Question about output buffers and impedance

Started by kraal, August 13, 2020, 04:20:55 PM

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kraal

Hello,

I tried to find answers to the following questions by myself, but with no great success (i.e. I'm still not sure to understand correctly). It would be great if you could help me.

When using aTL071 opamp in unity gain configuration as a buffer (Avd=200 V/mV, 160 Ohm internal output resistance) with no other series resistor, nor output pulldown resistor, with an electric guitar (roughly frequency range of 200Hz-2000kHz), TL071 datasheet states that the gain in this frequency range is ~10k-1k.

  • my understanding is that the output impedance (with no load in open loop) will be between ~160/10000 @ 200Hz and ~160/1000 @ 2000Hz. Question 1: Am I correct ?
  • If I'm correct, as this is a really low output impedance (negligible?). Question 2: why do some people add  a 100R resistor at the unity gain opamp buffer's output ?
  • when plugin' a cable, the impedance will raise with cable length. With the above Zout, when using the following calculator [1] (with 120pfF/m and a 30m long cable) the 3dB cutoff occurs at fc=276.3MHz which is ways above the audible range (1000x higher). According to the second calculator on the same page, in order to lose 3dB at 20kHz the cable should be 400+ km long (?!?) Even with 160 ohm at Zout, the cutoff would occur at 20khz. Question 3: Am I misunderstanding something (cable resistance per meter ?)(and back to question 2 this makes me wonder why some people add an output resistor in series to increase Zout as it lowers the cutoff frequency) ?
  • Question 4: I understand that adding a pulldown resistor at the output can lower Zout even more, but why even bothering adding one if Zout is low enough (to create a RC filter with the ouput capacitor in order to cut even more trebble at the effect's output, while they put a buffer to avoir this?) ?
  • Question 5: if the output is grounded in bypass, is it then required to have a pulldown resistor at the output of the effect ? (if yes this may answer question 4)

Thanks a lot in advance for your answers and kind regards,

Michel

[1] http://www.sengpielaudio.com/calculator-cable.htm

Edit: added missing link

antonis

#1
>When using aTL071 opamp in unity gain configuration as a buffer (Avd=200 V/mV, 160 Ohm internal output resistance) with no other series resistor, nor output pulldown resistor, with an electric guitar (roughly frequency range of 200Hz-2000kHz), TL071 datasheet states that the gain in this frequency range is ~10k-1k.<

You might read datasheet incorrectly.. :icon_wink:
(a unity gain buffer has gain of unity..)

>my understanding is that the output impedance (with no load in open loop) will be between ~160/10000 @ 200Hz and ~160/1000 @ 2000Hz. Question 1: Am I correct ?<

Generaly speaking, yes..
edit: Trying to get your point but..

>as this is a really low output impedance (negligible?)<

Compared to much higher impedance to be driven, yes..

>why do some people add  a 100R resistor at the unity gain opamp buffer's output ?<

For reasons other than impedance concern..

>when plugin' a cable, the impedance will raise with cable length. With the above Zout, when using the following calculator [1] (with 120pfF/m and a 30m long cable) the 3dB cutoff occurs at fc=276.3MHz which is ways above the audible range (1000x higher). According to the second calculator, in order to loose 3dB at 20kHz the cable should be 400+ km long (?!?) Even with 160 ohm at Zout, the cutoff would occur at 20khz. Question 3: Am I misunderstanding something (cable resistance per meter ?)(and back to question 2 this makes me wonder why some people add an output resistor in series to increase Zout) ?<

Sorry but I can't follow you..
You maybe mess with input/output impedances..

>I understand that adding a pulldown resistor at the output can lower Zout even more, but why even bothering adding one if Zout is low enough (to create a RC filter with the ouput capacitor in order to cut even more trebble at the effect's output, while they put a buffer to avoir this?) ?<

Zout is Zout and no external item, other than those involved in feedback loop, can lower it..
Anything added from Zout to ground simply lowers total load impedance hence raises current driving demands..
(from another point of view, pulldown resistor actually raises Zout/Load ratio..)

>if the output is grounded in bypass, is it then required to have a pulldown resistor at the output of the effect ? <

Not necessarily but many people do it..
(it has to do nothing with impedances - just for OUT cap popping issues..)

