Question

For the circuit above in question 1, what is the most negative value v_{s}v s
​ can take before the amplifier saturates? Express your answer in mV and omit units from your answer

Answer 1

Given that,

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

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

The open voltage gain is $2\times10^{4}$ where, $V_{0}=-A(V_{s})$

According to figure,

The negative output value is

$V_{0}=-10\ V$

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

Using given formula

$V_{0}=-A(V_{s})$

Where, $V_{0}$ = output value

A = voltage gain

Put the value into the formula

$-10=-2\times10^{4}\times V_{s}$

$V_{s}=\dfrac{10}{2\times10^{4}}$

$V_{s}=0.0005\ V$

$V_{s}=0.5\ mV$

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

Answer 2

Explanation:

The following two questions refer to the circuit below.

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

The open loop gain A = 2 x 10^{4}10

4

, where v_{o}v

o

=−Av_{s}v

s

.

The positive supply voltage for the op-amp is +V_S = 15+V

S

=15V. The negative supply voltage for the op-amp is -V_S = -10−V

S

=−10V.

What is the most positive value v_{s}v

s

can take before the amplifier saturates? Express your answer in mV and omit units from your answer.