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.

Answer :

igeclement43

Answer:

The most positive is value of Vs is 0.5mV

${teks-lihat-gambar} igeclement43

Other Questions