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An introduction to my question is that ultimately op-amps drive something with their output stage (i.e., they source or sink current). I am looking specifically at the current sourcing scenario for the remainder of this question.

The current that an op-amp can provide is sometimes shown on the datasheet as its Output Current and even more rarely a Short-Circuit to Ground Current is also provided. Quite often these values are very low, perhaps single digit mA or maybe 10-20 mA, maximum. Manufacturers are quick to point out how little current their op-amp uses, but it's the driving ability of the op-amps that is paramount for many circuits. Without drive ability, one has to add additional gain stages that reduce the wonderful characteristics of the op-amp (e.g., noise, offset, etc...).

This current drive ability seems very small and I am wondering why the op-amps are designed to source so little current, forcing one to add another active stage after the op-amp for current amplification in order to drive larger loads.

To get higher currents, one has to venture into the world of so called power amplifiers, but often these have much worse characteristics on their datasheet than good op-amps. Also, they are much more expensive for decent ones (e.g., 10x as expensive).

Is the low drive ability of op-amps a side effect of trying to give them their good characteristics? Do these characteristics get worse if an op-amp is designed to source more than say 20 mA or so of current, that being the reason why they are so limited in their drive ability?

Update Perhaps an aspect of my question is why there is such a large gap between op-amps and power-amps (i.e., the jump from 10-20 mA to 2 A).

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    \$\begingroup\$ opamps are designed for specific purposes. If you need a small signal amplifying opamp, it likely has limited output current capabilities, but if you look at power opamps, they can output several amps. But why pay for that capability in your design if you dn't need it? \$\endgroup\$ – PlasmaHH Oct 10 '17 at 14:37
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    \$\begingroup\$ For most applications that 10-20 mA is enough. If you need more there are solutions for that. Compare the datasheet of such a 10-20mA output current opamp to one that can drive 2A, like the opa544. Look at current consumption, gain BW etc. Also price. (2A transistors need silicon area, perhaps even extra bondwires, less chips will fit on a wafer, also few people need this, this all increases price). The opa544 can drive 2 A but at the cost of compromises elsewhere. \$\endgroup\$ – Bimpelrekkie Oct 10 '17 at 14:38
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    \$\begingroup\$ They make power opamps. Even the TCA0372 in a small package. But you then have to get rid of the heat! \$\endgroup\$ – George Herold Oct 10 '17 at 14:44
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    \$\begingroup\$ Regarding your update: there are plenty of op amps available with output currents between 20 mA and 2 A. TI has 678 op amps with a typical output current between 40 mA and 1 A. Granted, some of these are single/dual/quads of the same op amp design, but there are nonetheless plenty in that range. \$\endgroup\$ – Null Oct 10 '17 at 14:55
  • \$\begingroup\$ Check out apex. apexanalog.com \$\endgroup\$ – Gregory Kornblum Oct 10 '17 at 15:03
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It's simple: manufacturers make what customers will buy! It's the same reason why Ferrari won't put a trailer attachment at the rear of their cars...

Price is a very important part of this, of course, and price is tied to silicon area, process, yield, and of course packaging.

For example, an opamp with +/-12V supplies and 1A output current will dissipate several watts, so it can't be a standard opamp package like SO-8. Thus its target audience shrinks down to customers who are willing to use this specific package, which won't be standard so there won't be a second source in case the first manufacturer is out of stock. Also the non-standard package means it can't be marketed to customers who want a standard opamp, unless it's something like a SO-8 with a thermal pad, which has its own sets of issues, it's a lot worse thermally than TO-220-5 LM1875 for example.

Contrast this with, say, a NE5532. In the unlikely case it's out of stock, there are tons of equivalents. Package is standard... it's a jellybean part.

We haven't gotten into actual tech limitations yet, and already the economics favor the chip that will cater to the largest audience, in this case small signal, low current circuits. Even if the thermally enhanced package adds only 5c to the cost of the beefier opamp, it's 5c too much if you don't need the capability.

