Given the following circuit: (1) Non-inverting amplifier circuit with a feedback

Question

Given the following circuit: (1) Non-inverting amplifier circuit with a feedback resistor of 6 kohm and a dc closed-loop gain of +3 connected to a load resistor RL=0.5 kohm, (2) Op-amp is connected to +/- 12V supplies and has rail-to-rail output voltage swing, (3) Op-Amp can source or sink up to 25 mA - that is, io_max = +/-25 mA, (4) The slew rate of the op-amp is SR = 5 V/us, (5) The unity-gain frequency of the op-amp is 1 MHz. For a sinusoidal input of 100 kHz, what is the maximum peak input voltage possible without distortion? 4V 3.33 V 2.65 V 0.625 V

Explanation / Answer

THe op-amp is connecred to +/-12V supplies. WHich means the output can go max 12V to -12V. A 12V across 0.5kohm load implies 24mA source or sink capability which satisfies point 3. Now, the slew rate is 5v/uS which means in 1uS, the output can go 5V. In 100 KHz i.e., 10e(-5) sec, it can go maximum swing of 50V. Thus an output going from -12V to 12V, meets the criteria of slew rate as well. Since 100KHz < 1MHz, the unity gain of the opamp, it meets this criteria too.

Since the gain of the opamp is 3, an output of peak 12V and 100KHz will translate to an input of 12/3= 4V peak 100KHz input signal.

THus the answer is a.

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