The operational amplifiers used in Fig. 23-12 have an Rin of 3 MΩ, an RCM of 200 MΩ, and an open-loop voltage gain of 175,000. Solve for Zin(CL), AV(CL), and vref.

Question

The operational amplifiers used in Fig. 23-12 have an [latex]R_{in}[/latex] of 3 MΩ, an [latex]R_{CM}[/latex] of 200 MΩ, and an open-loop voltage gain of 175,000. Solve for [latex]Z_{in(CL)}, A_{V(CL)}[/latex], and [latex]v_{ref}.[/latex]

Accepted Answer

[latex]Z_{in(CL) }= (1 + A_{VOL}B)R_{in} ∥ R_{CM}[/latex]

In a voltage follower, the feedback fraction “B” is 1. This reduces the previous equation to:

[latex]Z_{in(CL) }= (1 + A_{VOL})R_{in} ∥ R_{CM}[/latex]

[latex]Z_{in(CL)}=\frac{1}{\frac{1}{525 \ G\Omega }+\frac{1}{200 \ M\Omega } } =199.9 \ M\Omega [/latex]

[latex]Z_{in(CL)} ≅ R_{CM}[/latex]

The closed-loop voltage gain of the amplifier is found using Eq. (16-12):

[latex]A_{V(CL)}=\frac{R_f}{R_1 } +1 [/latex]

[latex]A_{V(CL)}=\frac{100 \ k\Omega }{220 \ \Omega } +1 [/latex]

[latex]A_{V(CL)}= 455.5[/latex]

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 [latex]R_2[/latex] and [latex]R_3[/latex] provides the dc reference voltage. The equation to solve for the reference voltage was first introduced as Eq. (20-2):

[latex]v_{ref}=\frac{ R_2 }{R_1+R_2} V_{CC}[/latex]

Substituting in the resistor designations and values from the circuit in Fig. 23-12:

[latex]v_{ref}=\frac{ R_3 }{R_2+R_3} V_{CC}[/latex]

[latex]v_{ref}=\frac{ 1 \ k\Omega }{3.6 \ kΩ + 1 \ kΩ} (15 \ V)[/latex]

[latex]v_{ref}= 3.26 \ V[/latex]

Question 23.4: The operational amplifiers used in Fig. 23-12 have an Rin of...

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