Question 23.4: The operational amplifiers used in Fig. 23-12 have an Rin of...
The operational amplifiers used in Fig. 23-12 have an R_{in} of 3 MΩ, an R_{CM} of 200 MΩ, and an open-loop voltage gain of 175,000. Solve for Z_{in(CL)}, A_{V(CL)}, and v_{ref}.
Z_{in(CL) }= (1 + A_{VOL}B)R_{in} ∥ R_{CM}
In a voltage follower, the feedback fraction “B” is 1. This reduces the previous equation to:
Z_{in(CL) }= (1 + A_{VOL})R_{in} ∥ R_{CM}
Z_{in(CL)}=\frac{1}{\frac{1}{525 \ G\Omega }+\frac{1}{200 \ M\Omega } } =199.9 \ M\Omega
Z_{in(CL)} ≅ R_{CM}
The closed-loop voltage gain of the amplifier is found using Eq. (16-12):
A_{V(CL)}=\frac{R_f}{R_1 } +1
A_{V(CL)}=\frac{100 \ k\Omega }{220 \ \Omega } +1
A_{V(CL)}= 455.5
The trip point of the comparator is determined by the dc voltage level present at the inverting input of the operational amplifier. The voltage divider formed by R_2 and R_3 provides the dc reference voltage. The equation to solve for the reference voltage was first introduced as Eq. (20-2):
v_{ref}=\frac{ R_2 }{R_1+R_2} V_{CC}
Substituting in the resistor designations and values from the circuit in Fig. 23-12:
v_{ref}=\frac{ R_3 }{R_2+R_3} V_{CC}
v_{ref}=\frac{ 1 \ k\Omega }{3.6 \ kΩ + 1 \ kΩ} (15 \ V)
v_{ref}= 3.26 \ V