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Opamp LM108 failure_ drawing 60mA dc supply current

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dhan_pow

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Opamp LM108: Pin no. 3 is connected to 15 Return. Pin. no. 7 connected to 15V through 100E and pin no. 4 connected to -15V through another 100E. Between pins 1 and 8, 100pF capacitor is wired. All other pins are open.

+15V and -15V supply are dipping to 8.2V and -7.9V respectively,60mA current is drawn from each supply. Also the opamp is getting heated up. Opamp being a very rugged device , what may be the reason for such a failure?

[The maximum voltage available in my card is 40V. Even by shorting the output pin to 40V I could not make a good opamp fail.]
 

The maximum voltage available in my card is 40V. Even by shorting the output pin to 40V I could not make a good opamp fail.

No quite true. LM108 is like most general purpose OPs specified to tolerate continuous output short circuit. Driving any pin beyond the V+/V- supply rails with a higher current than a few mA can however cause latch-up and may destroy the device.

I don't believe that it causes the observed excessive current consumption, but supplying the OP through series resistors without bypass capacitors is a bad idea and might cause self oscillation. The same with floating inputs.
 

The above mentioned one is a test circuit. In the actual board, the opamp is configured as I to V converter at the output of DAC. The output of opamp was steady and not oscillating. The inputs to opamp come from DAC which is perfectly OK.
 

The plus and minus total supply is 16.1V then with a current of 60mA the IC is dissipating 966mW, but its absolute maximum is 500mW so it is frying away.
 

Hi,

Why are there 100R resistors into V+ and V- (case)? Wouldn't that mean pin 7 sees 150mA and likewise -150mA into pin 4? If so, that's a reason to get hot.
 

The 100 ohm resistors would get 150mA though them only if they had a voltage of 15V across each one, then the IC is simply a dead short and it would not even get warm.

The voltage at one supply pin was said to be +8.2V and -7.9V at the other. The main power supply is +/-15V so the IC has a total voltage of (8.2V + 7.9V) 16.1V and a current of 60mA so the IC dissipates (16.1V x 60mA) 966mW. Its maximum allowed dissipation is only 500mW so it is frying.
 
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