P.S.
https://www.youtube.com/watch?v=dbCcnnqQxpU
https://www.youtube.com/watch?v=gtJPeh3HvHU
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

kraal

Thank you for taking the time to answer my questions.

Quote from: antonis on August 13, 2020, 05:01:49 PM
You might read datasheet incorrectly.. :icon_wink:
(a unity gain buffer has gain of unity..)

Unity yes... (never ask questions when you're tired)  :icon_redface:
but then does it mean that the "raw" output impedance is 160 ohm over the full frequency range when using a unity gain buffer ?

Quote from: antonis on August 13, 2020, 05:01:49 PM
You maybe mess with input/output impedances..

There's definitively something I don't get about the role/impact of the cable, and which makes me unable to express my question correctly.
Maybe the best next question to ask is "what book, document can I read on this topic ?"

Again thanks.

antonis

#3
Quote from: kraal on August 13, 2020, 06:00:23 PM
Unity yes... (never ask questions when you're tired)  :icon_redface:
but then does it mean that the "raw" output impedance is 160 ohm over the full frequency range when using a unity gain buffer ?

Not quite..
For unity gain buffer, Zout is reduced by the loop gain (open loop gain X feedback factor) following the general formula: ZCL = ZOL / (1 + β*AOL), where ZCL is close loop output impedance, ZOL is open loop one, AOL is open loop gain and β is feedback factor (unity for buffer)

You can see that unity gain buffer configuration exhibits the lowest Zout of any other particular op-amp configuration (when bandwidth isn't exceeded) 'cause it has the highest β..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

ElectricDruid

Why do you *actually* want to know the answers to these particular questions? Is there a specific reason? Something you're trying to do that doesn't work because of some issue related to this?

Seems to me you're probably getting bogged down in technical details. Do you *need* to worry about any of this stuff? Op-amp output impedance is "very low" - like you said, "negligible" - so why consider it in such depth?

</curious>

kraal

Quote from: ElectricDruid on August 13, 2020, 07:54:29 PM
Why do you *actually* want to know the answers to these particular questions? Is there a specific reason? Something you're trying to do that doesn't work because of some issue related to this?

Seems to me you're probably getting bogged down in technical details. Do you *need* to worry about any of this stuff? Op-amp output impedance is "very low" - like you said, "negligible" - so why consider it in such depth?

</curious>
Is "curiosity" an acceptable enough reason ?

merlinb

Quote from: kraal on August 13, 2020, 04:20:55 PM
  • If I'm correct, as this is a really low output impedance (negligible?). Question 2: why do some people add  a 100R resistor at the unity gain opamp buffer's output ?
To isolate the opamp from load capacitance, i.e. to prevent oscillation.

kraal

Quote from: merlinb on August 14, 2020, 03:21:25 AM
Quote from: kraal on August 13, 2020, 04:20:55 PM
  • If I'm correct, as this is a really low output impedance (negligible?). Question 2: why do some people add  a 100R resistor at the unity gain opamp buffer's output ?
To isolate the opamp from load capacitance, i.e. to prevent oscillation.

I woke up less stupid than when I went to bed, thank you :-)
Based on your answer I found these articles [1,2] which explain a lot. Maybe these can help other people as well.

[1] https://www.analog.com/en/analog-dialogue/articles/techniques-to-avoid-instability-capacitive-loading.html
[2] https://www.analog.com/en/analog-dialogue/articles/ask-the-applications-engineer-25.html

kraal

Quote from: antonis on August 13, 2020, 06:46:11 PM
For unity gain buffer, Zout is reduced by the loop gain (open loop gain X feedback factor) following the general formula: ZCL = ZOL / (1 + β*AOL), where ZCL is close loop output impedance, ZOL is open loop one, AOL is open loop gain and β is feedback factor (unity for buffer)

You can see that unity gain buffer configuration exhibits the lowest Zout of any other particular op-amp configuration (when bandwidth isn't exceeded) 'cause it has the highest β..

