Your datasheet for your OpAmp explains this, and will also likely give you derating curve. furthermore, if you get a heatsink, it will have its own datasheet letting you know how much it will dissipate.Im trying to figure out how to calculate how much power an op amp dissipates.
Anyone lend a hand?
right , i agree unless connected to a buffer or something high impedance.Generally, if you are worried about it, then what's connected to the output will be responsible for the vast majority of it, and the internal supply current and feedback current won't make much difference.
If connected to something high impedance, you wouldn't worry about it!right , i agree unless connected to a buffer or something high impedance.
As others have stated the datasheet gives the required information for a particular OPA.
Knowing how the OPA is connected and configured in a circuit is required in order to determine the actual dissipation.
So that is the part i dont get. Why are we multiplying by the difference in the power supply voltage and the output voltage?The feedback resistor will probably be much bigger than the load resistor. Calculate the resistance of the two in parallel.
Use that value to calculate the output current. Then multiply by the difference between the power supply voltage and the output voltage as above. . . . .
When you say load resistor you mean just a resistor to ground at the end of an op amp?Let's say your op-amp runs off a 12V supply, and its output is at 9V and you have a 1k load resistor to 0V.
9mA flows in the load resistor. There is 3V between the 12V supply on pin 7 and the 9V output on pin 6.
9mA flows from the 12V supply. The supply supplies 9mA*12V = 108mV.
The load resistor dissipates 9V*9mA = 81mW
The remaining power (3V * 9mA = 27mW) is dissipated by the op-amp.
It might help to think that the op-amp is pretending to be a 333 ohm resistor.
Let's suppose that your op-amp is a "typical" 741 with a 1.7mA supply current, and that pin 4 is connected to -12V.
The dissipation due to the supply current is 24V * 1.7mA = 40.8mW
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by Jake Hertz