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I want to calculate the total supply current and power dissipation of an op amp in a non-inverting voltage follower configuration with a split supply:

schematic

simulate this circuit – Schematic created using CircuitLab

I know there are two components to the op amp's power consumption: quiescent current and current used to drive the output. The datasheet for the NJM4556A lists the quiescent current as ≤ 12 mA:

enter image description here

So for this portion I can easily calculate 24 V * 12 mA = 288 mW.

Now I'm struggling to calculate the additional current and corresponding power dissipation to drive the output, given a known input signal. For this purpose, let's say the input is a ground-referenced 1 kHz sine wave with an amplitude of ±5 V (10 VPP or ~3.54 VRMS). The power through (and dissipated by) R_LOAD is then (3.54 VRMS)2 / 150 Ω = 83 mW. The output current is sinusoidal at ±33 mA (plus a negligible amount for the negative input bias current: 50 nA for this part, let's ignore it).

But the op amp's output stage can't be 100% efficient, and so will draw more current than this, with the extra power that doesn't reach the load dissipated in the op amp itself. How do I calculate this? Should I expect to find curves for this in the datasheet (there aren't any), or is there a formula based only on the output stage topology?

Related but not duplicate (or at least I'm too dumb to find my answer in):

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3 Answers 3

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But the op amp's output stage can't be 100% efficient, and so will draw more current than this, with the extra power that doesn't reach the load dissipated in the op amp itself.

You can assume that the maximum quiescent current of 12 mA contains the current That activates the output transistors and not supplied to the load when driving the load. After all, 12 mA is a fair bit for an op-amp and it's not taking that current into the input stages of the device.

Then, knowing that most op-amps have a Class AB output stage and, knowing that the power efficiency of a Class AB stage is about 65% for the full permissible output voltage you can start to figure these things out.

let's say the input is a ground-referenced 1 kHz sine wave with an amplitude of ±5 V (10 VPP or ~3.54 VRMS)

OK, it's not going at the full permissible level so, the efficiency is going to be somewhat less than 65%. You could work this out of course but, it's probably easier just to use a simulator to check the numbers. I'm not going to do that; I'm going to assume a power efficiency of only 40%.

This means that the power taken from your supply rails will be 83 mW / 0.4 = 207 mW. 207 mW from a 24 volt supply, is a DC current of 8.63 mA.

Therefore I estimate you total current consumed from the power rails will be 12 mA (as per the quiescent in the data sheet) plus 8.63 mA = ~20 mA.

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  • \$\begingroup\$ Great info, thanks. The SPICE model for the NJM4556A is behavioral only and doesn't model output supply draw (I guess this is common for op amp models) but maybe I could find a full model for another op amp with a similar output stage and check against your estimation until I can try it on a protoboard. Thanks again. \$\endgroup\$
    – TypeIA
    Mar 16, 2022 at 15:33
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    \$\begingroup\$ Or, just make a trial class AB stage from an NPN and PNP transistor and see what the efficiency looks like. \$\endgroup\$
    – Andy aka
    Mar 16, 2022 at 16:43
  • \$\begingroup\$ Well, for that matter there's a full equivalent circuit in the datasheet showing the actual output stage (without component values of course)... \$\endgroup\$
    – TypeIA
    Mar 16, 2022 at 16:58
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If you want a quick figure then consider the supply current plus twice the average value of the half-wave rectified waveform of the load current. For example, if the (total) supply current is 12 mA and the load is 100 Ω with 1 V peak, then the positive supply will draw the positive alternances of the sine, 1 V / 100 Ω = 10 mA peak, and the negative source the negative alternances. The average of the half-wave rectified waveform is 1/π, so the total will be: 10 mA * 2/π + 12 mA = 18.37 mA. If the supply current is specified per rail, then multiply 12 mA by 2. Unless you're pushing the opamp to the limits, you won't be far.

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Because efficiency and power dissipation are not usually important parameters when designing small signal op amp circuits, you generally don't see this in datasheets. (If you were doing a micropower design you would typically look for parts optimised for such an application, and there you'd expect to see very low quiescent current quoted).

If this was a big consideration I would do the same analysis you already did, and then do some real world measurements on a protoboard, then see how much greater the current I measured was than the prediction that was calculated.

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  • \$\begingroup\$ Thanks. The real concern is power dissipation. The 288 mW worst-case quiescent power is already a large fraction of the packages' thermal ratings. In addition to ensuring adequate supply current headroom, I want to check if my design is actually limited by these ratings (I believe it is). \$\endgroup\$
    – TypeIA
    Mar 16, 2022 at 14:11
  • \$\begingroup\$ it's rather a low load for a general purpose opamp. the 5532 has one of the best current drivers out there, and that doesn't like much less than 600R. Have you considered adding drive transistors to boost the current drive? \$\endgroup\$
    – danmcb
    Mar 16, 2022 at 14:14
  • \$\begingroup\$ this gives some ideas : allaboutcircuits.com/technical-articles/… \$\endgroup\$
    – danmcb
    Mar 16, 2022 at 14:21
  • \$\begingroup\$ This is a high-current op amp and can definitely handle this, although I edited the question to increase R_LOAD to 150 R so that it matches the test conditions in the datasheet (70 mA into 150 R; from experience I know it can drive lower resistances too). \$\endgroup\$
    – TypeIA
    Mar 16, 2022 at 14:22
  • \$\begingroup\$ at low voltage yes, but as you have already calculated you are getting a bit close for comfort. Just a suggestion. The point is that if you are needing to do this kind of calculation that suggests that you are using the component a bit outside of the thing it was designed for. There are ways to mitigate that at low cost, saving yourself the headache. \$\endgroup\$
    – danmcb
    Mar 16, 2022 at 14:23

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