I’m using the LM7321MAX operational amplifier to supply a 12V digital signal from a 5V source. For the op-amp supply voltage: is it better to use something close/equal to my output value (like 12V) or further away (like 22V)? I know the internals of an op amp are mostly BJTs so am wondering if I’ll draw the same supply current in either case, with the 20V variant therefore using more power. Thinking the 20V->12V will happen across those BJTs

Thanks for your help!


2 Answers 2


There are a number of curves showing the supply current for different supply voltages and for different temperatures and common-mode voltages, so you can predict the typical supply current. It increases with increasing supply voltage, but fairly gently.

As far your your load current, you'll need enough swing to satisfy your actual requirements, and again the datasheet provides a great number of curves. Once you have enough voltage to (worst-case) supply your required output swing, then additional voltage simply results in more power dissipation in the chip, since the additional output current is drawn directly from the power supply terminals (and possibly a bit more).

The total current drawn is the sum of the two, so it's Is + k\$\cdot\$Iout(Vsupply) where k may be a bit higher than 1, and this the power dissipation in the chip is (Is \$\cdot\$ Vsupply) + k\$\cdot\$Iout(Vsupply-Vout).

  • \$\begingroup\$ Ok that makes sense. So if I use a 20V source for the op-amp and have a max output voltage of 12V, that 8V drop between supply & output is lost to heat dissipation? \$\endgroup\$
    – cdubs
    Jan 18, 2021 at 21:31
  • \$\begingroup\$ Yes. At 0V out, all of it is lost in the chip, and then some. \$\endgroup\$ Jan 18, 2021 at 21:34
  • \$\begingroup\$ Ok, what about for 12V out though: is the 8V difference between the 20V source and the output just lost in the op-amp’s internal circuitry? \$\endgroup\$
    – cdubs
    Jan 18, 2021 at 21:47
  • 1
    \$\begingroup\$ Yes, that is what happens. \$\endgroup\$ Jan 18, 2021 at 21:48
  • \$\begingroup\$ Ok cool, thanks. So in general is there no point in supplying an op-amp with a higher voltage than you’d ever want to output? (Like max output is 15V, for example, doesn’t make sense to use 30V supply) \$\endgroup\$
    – cdubs
    Jan 18, 2021 at 21:52

This is a Rail-to-Rail output op Amp and hence CMOS, not BJT’s.

  • The bias quiescent current is guaranteed in the tables
  • the curves are for typical results.
  • as 2.0mA max (and 1.0 mA typ) at 10V or +/-5V
  • as 2.4mA max at 30V or +/-15V

It also increases slightly with temperature due to ambient or load current.

Load current is up to you with the max currents specified. VI drop in the output stage determines the self heating power.


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