How do I calculate input impedance of this?

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I have this SE to balanced converter using signal transformer. It is buffered with an OPA551.
The circuit looks something like this.
IMG_20240826_110032.jpg
4.7uf series and a 10k shunt before the op amp.
Do I need to be looking at the OP551 DS to know this?
Thanks
 
Depends on how accurately you need to know it. The negative feedback makes the input impedance of the op-amp very high at audio frequencies and the coupling capacitor is supposed to have negligible impact over the audio band, so in first approximation, the input impedance is 10 kohm. It gets more complicated if you also care about frequencies outside the audio band, or need very accurate figures.
 
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That's a difficult question, because the value for such a potmeter is a compromise. I guess I would increase the 10 kohm resistors to 100 kohm and use a 20 or 25 kohm potmeter. The settling time constant of the 4.7 uF capacitor would then increase to about half a second, which is still reasonable, and the DC voltage drop across the resistor is still negligible, because the OPA551 has only 100 pA of input bias current (see https://www.ti.com/document-viewer/opa551/datasheet ).

A. For minimal noise, the lower the potmeter's resistance the better.
B. To make it easy to drive for whatever equipment will be connected to the input of the potmeter, the higher the input impedance including the potmeter the better.
B2. For minimal bass loss and bass phase shift across the output capacitor of the driving equipment, if any, the higher the input impedance including the potmeter the better.
C. I guess distortion of the wiper contact resistance will be less with lower-valued potmeters, but I'm not sure of that. It's very small anyway.
D. The deviation from the taper of the potmeter will be small when the potmeter's resistance is smaller than the input impedance of the circuit it drives.

A. When I look in the OPA551 datasheet, I see that its voltage noise is 14 nV/sqrt(Hz) and that its current noise is very small. Some calculations show that any potmeter of less than 47 kohm will produce less noise than the OPA551.

B and B2. I usually aim for an input impedance of at least 10 kohm for an unbalanced line input, but that would in this case lead to an infinite value for the potmeter as the potmeter and the resistor are in parallel when the gain is set to maximum. An infinite resistance potmeter is, of course, impractical. Increasing the 10 kohm to some higher value would solve this.

C, D: for this, a higher value of the resistor also helps.

If increasing the input resistors is impossible or impractical, I guess I would use a 10 kohm potmeter and hope the driving circuit can handle a 5 kohm load well.
 
Wow thank you for that very in-depth reply. I appreciate the time you have invested to do so.
It is a little 'beyond my pay grade ' in parts but I shall enjoy reading and learning more in those areas.
Kind regards
James
 
Thanks for the additional input. The initial 10k to ground before the (4.7uf) cap is easily done. And yes the pot will be wired to the board, not on the front panel. Probably just for initial set up purposes. Then a resistive divider could be used once the value is known. This is all part of a biamping/active project and so balancing the LF and HF.
 
IMG_20240828_060929.jpg


This is how I understand the recommendations given in this thread.

You got contradictory recommendations for the potmeter value from totally analogue and me, but both recommendations are perfecly reasonable. The large resistor seems superfluous to me, but it won't do any harm and it can be useful if the converter is also to be used without the potmeter (and without the fixed voltage divider that will replace the potmeter).
 
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I have not investigated the op-amp you intend to use and for what conditions it is unity gain stable.
Presenting a very low impedance, that might occur when potentiometer reaches end positions might make the op-amp oscillate.
A small series resistor can prevent that from happening.
Keep this in mind if you run into oscillation problems during trial and error testing.

Skärmbild 2024-08-28 071429.jpg
 
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This is how I understand the recommendations given in this thread.
To answer your initial question the input impedance of the op in this circuit is very high, meg ohms. So the active impedance will be set by the grounding resistor, in my suggestion 120K. It can be anything but should be at least 10x the value of the pot. What value pot you use depends on what you are feeding in. Most solid sate devices will be happy with 10k if you are feeding from a tube circuit you will likely need 100K or greater. The resistor from the wiper to ground is not strictly needed but it can help with noise as the pot wipers don't always make the best contact. If this resistor is made too small i.e. close to the value of the pot you will change the law of the pot and the range might feel a bit strange. You mentioned eventually making the input attenuator a divider so in that case it is definitely not needed.
 
@totally analogue , in post #9, you recommended 100 kohm, not 120 kohm. Not that it makes much of a difference, but at least 100 kohm matches what I recommended to @jimk04

With the values I recommended in post #4, the deviation from the potmeter's law is about 0.5 dB at the worst point, when the potmeter is set halfway. I doubt anyone would even notice that.

Regarding the big resistor, suppose that whatever will drive the input will have a small DC offset. There will then also be a small DC offset at the wiper of the potmeter. The AC coupling capacitor gets charged to this voltage, so it won't affect the buffer. So far, so good.

Now suppose the wiper of the potmeter loses contact for a short time. Without the big resistor, the voltage on the left side of the coupling capacitor will just stay constant, so apart from temporarily losing the signal, nothing happens. With big resistor, the resistor tries to pull down the left side of the capacitor, so you get a jump in the voltage at the buffer input. That is, the cracking sound you get due to poor wiper contact gets stronger.

Without DC offset, there is no such issue. With or without big resistor, the voltage at the left side of the potmeter then just stays zero when the wiper momentarily loses contact.

If the wiper would lose contact for hours, then without the big resistor, due to capacitor leakage, the left side of the capacitor would settle to the same DC voltage as there is on the right. I don't think that is a problem for two reasons: first, there hardly is any DC voltage on the right side: 100 pA of bias current times 100 kohm is only 10 uV. Second, if the potmeter is that rotten, it needs to be replaced anyway.