Simple(?) potmeter + opamp puzzle

Hi all,

The goal is simple: I want to build a non-inverting opamp circuit with a 10K linear potmeter (the MAX5389 digital potmeter, to be precise), with a gain between 1/10 and 10, depending on the potmeter's wiper position. Ideally, I get 1/10 with the potmeter in the lowest position, 1 (unity gain) in the middle position, and 10 in the highest position.

Problem: I have no negative supply rail, the input signal is always positive, and the ground reference is also the opamp's lower supply voltage, so any inverting stage is out of the question -- which is a bit of a shame, of course, since that would offer an almost trivially simple solution.

Of course I can simply build a (roughly) 10x amplifier with a 100K/10K resistor divider as a negative feedback loop, and feed the + input through the potmeter in standard volume control configuration (signal fed in the top, bottom to ground). This, however, has a severe drawback: for unity gain, the potmeter must already be turned nine-tenths down, and lower gains are even trickier to set. Also, the opamp itself is always set to the maximum gain of 10, which isn't optimal from a noise point of view.

I've been toying around with several configurations, but especially the less-than-unity gain range is presented me with a bit of a challenge. Can anyone offer a smart solution to this little puzzle?

Thanks in advance,

Best regards,

Richard Rasker

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Reply to
Richard Rasker
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A) A 1:10 divider into a non-inverting amplifier with the pot wiper to the opamp's (-) input, one end to ground and the other to the output?

B) The pot, connected in the rheostat configuration, in the low leg of a voltage divider. Set the upper resistor such that it gives a 10:1 divider at minimum pot value and 1:10 at maximum. Follow that with the non-inverting amplifier with Rf/Ri=9.

B1)Variation on B), where the pot is wired as a pot to divide the voltage and the opamp as a non-inverting amp with gain of 10 (Rf/Ri=9).

You really need to know the wiper resistance of the pot and resolution to get the values right.

Reply to
krw

That would be an interesting pot. Are you planning on having one custom made? I don't think any of the lstandard log curves come close to that ratio.

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Reply to
Michael A. Terrell

This would give a gain between 1/10 and infinity (or more precisely: the opamp's open loop gain) -- but this latter can be fixed with an extra 100R resistor to ground. This doesn't solve the mentioned drawbacks though:

- With the potmeter halfway, the gain would still be merely at 1/5, not 1.

- Noise-wise, this is still rather bad: first, the input signal is cut back to 1/10, so in order to get the desired highest 10x gain, the opamp circuit must now have a 100x maximum gain (hence the 100R resistor I mentioned).

Not a bad idea, but there's still the problem that with the potmeter in halfway position, the gain is nowhere near unity.

As I said, the potmeter is the MAX5389, a digital linear potmeter with a 10K overall resistance, and 256 taps.

Thanks for your effort anyway, and I'll keep your configurations in mind.

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

With two inverting opamp stages, a standard linear pot will do the trick just fine: _ POT ___ ___ ___ '\__ ___ In o---|___|---|___|---|___|---|_\_|--|___|----o Out R | R | 1K 10K | 1K | | |\ | | |\ | `-|-\ | `-|-\ | | >--' | >----' .-|+/ .-|+/ | |/ | |/ === ===

Pot to the left: gain Out/In = +10x; pot to the right: gain Out/In = +1/10 (OK, 11 and 1/11, but that's not the point). Pot halfway: gain = 1. Exactly what I need.

My problem is that I have this circuit running on a single +5V supply, and that I'd like to keep it that way, so I can't use inverting stages.

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

No, pots, particularly digital ones, have a non-zero wiper resistance.

Noise is an issue, sure. The midrange issue is there but as you note could easily be solved. Linearity is an issue.

You want an exponential transfer function with a linear pot. It won't be easy. ;-)

Reply to
krw

Never heard of a virtual ground? ...Jim Thompson

[On the Road, in New York]
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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Yes I have, and I'm thinking about going down that road -- as inverting stages make everything so darn easy. But I'm also curious if there are any neat solution with no inverting stages.

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

I haven't followed your thread closely... is the signal AC only, or is there a need to handle DC signals? ...Jim Thompson

[On the Road, in New York]
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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Run the op amp driving a matched NPN diff pair. Op amp output drives both emitters through a suitable resistor. First transistor has its base connected to a nice stable 1.2ish volt reference, and resistor from op amp output to the emitters, feedback from one collector and output from the other collector into the SJ of a second op amp as a TIA.

Drive the base of the second transistor with a voltage divider between the 1.2V reference and the pot. You should be able to get a 1000x range that way.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
Principal Consultant
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Reply to
Phil Hobbs

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Put a 2x 47k across your 10k pot to make it 9k, then you get 0.1 - 10.

NT

Reply to
NT

On a sunny day (Fri, 26 Aug 2011 22:28:04 +0200) it happened Richard Rasker wrote in :

It takes next to no parts to generate a low current capable negative supply.

Reply to
Jan Panteltje

It has to handle DC signals -- the input signal is always positive -- and I realize now that virtual ground isn't going to work.

I think I'll create -5V after all, e.g. with a MAX660.

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

I know, and I'm rather tempted to simply chuck in a MAX660 and have done with it.

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

OK, this I can follow ...

... but this is becoming a bit hazy. I know that TIA stands for transimpedance amplifier, but what is meant with SJ?

Even though I fail to fully envision your circuit's geometry, it does sound an awful lot like a basic multiplier setup -- is that right?

Anyway, there doesn't appear to be a simple non-inverting solution, so I think I'll make a -5V supply after all, and go with two subsequent inverting stages.

Thanks everyone for thinking along and the suggestions you came up with :-)

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

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There was a very comprehensive and useful manual covering opamp circuits that has been resurrected by Analog Devices. It is entitled 'Applications manual for operational amplifiers for modeling, measuring, manipulating and much else', originally published by Philbrick/Nexus Research. It may be downloaded from:

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mplifiers.html

as a whole or in parts. If it does not help with this particular problem it will still be a most useful general reference.

Reply to
Prof78

"Summing Junction", aka "summing node"

...Jim Thompson

[On the Road, in New York]
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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Did you think about a multiplying DAC? Since your control is already digital, this might be another option. This route may also eliminate all kinds of offset errors from the opamps.

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Failure does not prove something is impossible, failure simply
indicates you are not using the right tools...
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Reply to
Nico Coesel

The digital potmeter is controlled by some simple logic, and there's no microcontroller in the rest of the setup. Also, accuracy isn't really important. I can't go into details, but the aim is to pinpoint the wavelength of a resonance peak in a sort of optical spectrum analyzer, where the resonance peak can vary wildly in strength and width -- the only constant is that it's symmetrical. The idea is to increase or decrase the signal's gain until the peak just intersects a particular level; the wavelength at which the peak occurs is then found exactly halfway between these intersection points. Sure, all this can be done digitally too, but that would involve fast, accurate A/D-converters and a fair bit of programming. The people I'm building this for just want simple analog levels, so that's what they get.

Richard Rasker

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http://www.linetec.nl
Reply to
Richard Rasker

Hi Richard,

first you can move your reference level. There is no need to hold this on the level you get it. You just need three resistors to move your signal to an offset you desire and reduce the signal (for example move it to half the outputrange of the opamp and reduce the signal to half the voltage of this range). Now it is easy to use two inverting circuit stages. If there is a need to shift the level back to the original ground this could mostly be done in the last opamp stage.

Marte

Reply to
Marte Schwarz

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