Question

For the circuit above in question 1, what is the most negative value vs can take before the amplifier saturates? Express your answer in mV and omit units from your answer.

Answer 1

Answer:-

Given that,

The most negative value v_{s}vs  can take before the amplifier saturates.

Suppose, Consider a non-ideal op amp where the output can saturate.

The open voltage gain is 2\times10^{4}2×104 where, V_{0}=-A(V_{s})V0=−A(Vs)

According to figure,

The negative output value is

V_{0}=-10\ VV0=−10 V

We need to calculate the most negative value of v_{s}vs  

Using given formula

V_{0}=-A(V_{s})V0=−A(Vs)

Where, V_{0}V0 = output value

A = voltage gain

Put the value into the formula

-10=-2\times10^{4}\times V_{s}−10=−2×104×Vs

V_{s}=\dfrac{10}{2\times10^{4}}Vs=2×10410

V_{s}=0.0005\ VVs=0.0005 V

V_{s}=0.5\ mVVs=0.5 mV

Hence, The most negative value of v_{s}vs  is 0.5 mV.

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