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Old 19th September 2014, 11:46 PM   #1
ammel68 is offline ammel68  United States
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Default OPA2134 Input Stage...

Shown below is one channel of the input stage of a headphone amp PCB that I've tried my best to reverse engineer.

Can someone please explain how this works as I've never seen anything like this before.
I'm used to one end of the volume pot being connected to ground, but on this design it appears none of the pot's pins are connected to ground.
It seems as though the pot is only changing the resistance between pins 6 and 7 of the OPA2134.

Out on my drawing isn't out to the headphones, it's out to the next stage.

Thanks...
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Old 20th September 2014, 12:10 AM   #2
agdr is offline agdr  United States
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Quote:
Originally Posted by ammel68 View Post
Can someone please explain how this works
The circuit is just two inverting op amp stages, one after the other:

Operational amplifier applications - Wikipedia, the free encyclopedia

The pot is used to vary the amount of the feedback resistor on the second stage. The gain formula for inverting op amps is R(feedback) / R(input). So the gain of that last stage would be whatever resistance the pot is set to, divided by 4.7K.

Having the two inverting stages in series gives you two inversions, of course, resulting in a non-inverted output signal. The input impedance would be that of the first stage input resistor, 47K. The capacitors are all compensation caps.

Having a closed loop gain of less than one, which is what would happen if that pot is reduced to less than 4.7K ohms, may not lead to happy stability results. I would suggest adding a 4.7K resistor in series between the pot and the inverting input of the second stage. That way the lowest gain you would ever get for the second stage is unity: {Rfeedback[0R(pot) + 4.7K] / Rin(=4.7K)} = 1.

Same goes for the first stage. It looks like what they are trying to achieve is a unity gain (inverting) input buffer to drive that low 4.7K input resistance of the second stage so the audio source doesn't have to. But again less than one gain, 33K/47K, probably isn't the best of ideas. Use 47K for the first stage feedback resistor rather than 33K.

Last edited by agdr; 20th September 2014 at 12:21 AM.
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Old 20th September 2014, 01:13 AM   #3
agdr is offline agdr  United States
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Default An inverting attenuator IS stable

Whoops. Ignore my comments above about stability concerns with an inverting gain less than one. I was dead wrong about that one. Here is the correct answer:

The Inverting Attenuator, G = -0.1? is it unstable? - The Signal - Archives - TI E2E Community

An inverting closed loop gain of less than one is an "inverting attenuator" and IS stable. So it looks like both stages in your circuit are just fine as drawn. The first stage is still an inverting buffer, but with a small amount of attenuation: 33K/47K = 0.7x. The second stage voltage gain would go from nearly zero to 50K/4.7K = 10.6x with the pot rotation.

As for why I got this issue wrong, scroll down in that Texas Instruments link to the comment section and read the first two comments. Apparently "common knowlege" about inverting op-amp stability with gains less than one - and a certain section in a popular op-amp book - are just flat out wrong.

Last edited by agdr; 20th September 2014 at 01:32 AM.
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Old 20th September 2014, 03:22 AM   #4
ammel68 is offline ammel68  United States
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Thank you for your replies!

It looked like the pot was varying the feedback resistance, but I could not understand why in the world you would want to do such.
I typically work with non-inverting inputs, not inverting inputs, so everything you stated is new to me and very informative.

Are there any advantages to such a circuit vs. what I consider to be the norm of having the pot control how much signal reaches the input of an op amp?
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Old 20th September 2014, 04:24 AM   #5
agdr is offline agdr  United States
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The inverting configuration is better at cancelling out distortion products than non-inverting. But I never use them due to the resistor noise involved with the input resistor, which winds up in series with the signal path. With a non-inverting stage you get a very high input impedance that winds up in parallel with your (usually low) source impedance or the low output impedance of a previous stage. With the inverting circuit here you wind up with a 47K resistor in series with the input, which adds 3.9uVrms of noise at room temperature. Of course a 100K pot at the input of a non-inverting stage would do the same thing if turned to the middle.

