How to calculate an op-amp's current usage based on datasheet's supply current and its Vout terminal?

Question

This might be a dumb question so if yes, I apologize. I would like to power an op-amp off of a boost converter and I need to know how much current an op-amp would use. I'm using TI's web bench and there's a lot of cost / space differences depending on the max output current. So I have to pose this as a question even if it sounds dumb because it has a lot of space / cost ramifications.

On a datasheet, it says that the "supply current" for an op-amp like the MCP6002 (https://www.digikey.com/product-detail/en/microchip-technology/MCP6002T-I-MS/MCP6002T-I-MSCT-ND/669500) is 100 uA. The MCP6002 is actually two op-amps in one integrated component.

I will be outputting roughly ~1 mA from one of the op-amp's Vout terminal. For now just assume the other Vout terminal isn't used.

When I design my boost converter, I'm assuming the booster should only require a max output of 1.1 mA. This is for the 1mA coming from Vout and the 0.1 mA going to the "supply current". I can round up to 3 mA to be safe. Is this a correct assumption, however?

Another question: If I do wind up using the other output's Vout, do I need to also 2x the supply current (e.g. 200 uA).

Answer 1

The typical current is 200uA for the package, plus whatever else you draw from connections to the op-amp output(s). You should use the maximum of 340uA rather than the typical.

You can't stop the unused op-amp from drawing quiescent current, it does so whether you connect to the output or not, however if you connect the unused one incorrectly it might draw more current because it's oscillating or whatever.

If you draw 1mA from the output of the used op-amp, total current could be as high as 1.34mA.

Of course the current at the input of your boost current will have to include the voltage ratio (if you double the voltage, the current at the input will be double the output current, plus more current for the quiescent current of the switching regulator and the losses in the switching regulator.

Answer 2

On a datasheet, it says that the "supply current" for an op-amp like the MCP6002 (https://www.digikey.com/product-detail/en/microchip-technology/MCP6002T-I-MS/MCP6002T-I-MSCT-ND/669500) is 100 uA.

You didn't read carefully. The specification is "quiescent current per amplifier".

That means for the 6001 single op-amp, the quiescent current is 100 uA. But for the '6002 dual op-amp it's 200 uA.

When I design my boost converter, I'm assuming the booster should only require a max output of 1.1 mA. This is for the 1mA coming from Vout and the 0.1 mA going to the "supply current". I can round up to 3 mA to be safe. Is this a correct assumption, however?

Yes, except that it should be 1.2 mA to allow 200 uA quiescent current for your dual amplifier.

Another question: If I do wind up using the other output's Vout, do I need to also 2x the supply current (e.g. 200 uA).

You need to provide the quiescent current to the other amplifier whether you use it or not. ("quiescent" means essentially how much current it takes, even when not being used)

You should also be careful if you don't use the other amplifier to connect it in a way that guarantees it doesn't oscillate. If you use it totally unconnected (for example) its output could "flop around" and then the "unused" amplifier would consume much more current than the quiescent value. This can even lead to thermally overloading the chip in some cases.