Comparator question?

Question

An op amp has the following specifications:

op amp used as a comparator
–18 V < VSupply < +18 V
maximum difference in supply voltages = 32 V
supply voltages are +15.2 V and –15.2 V
rail headroom voltage = 1.1 V (for both + and – supplies)
A = 1.5*106
maximum output current = 21.9 mA.

The input voltages are Vni = 3.8 mV and Vinv = 13.7 mV.

The output voltage of the op amp is _______ V.

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completely lost

Answer 1

Differential input voltage is +3.8 mV - 13.7 mV = -9.9 mV

Amplification factor: A = 1.5 million

Vo = Vin_diff * A = -9.9 *10^-3 * 1.5 * 10^6 = -14850 volts

Clearly the output of the op amp comparator cannot produce -14 thousand volts, so it is saturated at the negative side.

The negative power supply is at -15.2 volts and the output transistors will not operate at the 'rail' (this is not a rail-to-rail amplifier). The head voltage is specified at 1.1 volts, so -15.2 volts + 1.1 volts is where the output of the amplifier will be: -14.1 volts.

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Answer 2

Since the voltage at non-inverting terminal is higher than that at inverting, the op-amp would saturate and the output would be close to the -ve bias supply. This is what a comparator is, an op-amp without feedback (open-loop)