How asymmetrical is an op-amp supply allowed to be?

Some op-amps are designed to run on single supply only - you can't get much more asymmetric than that! - It really depends what your application is, for most purposes good symmetry optimises the maximum output swing without asymmetric clipping.

"Graham W" wrote in news:4453d4d8$0$9259$ snipped-for-privacy@ptn-nntp-reader01.plus.net:

Ok, but if it's an op-amp made for +/- 15V (+/- 18V max), would that be a limit on the drivers for each polarity? I'm not sure that it's safe to make a swing beyond 15 to 18 V of ground. It might be in a single rail 36V supplied op-amp like the CA3140, but is it safe in an LF412, which isn't meant to be asymmmetrically supplied, so far as I know?

I actually think it might be, but I don't think I know enough to trust my judgement without asking people first on this. I hate burning op-amps. :)

be a

make

isn't

You didn't understand his response. All op-amps are "single supply" if you think about it. Since there is no "ground" pin on an op-amp, only

+Ve and -Ve. The "ground" that you are thinking about (the one half-way between +Ve and -Ve is purely conceptual, just like the one you create in a single supply setup by biasing the + input to 1/2 Vcc. So if you want your virtual ground to only be a bit above the negative supply, then have it your way, the op-amp shouldn't care at all. The point of having it in the middle is to maximize the output voltage swing without clipping. As long as you don't mind most of the lower half of your signal being clipped off, you should have no trouble.

my

Trust me, the things you learn by actually letting the smoke out stick with you allot better.

Is it the smell? Seriously, most op-amps are fairly durable in my experience. What are you doing to them? The only thing that I can see killing them is exceeding the voltage ratings. Most outputs are short-circuit proof.

"Anthony Fremont" wrote in news: snipped-for-privacy@news.supernews.com:

You think I didn't understand his response. :) It might not be an ideal concept, that's just how op-amps appear, it's why they're easy to use. The innards might be something else entirely. You mention voltage ratings, and that's exactly what I'm getting at...

sAn op-amp has a push-pull output stage, and if either pole's transistors had a voltage maximum of around 20 volts, you'd get away with symmetrical rail supply but not with approaching the far rail in a strongly asymmetric one. CA3130's have limits on safe input and I wondered if limits like that are part of the usual dual rail op-amp design's output stage. I've never seen a circuit advocating extreme asymmetry, so it seemed wise to ask here and not to make assumptions. I can learn as much here, as by trying stuff, which I do anyway. I can do that alone, it's not why I posted here. I don't mind taking risks with laser diodes to see what they can do beyond spec, but op-amps are more complex beast even if they're cheaper, so asking for informed comment seems appropriate to me. Burning an op-amp might tell me not to repeat my action, but it won't explain why it happened.

And yes, the smell sucks. :) LED's are worse though, the old ones would smell like burning shit.

Nope.

Think about it, and follow the voltage. If you have a +15 and -15V supply, with the output biased to 0V, that's the EXACT SAME as calling the negative rail zero, in which case the output is biased at +15V and the supply is

+30V. A voltage swing produced by the amplifier might run near the rails, which could be 14V to -14V in the first case, or 29 to 1V in the second. In either case, one transistor has (30-29) = (15-14) = 1V across it, while the other has (30-1) = (+15 - (-15) -1) = 29V across it. Thus, both transistors in the output stage must be rated for at least 30Vceo, if the op-amp is to withstand a total 30V supply voltage.

Anthony: as a matter of fact, my LM311's have a "GND" pin marked. However, the internal schematic shows this is the output ground *return*, so you can, in effect, use the OUT and GND pins as a switch, rather than the more common single output most op-amps provide.

Wait, LM311 is a comparator.. no matter, there's probably a slew-rate-limited (i.e. op-amp) copy of it out there...

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

"Tim Williams" wrote in news:bVV4g.197$ snipped-for-privacy@fe07.lga:

Ok, thanks, I'm convinced. :) That's what I'd hoped, actually.

Would it follow then, that my idea is sound? To have the ground only far enough above negative to allow through-zero offset correction (and to bring it in range of the lowest output value) and a positive headroom of over 30 volts? As the single rail use of a CA3140 can't allow the through- zero offset correction if the ground is the same as the negative, that would rule out any advantage of using the CA3140's output swing to negative at all, if you solve the problems with the new ground vs the 'real' negative one. Which in turn allows me to use nice LF412's, which I like a whole lot better anyway.

Basically, if you see any glaring problems in my scenario, please let me know. :)

I still don't know what you want. If you want to linearly amplify a small voltage, you only need to bias a few volts above ground (or below +V as the case may be), to avoid clipping. If you want to produce a large swing, say

15V peak, you MUST bias it at the half-way point to avoid clipping.

If you specifically *want* clipping, there are better ways than saturating an op-amp. (When you push an op-amp too far and it smacks into the supply rail, that's called saturation.)