So long story short, if you were to measure THD+N of your 2 stage inverting circuit the THD might be a little less than a 2 stage non-inverting with the pot in front, but the noise (the "N") might be a bit higher. Tradeoffs...

That second stage in your circuit is interesting though, knowing that it is actually stable. I've always assumed a closed loop gain less than 1 wouldn't be, even if the op-amp involved were unity gain stable. That second stage might be a slick way to do attenuation, if using much lower resistor values though to lower the Johnson noise. Like a 600 ohm input resistor instead of 4.7K, if the driving buffer stage (non-inverting in this case) can handle a 600 ohm load at low distortion. Some op-amps can. Then a 5K pot instead of 50K, which would result in a maximum stage gain of 8x with the pot fully rotated. The net result should be a (nearly) 0x to 8x attenuator / gain stage.

Last edited by agdr; 20th September 2014 at 04:39 AM.
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Old 20th September 2014, 05:32 AM   #6
ammel68 is offline ammel68  United States
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Are you suggesting that if I use like a 10K pot(instead of a 50K), along with lower value resistors, that it will result in lower noise?
Those are the values on the board, but I'm always open to better suggestions/improvements as I haven't soldered anything in place.

Another issue here, I really don't 10X gain. Is this 10X gain with volume pot near its lowest level(fully CCW) or with it near its max. setting? I would assume the gain is increasing as it's turned CW.

I usually build op amp circuits with only a gain of 2 or 3.
Can I increase the 4.7K resistor to like 22K, to reduce the gain, or is that going to raise the noise level?

Guess what I'm asking here is if I can use/modify the circuit to have only a gain of 2-3 without using large resistors and raising the noise level?

Last edited by ammel68; 20th September 2014 at 05:37 AM.
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Old 20th September 2014, 03:19 PM   #7
00940 is offline 00940  Belgium
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The whole "pot in the feedback loop" idea is a bad idea. First reason is that DC current will flow through the wiper, leading to noise. Second reason is : what happens if the pot gets dirty and loose contact ? Third reason is that it makes the feedback loop huge.

In your situation, the best solution wrt noise is to have the first stage non inverting, with gain in the first stage. Then a 10K pot. Then a non inverting buffer. This last buffer might or might not be necessary depending on the next stage. Of course, this only works if you know what kind of voltage your sources are pushing; the risk being of clipping the first opamp. If you don't know, go for a non inverting buffer, then 10K pot then non inverting gain stage.

If you insist on using gain in the feedback loop: put a cap in serie with the potentiometer (so it only works at AC) and a resistor in // with the pot and cap to set the DC working point. If this resistor isn't much bigger that the pot, it will interact with it and mess up your attenuation curves. But if it isn't small, it won't protect you against excessive gain if the pot goes open.
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Last edited by 00940; 20th September 2014 at 03:22 PM.
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Old 20th September 2014, 05:47 PM   #8
ammel68 is offline ammel68  United States
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Hi Ben,

No, I'm not insisting on using this design at all. This is just the design that's already on the PCB. I'm always open to better ideas/suggestions.

To answer some of your questions:

Most of my sources are higher output sources around 2V, or so. With the power amp that I have, I simply don't need hardly any gain since it can make the volume pot overly sensitive.
I would be quite happy using just a single dual-channel op amp along with a volume pot for this first stage.

The next stage of this design is the signal passing through two 47 ohm resistors into a TPA6120.
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Old 20th September 2014, 05:55 PM   #9
00940 is offline 00940  Belgium
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Ok. You cannot have the pot in front of the tpa6120; you need an opamp there. The most simple option would be: 10K to 50K pot, opa134 in non-inverting topology, tpa6120. The gain for the input depends on the gain developped by the tpa and your needs.

The following thread is quite useful to determine what you need. I would add a bit more gain for the quiet recording that pops up now and then, especially if you listen to classical music.
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Old 20th September 2014, 06:13 PM   #10
Mooly is offline Mooly  United Kingdom
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Have a look at this for a proper implementation of an active volume control.

Despite it being properly done there is still controversy over its subjective performance, particularly at high levels of attenuation.
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