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Background: I'm often building audio op-amp circuits for various purposes, and for small projects usually prefer single "wall adapter" supply. So for a ground reference I'll either use a simple resistor pair with a capacitor, or better, a spare OP amp configured as a virtual ground, to make my 1/2V point point. But then there's the external ground referencing issue. Since there is only one supply, all ins and outs need blocking capacitors. If I use the "stronger" op-amp developed virtual ground, I can let that ground be the reference for all I/O, thus eliminating lots of capacitors. The downside there is that since ground is really 1/2V, doing this means I can forget about sharing the power supply with any other connected devices. But recently I've been toying with the idea of using a charge pump to create a V- supply, using a separate op amp package. Typically like this...

enter image description here

This would then allow me to use one side of the single supply (negative in this case) as my reference ground again. And so in larger circuit where there are a LOT of ins and outs (perhaps an audio mixer), it would seem using a stand alone charge pump for my negative supply should eliminate the need for a great many capacitors on all the I/O points, and I'd still be able to share the power supply with other small circuits (provided they use a negative ground scheme).

So my question is, is there any major downside I'm missing here? It would seem that as long s the circuits don't need to drive significant loads (that would pull at the V- supply), there might be a lot of advantages to this method. I know there is some issue because with the diode based charge pump, -V won't exactly equal +V. But there are ICs like the CMOS based TI LMC7660 which will do the charge pump function for me with a more accurate -V output and a lot fewer parts. Opinions?

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A "virtual ground" IS NOT a mid-rail generator. – Andy aka Jan 8 at 16:46
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Also I noticed you did not formally accept my answer on this question: electronics.stackexchange.com/questions/202570/… - then I noticed that you'd only ever accepted one answer for all your questions raised by yourself. There is also this answer/question: electronics.stackexchange.com/questions/135255/…. It's a small price that is asked to be paid. – Andy aka Jan 8 at 16:51
    
There are quite a few reasonably good answers on other questions you've raised that haven't been accepted either. Like I say it's the small "price" that is paid for getting good and free advice. – Andy aka Jan 8 at 16:58
    
I don't understand Any. I did "+1" your answer in both of those cases, and I always do that for multiple answers if I find them useful. I just checked both posts you mentioned, and upon trying to up-vote the answer again the system reported I had already done so. I always appreciate advice and try to help others when I think i have something to offer. If there is something I'm not doing, I'm not sure what "formally accept" means. Trust me, I'm not meaning to step on toes, nor deny credit where it is due. – Randy Jan 8 at 16:59
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It's the symbol below the up down arrows on the relevant answer I'm referring to. Great, I'm glad you have a good attitude towards this. BTW you must have managed to "accept" on your very first question by luck!! – Andy aka Jan 8 at 17:01
up vote 4 down vote accepted

Creating a negative supply from a positive one is OK but for audio you may want to use a negative linear regulator after it - this will remove any switching artefacts that might be heard on the audio. For an op-amp it's called power supply rejection: -

enter image description here

This is for the fairly good OP4177 and what it tells you is that (say at 10kHz) the effect of noise from the negative rail is quite small at about 107 dB. This means that if you have 10 kHz ripple on the negative supply, it will become an interfering noise on your inputs that is about 251,000 times smaller. Clearly, if your circuit is low gain this isn't likely to become a noticeable problem.

However, some switching circuits can generate ripple at (say) 1MHz and although this frequency cannot be heard (on the face of it) there could be a signal that is only 560 times lower at the inputs. Because it is at 1 MHz you can find that some op-amps will (due to their input circuits) demodulate this to baseband and you end up with a few back ground howls and whistles.

So this is what you should be wary of and there's nothing as good as a real test.

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Appreciated. I have been a little concerned about the supply rejection of that bit of switching noise at 10K. That LM7660 will be good in many cases where I'm dealing with a +9V or better supply to start with, though it WOULD be nice if they made a little switcher chip like this with a built in oscillator around 20Khz, just to settle that point. :-) – Randy Jan 8 at 19:36
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It's a bit of a slippery slope this one. On one hand having a frequency that is quite low means it's audible but the opamp CMRR can probably handle it. On the other hand, as f rises so CMRR falls and so does the opamp's ability to be dignified in the presence of what you'd normally expect to be inaudible but probably would be detectable due to input imperfections in the opamp. My best guess is 25 kHz for best operating area. However there's nothing to beat a really good opamp with maybe 10 ohm resistors in the power leads. – Andy aka Jan 8 at 21:21
    
+1, completely agree here. Opamps typically use the "analog" part of the PCB and introducing a switching power supply there could cause noise issues. – SunnyBoyNY Jan 9 at 3:38

Yes, using a charge pump to create the negative supply for opamps can be a useful thing to do.

