Opamp LM108 failure_ drawing 60mA dc supply current

[dhan_pow]

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.]

 

[betwixt]

The LM108 was released in 1969 - that's 48 years ago, are you still using them??

Brian.

 

[FvM]

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.

 

[dhan_pow]

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.

 

[Audioguru]

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.

 

[d123]

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.

 

[Audioguru]

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.