Opamp + digital potentiometer = no zero

[JK Sleepling]

Hi,

I have a buffer/splitter circuit with boost controls. Each send is controlled by hi-V digital potentiometers (AD5290).

My problem is that these (like most) digipots cannot go to zero ohms because of the wiper resistance. I measure this to somewhere between 500 and 800 ohms (even though the datasheet says 150 ohms). This means I cannot turn a channel all the way down in this inverting op-amp setup. And there's a slight audible signal in my breadboard setup.

Running a freq-domain simulation shows that minimum is -32dB. Any ideas for an op-amp configuration to get this minimum somewhat lower?

https://www.circuitlab.com/editor/#?id=ehsyw37x9qj8

[antonis]

Make R3 as big as you can afford and R15 zero (short)..

e.g. for R3 100k, R15 zero and pot min 800R you'll get about -42dB..

[JK Sleepling]

DIGIPOT1 is the AD5290. The maximum resistance is 100K.
The R15 is my way of adding the messured wiper resistance to the simulations. It is not there in reality.
The 0-100k / 27K setup gives me a gain of 11.4 dB which is great for what I need.

[antonis]

The 0-100k / 27K setup gives me a gain of 11.4 dB which is great for what I need.

If so,  make R3 bigger and  implement a T network into NFB gain stage..
(you'll have to trim feedback resistors, of course..)



The trick here is to have high gain by using low value feedback resistor..

[R.G.]

Use a sledgehammer.  If you have a digital anything in there, sense when you want zero, and turn on a muting FET when you get there.

[Vivek]

Would this work in your application ?


First Buffer feeds 3 digipots to ground

The 3 wipers are connected to 3 Opamps with as much gain as you require

[JK Sleepling]

Use a sledgehammer.  If you have a digital anything in there, sense when you want zero, and turn on a muting FET when you get there.

Yup - that one also crossed my mind

[JK Sleepling]

Would this work in your application ?


First Buffer feeds 3 digipots to ground

The 3 wipers are connected to 3 Opamps with as much gain as you require

I'm too much of a noob to see it. Might need a drawing to get it right 

[ElectricDruid]

Would this work in your application ?


First Buffer feeds 3 digipots to ground

The 3 wipers are connected to 3 Opamps with as much gain as you require

+1 agree. Wire the digipot as a volume control on the output of an op-amp with fixed gain of whatever the maximum you need is.

A little bit of series resistance going out of the output won't be a big deal, and I bet the digipot "value zero" is right at the bottom of the resistor chain, so you should get solid silence.

[JK Sleepling]

Would this work in your application ?


First Buffer feeds 3 digipots to ground

The 3 wipers are connected to 3 Opamps with as much gain as you require

+1 agree. Wire the digipot as a volume control on the output of an op-amp with fixed gain of whatever the maximum you need is.

A little bit of series resistance going out of the output won't be a big deal, and I bet the digipot "value zero" is right at the bottom of the resistor chain, so you should get solid silence.

Thanks guys!
I made the change to the "Send 2" - channel in circuitLab. And it looks great when simulating the output!
Is this also the circuit you have in mind?

[PRR]

This may give 1st-order cancellation of the leak.

[ElectricDruid]

Thanks guys!
I made the change to the "Send 2" - channel in circuitLab. And it looks great when simulating the output!
Is this also the circuit you have in mind?

Yes. You could also do it with a non-inverting op-amp, if that suits better.

[Vivek]

You placed digipot after 2 Opamps

If send 2 has a 100K digipot at the very output, maybe the impedance seen by the next stage after wiper is 25K

Maybe that is sufficiently low of an output impedance and will work


However if you place the digipot in between the two opamps, the output impedance will be effectively zero (You might add a 2K in series with right opamp, as current limit resistor)


Please try to simulate :

Leftmost Opamp is input buffer
It feeds digipot(s) as a volume control(s)
wiper of digipot goes to right opamp (buffer + gain). (This Opamp has 2K CLR)

[Rob Strand]

I can't see how the divider method works any better than the feedback resistor method.   Am I missing something?

For the inverting amp with feedback method if wire like this,

   Input R to   Digital pot "cw"
   Digital pot "wiper" to opamp - input
   opamp output to Digital pot "ccw"

Digital pot set-up as 3-wires.  Then when the pot is set to minimum the input resistor increases and that will increase the amount of attenuation in the minimum gain setting.

The hard mute RG suggest is probably the best.

For a one off project I don't see what's wrong with PRR's null idea.  You might need to tweak the resistor value to get a deep null.

[Vivek]

Maybe divider method can give minimum gain of Zero

and feedback resistor method can give minimum gain of One ie some leakage always present.

[Rob Strand]

Maybe divider method can give minimum gain of Zero

I had a look through the datasheet before but I couldn't see it.   It could give zero output if the "three wire" pot mode
opened the pot but I can't imagine that happening.  (If so it might also work in the three wire pot feedback connection I outlined.)

[Vivek]

I did not read specs,

But suppose digipot has 256 steps

and zero is not included as smallest step

Then smallest step could be 1/256th of total

ie -48dB when used as volume control

[PRR]

I did not read specs,....

You maybe should. It helps prevent going off on a wrong path.

This "pot" has three layers of switching. So it has a significant "wiper resistance". This happens in either rheostat or potentiometer mode.

There are other digi-pots but this year we may have to use what is in-hand, cuz so much has become hard to get at any price.

[ElectricDruid]

Here's the datasheet:

https://www.analog.com/media/en/technical-documentation/data-sheets/AD5290.pdf

The most significant thing in my view is figure 27 on page 15, which shows how the digipot is organised. As Paul said, it's got three layers of switching, which makes it somewhat less than ideal for this particular job.

On simpler digipots where the "pot" itself is simply a chain of resistors from Va to Vb and the wiper is a 256-position switch that connects to one of those points, the wiper resistance is a lot less relevant. At value zero, the switch is connected directed to the bottom of the pot, whatever voltage that is. If that's grounded, it's ground. And 150R of resistance in the wiper doesn't make any odds, because you won't hear a signal if your output is connected to ground. Imagine it as a volume control with a small series resistor on the wiper. If you turn it down to zero, it's zero, and it doesn't really matter what the series resistor is - it could be 1K and we still wouldn't care.
That doesn't work if you use the digipot as a  variable resistor in a feedback loop like that because for that to produce no sound, the value has to go to zero, and since you've always got that wiper resistance, it can't.


In this case, even the "volume control" method doesn't work because we don't have a direct connection to the "bottom" of the pot. Instead we have a series of switches between us and ground, so our output never is directly tied to ground. Plus the three layers of switches make the wiper resistance larger (it's quoted as "3 x Rw" in the equation, since it is literally three lots of switch resistance).

So...dunno how you can do it with this switch. Using the 100K version so the wiper resistance isn't such a big percentage of the whole value would help a bit.

[Vivek]

Tom, thanks for posting the spec sheet

Figure 27 is great !

It seems to explain internal architecture and how it affects the rheostat mode.

The voltage divider formula is on the next page.

And it shows that in almost all cases, wiper resistance is immaterial in voltage divider application. And it shows you can get 0V on wiper if B is tied to ground.