Often you can arrange for your internal signals to be positive, and run most things from a single supply. Sometimes you need just a little headroom below ground. Maybe the opamp you want to use for other reasons doesn't go all the way to the negative supply, maybe some input signal is small but ground-centered, or something else.

I have used a charge pump in such cases several times. Nowadays just about any circuit will have a microcontroller in it, and these things often have clock outputs. Instead of a separate oscillator for the charge pump, I usually use the clock output of the micro to drive a PNP/NPN emitter follower pair, which drives the charge pump. You only get a couple of volts or so this way, but that's still enough to get around the headroom problem of most opamps.

I've done this where the cost of the two transistors, diodes, and caps to make the charge pump allowed using a much cheaper opamp, like the LM324. Despite what some datasheets claim, they get flaky when their inputs get to within a diode drop of the negative supply.

As a example, take a look at the schematic of my USBProg2 PIC programmer. You can see a charge pump in the lower right corner of page 3. The GP0 processor pin was not otherwise used, so I assigned it to the clock out function. This resulted in about -2.2 V, which was plenty to keep 0 V well within the proper operating range of the opamps.

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Well thanks Olin. Really you hit on the main point, which was to help me discover whether there were some unseen "gotchas" I hadn't though about in doing this. And while my need today is unrelated, I have indeed had other posts recently for more of a DC situation, where indeed a few negative volts would have made cheaper op-amps perform where they were working badly (like you said, near the negative rail). I'm going to try this with that LM7660 or something similar, as it should allow the minimum parts when no other clock is available, and get me pretty close to -V – Randy Jan 8 at 19:30
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@OlinLathrop - google safe browsing seems to be red flagging your site. – Scott Seidman Jan 8 at 23:56
    
@Scott: Yes, I know about that. It's a false positive. They do a very simple pattern match, and something in our low level libraries matches one of the patterns they are looking for. I hope they do this to someone big enough to fight back, to sue them for defamation or libel or something. – Olin Lathrop Jan 9 at 14:01

Depending on the op-amp type the output may not be capable of all that much current compared to a circuit that is designed to switch hard. There is also drop in the diodes and impedance due to the capacitors. You might get -1V or -2V out from a +5 supply.

The circuit you show will oscillate somewhere in the audible range, so if your project includes audio that might cause a noticeable whine in the output, from power supply rejection imperfection or from interaction between the op-amp you show and another one in the same package.

Having bipolar supplies is not a panacea for amplifiers- if you DC-couple everything then input offsets will be amplified, so you probably still want some coupling capacitors.

The easiest solution is an isolated DC-DC converter (one part) but it's also probably the most expensive.

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Thanks. About 1 part converters, any thoughts on that LMC7660 I mentioned? Its less then a buck and it will only lose about a volt when used with a +%V supply, and outputting 20mA. That ought to be enough drive for most of my apps. By the way, I've yet to find any scheme that would allow me to use one op amp charge pump to supplly -V to itself anyway, so it would always be a stand alone package. But yeah... I guess I'll have to be careful to isolate any PS coupling of that oscillator. – Randy Jan 8 at 17:42
    
The LM7660 is a really old chip - it's okay for higher supply voltages. For +5 there are better parts available such as the LM2664 which has only 12\$\Omega\$ typical output impedance and works at a reasonable and inaudible 160kHz. Fairly cheap too - about 25 cents in quantity. If you only need a small current and can live with some voltage drop, you can often use any old available clock (with a strong enough output), a dual diode (perhaps a dual Schottky) and a couple tiny ceramic caps. – Spehro Pefhany Jan 8 at 18:21
    
If you can't live without current (most audio draws ~1mA per op amp in just standing power) you could just use a DC-DC converter. The LT8582 can provide a SEPIC and CUK (inverting sepic) converter in one package. With that you could get about 90x more current than from a charge pump. – Dave Jan 8 at 22:30

When you have a charge pump to make the neg rail there are some snags .Firstly the charge pump is only really suitable for low currents which precludes say a speaker output . Secondly the neg rail will always be lower than the pos rail that it is derived from which may be an issue depending on your application .The third issue is power rail sequencing ,At power up the neg rail always arrives after the pos rail and at powerdown the neg rail falls after the pos rail.This powerup thing could cause problems with some chips .

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