Calculate power dissipation of split-supply op amp for sinusoidal input

Question

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:

^{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:

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 V_PP or ~3.54 V_RMS). The power through (and dissipated by) R_LOAD is then (3.54 V_RMS)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):

• Opamp power consumption
• How to find the operating current consumption of the op-amp?

Answer 1

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.

Answer 2

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.

Answer 3

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.