If the open-loop gain of an op-amp is 105 and the input voltage is 100 mV, the value of output voltage is ____________ assuming a supply of ±15 V.

Explanation:

The open-loop gain of an operational amplifier (op-amp) refers to the amplification factor it provides without any external feedback applied to the circuit. This gain is typically very high for op-amps, which allows them to amplify very small input signals to a much larger output signal. In the given problem, the open-loop gain (A_OL) of the op-amp is 10⁵ (100,000), and the input voltage (V_in) is 100 millivolts (0.1 volts).

The formula to calculate the output voltage (V_out) in an open-loop configuration is given by:

V_out = A_OL × V_in

Substituting the given values:

V_out = 10⁵ × 0.1 V

V_out = 100,000 × 0.1 V

V_out = 10,000 V

However, it is essential to consider the power supply voltage of the op-amp. In this case, the op-amp has a power supply of ±15 volts. This means that the maximum output voltage the op-amp can provide is limited to +15 volts and the minimum output voltage is limited to -15 volts, regardless of the calculated value.

Since the calculated output voltage (10,000 V) far exceeds the maximum supply voltage of the op-amp, the actual output voltage will be limited to the maximum supply voltage, which is +15 volts.

Therefore, the correct value of the output voltage is 15 V.

Important Information:

Analyzing the other options:

• Option 1 (10⁴ V): This is the result of the calculation without considering the power supply limitation. This value is impractical because it exceeds the op-amp's supply voltage.
• Option 3 (0 V): This value would be correct if the input voltage was 0 V or if the op-amp was not powered. However, neither condition applies here.
• Option 4 (100 V): This value does not consider the power supply limitation and is also much higher than the maximum output voltage the op-amp can provide.