Thank you, I'm still not sure to understand how to exactly use the formula but I'll find out :-)

antonis

#9
You don't have to use it at all..!!!  :icon_wink:
(just keep considering Zout of voltage series feedback amplifier negligible..)

e.g.
Consider an op-amp with open loop gain of 105, Zout-open of 160ohms and close loop gain of a) 100   b) 1 (buffer)..
in case of a) Zout is 160/1000    -> 160mohms
//   //     b) Zout is 160/100000 -> 1.6mohms

You can clearly see that for both cases, what counts more is wiring/tracks/solder joints resistance than Zout itself.. :icon_wink:

P.S.
1000 & 100000 in denominators should actually be 1001 & 100001 respectively..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

kraal

Quote from: antonis on August 14, 2020, 06:07:31 AM
You don't have to use it at all..!!!  :icon_wink:
(just keep considering Zout of voltage series feedback amplifier negligible..)


You know some people are satisfied with a raw "it's negligible" fact. I'm not :icon_lol:
I prefer to use it at least once with different values, understand the output, document it for myself then say "ok that's how it works".
(I just imagined myself writing "negligible" in my schematics documentation, having my daugther look at it, ask me "why" and not being able to even start answering the question...)

Quote from: antonis on August 14, 2020, 06:07:31 AM
1000 & 100000 in denominators should actually be 1001 & 100001 respectively..

So it is the part I was missing (the "+1") and thus it is the formula I used in my first message (while omitting the "+1"), and now I understand it better,
Is, for non-unity gain, the value of β calculated using RG/(RF+RG) (where F stands for feedback resitance in the closed loop, and G resistance to ground in the closed loop) ?

Thanks again and cheers,

antonis

#11
Although your daughter should be more opportune to answer (considering she is already Blackman's theorem & impedance formula connoisseur..), yes..!! :icon_wink:

You can remember β as the inverse of Gain formula for inverting input configuration..
(which Gain actually derives from β delineation but let it be..)

P.S.
And DO NOT say " I'm not letting it be in cold blood 'cause if we deal charitably with it we'll need a better forum server..)
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

ElectricDruid

Quote from: kraal on August 13, 2020, 08:18:27 PM
Quote from: ElectricDruid on August 13, 2020, 07:54:29 PM
Why do you *actually* want to know the answers to these particular questions? Is there a specific reason? Something you're trying to do that doesn't work because of some issue related to this?

Seems to me you're probably getting bogged down in technical details. Do you *need* to worry about any of this stuff? Op-amp output impedance is "very low" - like you said, "negligible" - so why consider it in such depth?

</curious>
Is "curiosity" an acceptable enough reason ?

Of course! If you just want to know, go for it.

I only ask because sometimes people pop up around here asking very complicated questions, but in reality they only need quite a simple answer, but then they started reading a datasheet and became convinced that nV/Hz and uSec/rBit and unwanted oscillations on the bifurcated flange sprocket have all definitely got something to do with it!! ;)

antonis

Shouldn't nV/sqrt*Hz be more precise noise term, Tom..??  :icon_lol:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

kraal

Quote from: antonis on August 14, 2020, 07:25:43 AM
You can remember β as the inverse of Gain formula for inverting input configuration..
(which Gain actually derives from β delineation but let it be..)

Thanks for the answer (and for sharing knowledge).

PRR

TL07* amp has naked Zout about 160r and GBW about 6MHz(?), so gain at 6kHz is about 1,000. Then closed-loop Zout is 160/1000 or less than an ohm.

However a TL07* is not happy driving loads under a few hundred ohms. And your 100 foot cable is 200r at 27kHz. While your guitar won't make 27kHz, the opamp will gain-up its own 27kHz noise and squeal supersonics. With a few hundred ohms between chip and world, the chip can do its thing and never see less than a few hundred ohms.

Pulldown has no effect unless the loss of level is large (so not a pull-down but an attenuator).

Your "fc=276.3MHz" probably has no physical meaning. (This seems to be Zout for a high audio frequency, against 3600pF, which does figure to hundreds of MHz; but Zout will be rising and the TL07* gets totally lame at a few MHz, so hundreds MHz is out-of-sight.)
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kraal

Thank you Paul,
Quote from: PRR on August 16, 2020, 12:57:26 AM
However a TL07* is not happy driving loads under a few hundred ohms.
That's for the TL07*. How can I know if this is the case for other op amps (and if it is the case, how can I know the minimal load to be driven) ?
What would be the information I should be looking for in datasheets (the graphs in TL07*'d datasheet don't show values below 100ohm, but I don't think that it is a good indicator :-) )?