You keep speaking of offset, but offset is only what you make of it. If your entire circuit were designed to operate at a bias of say +10V, then it's an offset in the absolute sense, but for your circuit, something like

9.6V somewhere would be an error offset of 0.4V.

Same thing for ground reference, you might have the entire circuit biased to

0V, but errors could push that up or down a little.

Note that, with respect to the negative rail, this ground-referenced signal has an absolute offset of as much.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

"Lostgallifreyan"  wrote in message 
news:Xns97B540EF0CBD7lostgallifreyangmail@140.99.99.130...
> Ok, thanks, I'm convinced. :) That's what I'd hoped, actually.
>
> Would it follow then, that my idea is sound? To have the ground only far
> enough above negative to allow through-zero offset correction (and to
> bring it in range of the lowest output value) and a positive headroom of
> over 30 volts? As the single rail use of a CA3140 can't allow the through-
> zero offset correction if the ground is the same as the negative, that
> would rule out any advantage of using the CA3140's output swing to 
> negative
> at all, if you solve the problems with the new ground vs the 'real'
> negative one. Which in turn allows me to use nice LF412's, which I like a
> whole lot better anyway.
>
> Basically, if you see any glaring problems in my scenario, please let me
> know. :)

"Tim Williams" wrote in news:mh75g.528$ snipped-for-privacy@fe02.lga:

I kept the first post VERY short because I'd twice posted a thread explaining exactly what I want, and it was ignored each time. That's not your fault, but you're still assuming I want equal swing across zero, and I never said that anywhere. The main paragraph in my last post says what I want.

I think the LF412 will do it, and I'm questioning the usefulness of the CA3140's famous to-negative output capability when its offset can't be tweaked truly to zero on a single rail supply anyway.

To summarise: I want to measure +30V in a single range, with accurate offset correction. As an LF412 won't swing to negative, a ground slightly above negative should solve the to-zero output, and the through-zero offset tweak, and the +30V range, given a supply of >32V, based on what you said about both transistors in the output being able to stand the full supply voltage.

I've not seen this idea mentioned in books or data sheets, so I thought there might be some ghastly flaw I can't detect in its logic, which is why I ask rather than just try it. I want to see what pther people have to say about this.

You didn't mention what kind of signal you're amplifying. A signal is assumed to be quickly changing and need a low average DC (offset) component. If you meant a DC signal, then yes, you can easily need a single supply amplifier to cover it.

Ok then, that's just a DC amplifier, with an extra couple volts negative (depending on your point of view!) to account for op-amp limitations.

I'm building a 5V 50A supply and I'm doing roughly the same thing for current sense. I want to use a 1 miliohm resistor (6" of 12 AWG ;) to give

5V full scale, so I need a DC gain of 5V / (50A * 0.001 ohm) = 100, plus a nifty way to provide that offset, I'm thinking current mirror and resistors, if not a plain trimmer adjustment to tweak it even (how unsophisticated!).

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

"Tim Williams" wrote in news:%ge5g.2301$ snipped-for-privacy@fe03.lga:

Exactly. But, the CA3140 is a single supply amp. And it can't do through- zero offset unless I use the small negative voltage. So I decided, the LF412 being a better device, would have an output all the way to zero, the new zero slightly above negative.

But the LF412 is not a single rail device. You said the two halves of the output must both be able to take the full supply voltage, so I guess I can have my full 30VDC range, but would there be any gotchas that would get me if I tried this with the LF412 or other standard dual rail op-amp?

(Note: I know the LF412 is a dual op-amp, but I want two stages, one for the split rail maker, and I'll add external offset tweak for the gain stage).

-----------

I was wondering about your own project, isn't the x100 gain living a tad dangerously? It would amplify offset error hugely, and also noise. What about using two parallel current carriers, both with known character, the main one carrying maybe 90% of the total (or even 99%), and measuring the remainder through a higher resistance? That way you can have easier control of the measurement, maybe. Certainly a simpler circuit might be possible. (I always like those, less parts and less work. :)

Lostgallifreyan wrote in news:Xns97B65002983BClostgallifreyangmail@140.99.99.130:

Never mind, ignore that, it's stupid, >:) I've been working all night and I'm not thinking well.

Well. The exact thing that's been yelled and spoken calmly throughout this thread is it's just point of view.

Biasing the amp at say, +2V, is the same as using a -2V supply, as far as

*any* op-amp is concerned.

Well, that's patently impossible in a chip, since to reach 0V, you must have

0 ohms. Any practical circuit has some voltage drop. Rail-to-rail op-amps get within fractional volts, but not zero (and certainly not that small negative amount you need for the op-amp to self-compensate at zero output).

Meh, it's a 1mohm resistor...

Offset will have to be trimmed of course, yes.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

"Tim Williams" wrote in news:gNo5g.9$ snipped-for-privacy@fe06.lga:

You didn't read the context....

"So I decided, the LF412 being a better device, would have an output all the way to zero, [UNDERLINE]the new zero slightly above negative.[UNDERLINE]"

It's easy to see faults in the logic of a statement if you don't read the whole statement.