The OTA in Figure 12–20 is connected as an inverting fixed-gain amplifier. Determine the voltage gain.

Question

Accepted Answer

Calculate the bias current as follows:

[latex]I_{\mathrm{BIAS}}=\frac{+V_{\mathrm{BIAS}}-(-V)-1.4 \mathrm{~V}}{R_{\mathrm{BIAS}}}=\frac{9 \mathrm{~V}-(-9 \mathrm{~V})-1.4 \mathrm{~V}}{33 \mathrm{k} \Omega}=503 \mu \mathrm{A}[/latex]

Using [latex]K \cong 16 \mu \mathrm{S} / \mu \mathrm{A}[/latex] from the graph in Figure 12–14 , the value of transconductance corresponding to [latex]I_{\mathrm{BIAS}}=503 \mu \mathrm{A}[/latex] is approximately

[latex]g_m=K I_{\mathrm{BIAS}} \cong(16 \mu \mathrm{S} / \mu \mathrm{A})(503 \mu \mathrm{A})=8.05 imes 10^3 \mu \mathrm{S}[/latex]

Using this value of [latex]g_m[/latex] , calculate the voltage gain.

[latex]A_v=g_m R_L \cong\left(8.05 imes 10^3 \mu \mathrm{S} ight)(10 \mathrm{k} \Omega)=\mathbf{8 0 . 5}[/latex]

P R A C T I C E EXERCISE

If the OTA in Figure 12–20 is operated with dc supply voltages of [latex]\pm 12 \mathrm{~V}[/latex], will this change the voltage gain and, if so, to what value?

Question 12.6: The OTA in Figure 12–20 is connected as an inverting fixed-g...

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