Regards,

Rob Strand

#17
QuoteThat's for the TL07*. How can I know if this is the case for other op amps (and if it is the case, how can I know the minimal load to be driven) ?
What would be the information I should be looking for in datasheets (the graphs in TL07*'d datasheet don't show values below 100ohm, but I don't think that it is a good indicator :-) )?
If it's not stated in the data sheet then there's no way to know other than by experience or experiment.

The minimum load and capacitive loads are often not specified.

You can work out minimum resistive load from datasheet from current limit and power dissipation info in the datasheet.    You have to consider output voltage swing reduction as well.

It's quite common to put something between 50 ohm and 1k in series with the output.   Exactly what value the minimum is for a given opamp is not known.

The output impedance of the opamp is a possible guide but really it's presence is more of a guide to explain why adding *a* resistor on the output helps.

If you look at this diagram R1 is the output impedance of the opamp,  R2 is the added series resistor, C is the capacitive load.



The load connects where C is but the feedback connects to where R1 and R2 join.    Where the feedback connects has extra phase-shift due to the capacitor.    The extra phase shift can make the opamp oscillate.  You often have somewhere around 30 deg to 60 degrees of slack before you get oscillation.   Making R2 larger limits the maximum phase-shift.  If you choose R1 = R2 the maximum phase shift is about 20 degrees so that usually covers it.   The exact numbers aren't important, the point is it helps.

The maximum phase shift is covered here,
https://www.electrical4u.com/compensation-in-control-system-lag-lead-compensation/

There other tricks about adding a feedback cap which helps stability with capacitive loads on opamps, do a google search.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

kraal

Quote from: Rob Strand on August 16, 2020, 04:13:43 AM
If it's not stated in the data sheet then there's no way to know other than by experience or experiment.

The minimum load and capacitive loads are often not specified.

You can work out minimum resistive load from datasheet from current limit and power dissipation info in the datasheet.    You have to consider output voltage swing reduction as well.

It's quite common to put something between 50 ohm and 1k in series with the output.   Exactly what value the minimum is for a given opamp is not known.

The output impedance of the opamp is a possible guide but really it's presence is more of a guide to explain why adding *a* resistor on the output helps.

If you look at this diagram R1 is the output impedance of the opamp,  R2 is the added series resistor, C is the capacitive load.



The load connects where C is but the feedback connects to where R1 and R2 join.    Where the feedback connects has extra phase-shift due to the capacitor.    The extra phase shift can make the opamp oscillate.  You often have somewhere around 30 deg to 60 degrees of slack before you get oscillation.   Making R2 larger limits the maximum phase-shift.  If you choose R1 = R2 the maximum phase shift is about 20 degrees so that usually covers it.   The exact numbers aren't important, the point is it helps.

The maximum phase shift is covered here,
https://www.electrical4u.com/compensation-in-control-system-lag-lead-compensation/

There other tricks about adding a feedback cap which helps stability with capacitive loads on opamps, do a google search.

Thank you a lot for your answer, the link and reading pointer.
I think that I have now all the answers I needed :-)

Kind regards

PRR

#19
> how can I know the minimal load to be driven

Any "normal" opamp will drive 10k. Nearly all will drive 2k. Or lower but performance falls off.

The TL07x chips are nearly as old as, and suggested to replace, '741 and '709 chips, and these specified 2k load at good performance.

The datasheet is a sales-tool, and avoids tough details. However you can see that it swings plenty in 10k load, less but still good in 2k.

The schematic suggests 150-200r dead resistance in the output. We usually like to load with >10X the internal resistance (so we get 90% voltage), and this too suggests >2K.

But do not trust the datasheet further than you can throw it. The junior draftsman plotted for 0.1K (100r) load, just zero volts. Got lin confused with log. While not supported by datashet, I "know" the TL0s will put a couple or 3 volts into 100r, and the plot should be more like the red line.

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