Question 12.6: The OTA in Figure 12–20 is connected as an inverting fixed-g...
The OTA in Figure 12–20 is connected as an inverting fixed-gain amplifier. Determine the voltage gain.
Calculate the bias current as follows:
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}
Using K \cong 16 \mu \mathrm{S} / \mu \mathrm{A} from the graph in Figure 12–14 , the value of transconductance corresponding to I_{\mathrm{BIAS}}=503 \mu \mathrm{A} is approximately
g_m=K I_{\mathrm{BIAS}} \cong(16 \mu \mathrm{S} / \mu \mathrm{A})(503 \mu \mathrm{A})=8.05 \times 10^3 \mu \mathrm{S}
Using this value of g_m , calculate the voltage gain.
A_v=g_m R_L \cong\left(8.05 \times 10^3 \mu \mathrm{S}\right)(10 \mathrm{k} \Omega)=\mathbf{8 0 . 5}
P R A C T I C E EXERCISE
If the OTA in Figure 12–20 is operated with dc supply voltages of \pm 12 \mathrm{~V}, will this change the voltage gain and, if so, to what value?
