Question

Inverting Amplifier Figure 4.2 shows the fundamental configuration of Op-Amp in which it is used as an inverting amplifier. I
2. Apply a sinusoidal input (Vin.) of 1Vp-p, 10 kHz from a function generator. 3. Observe the input and output waveforms on t
0 0
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Answer #1

0 V4 R2) 70 PRT V -1.8787V] 100k 3 1 output PR2 R1 U1 20ko input v 376.14my V3 2. 0.5V 10kHz 0 V2 0 10V 0Interactive 1 3 -PR1: V(output) -PR2: V(input) 2.5 2. 1.5 1 500m Voltage (V) 0 -500m - 1 -1.5 -2 -2.5 -3 181.8m 181.85m 181.9

The gain is 5

Phase is 180°

V1 I Pov 2 R2 500ko PR1 1 output PR2 R1 U1 V -859.00mV 20ko input v 36.290mV V3 0 0.5V 10kHz 0° V2 0 10VInteractive 1 12.5 -PR1: V(output) -PR2: V(input) 10 7.5 5 2.5 Voltage (V) o -2.5 -5 -7.5 - 10 - 12.5 50us/div

As VCC is 10v only . And the output is greater than 10 and hence saturated.

The phase is 180°.

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