Relate to Opamp Basics

Can anybody please let me know what output we should get when power supply of Opamp is grounded?

You would likely get some voltage.

That's all the detail I can give you for the small amount of detail you gave us. Need a schematic of the op amp connections if you want more.

Related to Opamp Basics

Please find the attachment & then let me know what output we should get when Power suppply is grounded?

Thanks in advance

These are open collector comparators. I'm going to guess that they did not include a clamping diode to V+ from the output as this would not allow the device to interface to higher voltage circuits, which is a common use of open collector outputs. Since the output transistor is not driven, the impedance looking into the output is extremely high, so the voltage at pin 14 will be a function of the voltage divider from source1 through R237 and R236 to 5.7 volts. The 5.7 volts is the result of V16 causing U28 to conduct which locks the voltage at U56C pin 9 at about 5.7 volts (assuming U28 is generic silicon). So V at U56C pin 14 would be approximately 5.007 volts. But that's just a quick estimate, so let's just call it 5V.

I am doing FMEA analysis on this circuit.
Expected O/P of this cicuit is when I/P is 28V then O/P should be +5V(Logic 1) & when input is open (28V source is not connected) then I should get O/P as 50.16mV(Logic 0)
My exact question is when I am connecting power supply of OPAMP to GND then also I am getting expected O/P.
How is it possible? By connecting power supply of opamp to GND does it gives its functionality?

Thanks,
Radhika

radhikapatil said:

My exact question is when I am connecting power supply of OPAMP to GND then also I am getting expected O/P.
How is it possible? By connecting power supply of opamp to GND does it gives its functionality?

Thanks,
Radhika

Click to expand...

hi,
The OPA is constructed from transistors/diodes
Transistors have base to emitter junctions, just like a diode.

When you ground the power rails of the OPA you are removing the internal biasing of the transistors, so if you apply a voltage to the input of the OPA you are forward biasing the transistor/diode network, so you could measure a low voltage on the OPA output under these conditions.

I would ask, why would anyone want to do this in a practical application.?

Thanks for your answer. I am doing FMEA analysis & When Failure Mode - Bypass capacitor is short has occured which is connected to supply then what o/p should I get that i am checking.

radhikapatil said:

Thanks for your answer. I am doing FMEA analysis & When Failure Mode - Bypass capacitor is short has occured which is connected to supply then what o/p should I get that i am checking.

Click to expand...

hi,
I always add a series resistor to the inputs of an OPA, where possible, especially when when the input voltage sources to the OPA are high and have a low impedance and the external source is powered from a seperate psu.
Under power supply failure conditions, fairly high input currents can flow into the OPA damaging the inputs.

If you look at OPA datasheets this limitation of input current is recommended.