1
\$\begingroup\$

I am trying understand the worst power consumption of an op-amp. Lets take this part: LM833. Supply voltage +/- 5 V and the supply current is 8 mA. I am considering that the total operating power as 5*8 + -5*-8 = 80 mW.

How this can be wrong?

LM833 Datsheet

\$\endgroup\$
8
  • \$\begingroup\$ You also need to consider the current delivered to the load. \$\endgroup\$
    – The Photon
    Feb 15, 2018 at 6:26
  • 2
    \$\begingroup\$ Also, this device is not a rail-to-rail op-amp, neither on the input nor the output. \$\endgroup\$
    – The Photon
    Feb 15, 2018 at 6:29
  • 1
    \$\begingroup\$ 50 to 80 mW + load is correct. \$\endgroup\$ Feb 15, 2018 at 6:57
  • 1
    \$\begingroup\$ i agree with your max. I just included typ. \$\endgroup\$ Feb 15, 2018 at 7:06
  • 1
    \$\begingroup\$ Some terminology might help you in understanding power budgets. The current that an op-amp (and other active devices) just uses to run, without considering the current it needs to drive something, is called the quiescent current. That would be your 8 mA which is noted by \$I_\mathrm{Q}\$ in the datasheet. I only mention because "supply current" is a terrible term because as others said, there will be more draw on the supply if you are driving a load. \$\endgroup\$
    – loudnoises
    Feb 15, 2018 at 8:31

1 Answer 1

4
\$\begingroup\$

The power an op-amp dissipates is from two parts.

1, power to run the device. The 8mA value listed is the current the device will consume with the output open circuit and applies between the rail voltages. So in your case, power will be the voltage range 10V times the quiescent current, here 8mA, So it will dissipate 80mW just sitting there.

2, Power to run the load.

This gets a bit more complicated.

If you are driving a load to ground at say 10mA, then that current will either be sourced from the +5v rail or soaked up by the -5V rail, so you need to add 10mA * 5V, another 50mW, in this example.

If the load you are driving goes to the bottom or top rail, then the worst case would be 10mA times your rail range, 10V. so you would need to add in 100mW.

Of course the latter is worst case with a given load.

If you are using the op-amp to buffer a DC value this worst case value may apply. If you are using the op-amp to drive some AC signal you can use the RMS value of the signal in the above math.

However, be aware, when you undersize the device based on some estimated average signal condition, if there is a failure or wrong connection that changes that signal, your device can cook. If there is ever any doubt on what the signal will be, use the worst case numbers.

When it comes to power dissipation, you really want to derate the devices specified power rating by a good margin, perhaps even less than 50% of that. Other-wise the device will be very hot and you will need to add heatsinking.

Further, you really need to factor in the tolerances of your design. For example, if your power supplies are rated at 5% your 10V becomes 10.5V. You need to use the worst case numbers here.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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