How much power does an op amp dissipate?

crutschow

Joined Mar 14, 2008
25,986
Using Watt's Law, you measure the current at each of the two op amp power terminals and multiply that by the supply voltage at each terminal.
This current is typically given in the op amp data sheet.
 

BobaMosfet

Joined Jul 1, 2009
1,292
Im trying to figure out how to calculate how much power an op amp dissipates.

Anyone lend a hand?

Thanks
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.
 

Audioguru again

Joined Oct 21, 2019
2,358
The datasheet of an opamp shows its idle dissipation and Ohm's Law and a power calculation can calculate how much more current and power the opamp uses when it drives a load.
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
186
Right, so the key point is, its not only the supply current but most also include either the source or sink current and the feedback current
 

Ian0

Joined Aug 7, 2020
1,113
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.
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
186
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.
right , i agree unless connected to a buffer or something high impedance.
 

ericgibbs

Joined Jan 29, 2010
11,564
HI,
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.

E
 

Ian0

Joined Aug 7, 2020
1,113
right , i agree unless connected to a buffer or something high impedance.
If connected to something high impedance, you wouldn't worry about it!

So what's the output voltage? (either DC, or RMS if it's AC).
And the resistance between output and ground?
Then work output the output current.
Take the difference between the supply voltage and the output voltage and multiply it by the output current. That's the power dissipation due to the load.
Then, multiply the TOTAL supply voltage, between V+ and V-,( not V+ and ground) by the supply current quoted in the datasheet. Add it to the number you worked out above. That's the TOTAL power dissipation. It won't be much different to the power dissipation due to the load.
(Next time, you'll probably only work out the power dissipation due to the load and say "that's near enough", unless it's pretty damn close to the maximum dissipation quoted in the datasheet)
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
186
HI,
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.

E
For sure,

So i guess my question then becomes mathematically what are we doing to obtain the power dissipation.

I get the fact that if you had an op amp with no loading and say no feedback then the power is the just the Q current * the supply voltage
but what is it when you have both feedback current and loading mathematically?
 

Ian0

Joined Aug 7, 2020
1,113
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. . . . .
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
186
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. . . . .
So that is the part i dont get. Why are we multiplying by the difference in the power supply voltage and the output voltage?
 

Ian0

Joined Aug 7, 2020
1,113
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
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
186
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
When you say load resistor you mean just a resistor to ground at the end of an op amp?
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
186
so how does this work with like say an op amp or comparator that is sinking current to ground?

Your above example makes sense for sourcing current.
Thanks for your assistance in adavance
 

Ian0

Joined Aug 7, 2020
1,113
Assuming the good old LM339/LM393.
Assuming that the negative supply is connected to ground.
Assuming a 12V power supply.
Assuming you have a load resistor (let's say 1k) connected to V+ at the other end
The output voltage is very low, probably <0.2V.
So there is 11.8V across the resistor, so 11.8mA will flow.
11.8V * 11.8mA = 139mW in the resistor.
0.2V between output and ground, 11.8mA flowing so the comparator dissipates 0.2V * 11.8mA = 2.36mW
 
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