoug" - using resistor across the output - sending a signal and increase the value of the resistor until the signal drops 6dB the measured impedance would be 270R.


This is confusing to me… please correct me here


Best regards

/John

Click to expand...

With this method you feed a signal into a 10K load and measure the level. You then change the load to 1K and measure again. The level with the 1K load should be lower than the the one with the 10K load. This means the value X should always be greater than 1. I am not sure how you measured you values A and B but they seem wrong to me.

Cheers

Ian

abbey road d enfer said:

Maybe the generator you used has a too large output impedance?

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You can chose between 50R or 600R… two separate connections. If it is bad or misleading I do not know


best regards

/John

ruffrecords said:

With this method you feed a signal into a 10K load and measure the level. You then change the load to 1K and measure again. The level with the 1K load should be lower than the the one with the 10K load. This means the value X should always be greater than 1. I am not sure how you measured you values A and B but they seem wrong to me.

Cheers

Ian

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Yes, that is what I thought too, but I just wrote the values down on a piece of paper and did the math. The formula is pretty straight forward and not confusing… it must be my way of measuring.

If you take a look at the attached picture where I have drawn how I connected it all… maybe there is something wrong there?


Best regards

/John

ruffrecords said:

Simplified direct method output impedance spreadsheet is attached. It contains two version: the one on the left expects you to input measurements in dBu and the one on the right accepts voltages.

You can input the the drive level and the value of the series resistor or use the default values. Then you just input your measured value and the spreadsheet works out the output resistance. V1 and V2 should be left alone. They are just the calculated values of the drive level and the measured level in volts. The formula used to calculate the output resistance is:

Rout = V2 *1000/(V1-V2)

I had to change the extension of the file to .txt in order to attach it. You need to change it to .xlsx

Enjoy

Ian

Click to expand...

The file is not opening


Best regards

/John

@ Ian

It just occurred to me that I used the "new" voltage measured after the 1k and 10k load… therefore the 10k value is a lot lower than the 1k load value.

Should I calculate on the voltage drop instead?

If I do so the calculation is:

Rout = 10 x (x-1)/(10-x) = 10 x (1.191-1)/(10-1,191) = (10 x 1.191)/8.809 = 0.217kohm… 217R

I am beginning to think I am just stupid  ???


Best regards

/John

johnheath said:

@ Ian

It just occurred to me that I used the "new" voltage measured after the 1k and 10k load… therefore the 10k value is a lot lower than the 1k load value.

Should I calculate on the voltage drop instead?

If I do so the calculation is:

Rout = 10 x (x-1)/(10-x) = 10 x (1.191-1)/(10-1,191) = (10 x 1.191)/8.809 = 0.217kohm… 217R

I am beginning to think I am just stupid  ???


Best regards

/John

Click to expand...

Rout = 10 x (x-1)/(10-x) = 10 x (1.191-1)/(10-1,191) = (10 x 0.191)/8.809 = 0.217kohm… 217R

Sorry for the typo


Best regards

/John

@JOhn,

I think you are getting confused with the two methods. The formula:

Rout = 10 x (x-1)/(10-x)

is to be used with the Indirect method. Here you send a signal through the amplifier you are testing and measure the output level with a 10K and then a 1K load. You are NOT injecying a signal into the output.

With the Direct method, you have no signal going through the amplifier. Instead you inject a signal from a generator into the output via a 1K resistor. You measure the voltage between the generator and 0V (V1) and between the amplifier output and 0V (V2). The output impedance is:

Rout = V2 *1000/(V1-V2) and the answer is in ohms.

To open the spreadsheet you need to change the extension to .xlsx .

Cheers

Ian

ruffrecords said:

@JOhn,

I think you are getting confused with the two methods. The formula:

Rout = 10 x (x-1)/(10-x)

is to be used with the Indirect method. Here you send a signal through the amplifier you are testing and measure the output level with a 10K and then a 1K load. You are NOT injecying a signal into the output.

Cheers

Ian

Click to expand...

Aah… If that is the way I have been doing it wrong.

Back to the bench

Best regards

/John

I hope this is as important as the time and effort invested.

Generally wrt source impedance it is either low enough or not.

JR

JohnRoberts said:

I hope this is as important as the time and effort invested.

Generally wrt source impedance it is either low enough or not.

JR

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Yees, it is a good question isn't it?

But in fact it is just what you imply, low enough or not… me being able to find out without having to send schematics all over and ask somebody else to do the math for me.

Now, with a few methods on hand (thank you all) I can in fact do this… measuring with both the "direct" and the "indirect" method given here at the forum gives the same result (Zout) so I guess I can use this in the future to mess around with gain and transformer ratios and all to get a good device.

Best regards

/John

@ Ian

Thank you sir

It seems to work now… both methods gives the same result so I guess that I am in the ballpark now.

Thanks again

Best regards

/John

ruffrecords said:

This already accounted for by measuring the the voltage on either side of the the 1K resistor, The current is V2/Rout.

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OK, got it.

> either low enough or not.

Which is why I suggested (a variant of): try 10K and 600r. If level does not drop "much", Zout is low enough for rock-n-roll. A 1dB drop is fine but must be accounted when doing precision gain-tests (which I gather nobody does anymore).

PRR said:

> either low enough or not.

Which is why I suggested (a variant of): try 10K and 600r. If level does not drop "much", Zout is low enough for rock-n-roll. A 1dB drop is fine but must be accounted when doing precision gain-tests (which I gather nobody does anymore).

Click to expand...

Yes… thank you

During this process I experimented with different load… both before and after the output transformer and I realized that load somehow alter the possibility to use maximum gain… meaning that the total output changed but also the clipping of the signal.

I guess that you could calculate the maximum gain before clipping depending on the load? And 10k and 600R load could be useful loads to compare I guess?


Best regards

/John

johnheath said:

Yes… thank you

During this process I experimented with different load… both before and after the output transformer and I realized that load somehow alter the possibility to use maximum gain… meaning that the total output changed but also the clipping of the signal.

I guess that you could calculate the maximum gain before clipping depending on the load? And 10k and 600R load could be useful loads to compare I guess?


Best regards

/John

Click to expand...

as  I mentioned several days ago

There is a difference between a nominal output termination impedance,  and output source impedance, and an output drive capability, which can be three completely different numbers.

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Output impedance is different than drive capability.

JR

JohnRoberts said:

as  I mentioned several days ago

Output impedance is different than drive capability.

JR

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I am sorry for not understanding it all at the first time, but I hope it all will fall in place one time or another.

I think I have grasped some knowledge about output impedance and drive capability… but am I wrong if I say that the output impedance i relation to the load affects the drive capability?

Best regards

/John

johnheath said:

Yes… thank you

During this process I experimented with different load… both before and after the output transformer and I realized that load somehow alter the possibility to use maximum gain… meaning that the total output changed but also the clipping of the signal.

I guess that you could calculate the maximum gain before clipping depending on the load? And 10k and 600R load could be useful loads to compare I guess?


Best regards

/John

Click to expand...

Now you are beginning to see the difference between output impedance and drive capability.

Cheers

Ian

ruffrecords said:

Now you are beginning to see the difference between output impedance and drive capability.

Cheers

Ian

Click to expand...

Yes thank you sir

It is all due to people like you and many more very competent people here at this forum… I believe that I owe a lot

Best regards

/John