Now, higher output current requires larger output transistors, thus more silicon area: it will be more expensive. It will also be slower.

Say you make an ADA4898, top of the line, ultra-everything opamp. The process for that is likely to be damn expensive, also all the chips that fail the stringent spec tests are trashed, so yield can be an issue. The folks at AD aren't going to enlarge the output transistors if this only interests 5% of the customers for this product... because this would make it more expensive for the others, so they'd pick another opamp...

Maybe the manufacturers can't do it? Well, nope. If there is a market, they will rush into it. Check ADA4311 for example, it's a driver for PLC/DSL/Whatever twisted pair, it's fast, low noise, high current, etc. But it's a current feedback amp with high offset and crummy DC specs because these don't matter to the application, so going for a precision design would only increase cost.

Now check LM1875, an audio power opamp. It's old, crummy and slow, so it can be probably be manufactured on a cheap old process with high yield. It could be better, sure... but it's good enough for the application.

If you want an opamp with good small-signal specs (DC offset, noise, etc) and high current drive, the simplest is to stick a power output stage on it, or make a compound with the first opamp driving a beefier one. If the second opamp is faster, no extra compensation is required. If it is slower, compensation is required.

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  • \$\begingroup\$ A wonderful response. I just have one question regarding your ansewr: You mentioned that if I want good small-signal specs with a high current drive to add a power output stage. But, wouldn't a power output stage completely degrade those good small-signal specs in trade for more current sourcing ability? In other words, wouldn't adding that extra current gain stage via say a pass transistor or a power op-amp, completely negate the benefits of the precision op-amp in front of it? \$\endgroup\$ – Michael Goldshteyn Oct 10 '17 at 18:47
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    \$\begingroup\$ Usually the first opamp will determine DC offset, noise, and basically everything except speed, distortion, and whether the output is rail to rail. For example if the output stage you add has say 1V of DC offset because it's just an emitter follower, the offset will be reduced by the open loop gain of the opamp, which at DC is usually "a lot". And most of the noise will come from the input stage of the opamp, too, since noise in the output stage is reduced by feedback. \$\endgroup\$ – peufeu Oct 10 '17 at 19:01
  • \$\begingroup\$ The gotcha is crossover distortion if the output stage is badly designed, and speed if it has lower bandwidth than the opamp, then it will add phase shift which makes the opamp unstable, this needs additional compensation caps. Since high current transistors are usually slower you should watch out for this. \$\endgroup\$ – peufeu Oct 10 '17 at 19:03
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    \$\begingroup\$ @MichaelGoldshteyn: The first stage in an amplifier is responsible for almost all the SnR. If it has a gain of 1000, the input to the next stage is almost all signal (and amplified noise). Actual noise added to the output of the first stage is being added to an already-large signal. See electronics.stackexchange.com/questions/305128/… and e2e.ti.com/blogs_/archives/b/thesignal/archive/2013/01/21/… for more about how a high-quality first stage is a Good Thing. \$\endgroup\$ – Peter Cordes Oct 10 '17 at 22:38
  • \$\begingroup\$ Ferrari may not offer it, but that doesn't stop customers from doing it themselves. Though I imagine the end result would be the same as overloading an electronic component -- do it too much and the magic smoke will be released from the car and it will stop working. \$\endgroup\$ – Johnny Oct 11 '17 at 0:56
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High output currents mean high dissipation and high thermal distortion, because the input diffpair will experience slightly different temperatures. How possible, if "symmetric layout"? Because the metallization will not be balanced near the diffpairs and the metal, as heat path, produces imbalanced temperatures at the diffpairs.

If thermal distortion is not important, if that slight 0.01% undershoot due to thermal feedback, or that not-so-slight 0.1% overshoot due to thermal feedback, are not important, your design task is easier.

But beware of PAM multiple-sensor signal processing, in any opamp.

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