Question

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

Answer 1

Given that,

Gain $A=2\times10^{4}$

$V_{o}=-10$

$V_{o}=15$

$V_{o}=-AV_{s}$

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

Using given formula

$V_{o}=-AV_{s}$

Put the value into the formula

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

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

$V_{s}= 0.0005\ V$

$V_{s}=0.5\ mV$

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

Answer 2

Answer

Vs=-Vo/A=10/(2*10^4)=0.5mV most +ve value for Vs

Similarly,

Vs=-15/(2*10^4)=-0.75mV for -ve value before saturation