I am simulating an op-amp circuit of which a small section is shown in the image. As per power consumption formula,

P = Icc * (V+ - V-) + Iout * (V+ - Vout)

I tried to calculate the power. I have put the values shown in the image.

P = 1.17mA (I have confusion in this value) * (5V-(-5v)) + 0.36 mA (5V-2.16V)

P = 11.7 mW + 1.0224 mW = 12.72mW. But the Watt meter probe in simulation shows 4.89mW .

I got a confusion that whether I am missing any point or doing wrong in the calculation?

Clarify this!


Power consumption of the op amp is different than the power drawn from the supply.

The equation below should give you the power consumed by the op amp I think.


  • \$\begingroup\$ Tip: You can use HTML entities such as &times; (\$ \times \$) and some markdown such as <sub>...</sub> and <sup>...</sup> for sub- and superscript. \$\endgroup\$
    – Transistor
    Feb 7 '20 at 20:47
  • \$\begingroup\$ Thank you I was wondering how people were doing that \$\endgroup\$
    – dsizzle83
    Feb 10 '20 at 18:27

I'm confused by the various measurements - DC, p-p and RMS - but just working with the DC I get:

enter image description here

  1. 5.5 mW is provided by VEE.
  2. -0.49 mW is provided by VDC2. I'm not sure how a circuit could do this.
  3. The output current is negative so the op-amp is sinking current so power out is -0.41 mW.

Adding them up I get PIC = PIN - POUT = (5.5 - 0.49) + (-0.41) = 4.6 mW.

That's the best I can make of it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.