Question

Consider a non-ideal op amp where the
output can saturate.
The open loop gain A-2x10' where --A
Us
The positive supply voltage for the op-amp is
*V, 15V. The negative supply voltage for
the op-amp is-V -10v.
What is the most positive value w can take
before the amplifier saturates? Express your
answer in mV and omit units from your
answer.
15
V
b+
+ 210
=
=​

Answer 1

Given that,

Loop gain $A=2\times10^{4}$

Positive voltage = 15 V

Negative voltage = -10 V

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

Using formula of voltage

$v_{0}=-Av_{s}$

$-10 \leq v_{0} \leq 15$

$-10 \leq -Av_{s} \leq 15$

$\dfrac{-15}{A} \leq v_{s} \leq \dfrac{10}{A}$

Put the value into the formula

$\dfrac{-15}{2\times10^{4}} \leq v_{s} \leq \dfrac{10}{2\times10^{4}}$

The most positive value of voltage is

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

$v_{s}=0.0005\ V$

$v_{s}=0.5\ mV$

The most negative value of voltage is

$v_{s}=\dfrac{-15}{2\times10^{4}}$

$v_{s}=-0.00075\ V$

$v_{s}=-0.75\ mV$

Hence, The most positive value of voltage is 0.5 mV.

The most negative value of voltage is -0.75 mV.