Question 12.7: The input to the OTA amplitude modulator in Figure 12–22 is ...
The input to the OTA amplitude modulator in Figure 12–22 is a 50 mV peak-to-peak, 1 MHz sine wave. Determine the output signal, given the modulation voltage shown is applied to the bias input.
The maximum voltage gain is when I_{\mathrm{BIAS}}, and thus g_{\mathrm{m}} , is maximum. This occurs at the maximum peak of the modulating voltage, V_{\text {mod }} .
I_{\mathrm{BIAS}(\max )}=\frac{V_{\bmod (\max )}-(-V)-1.4 \mathrm{~V}}{R_{\mathrm{BIAS}}}=\frac{10 \mathrm{~V}-(-9 \mathrm{~V})-1.4 \mathrm{~V}}{56 \mathrm{k} \Omega}=314 \mu \mathrm{A}
From the graph in Figure 12–15 , the constant K is approximately 16 \mu \mathrm{S} / \mu \mathrm{A}.
g_m=K I_{\mathrm{BIAS}(\max )}=(16 \mu \mathrm{S} / \mu \mathrm{A})(314 \mu \mathrm{A})=5.02 \mathrm{mS}
\\A_{v(\max )}=g_m R_L=(5.02 \mathrm{mS})(10 \mathrm{k} \Omega)=50.2
\\V_{\text {out }(\max )}=A_{v(\min )} V_{\text {in }}=(50.2)(50 \mathrm{mV})=2.51 \mathrm{~V}
The minimum bias current is
I_{\mathrm{BIAS}(\min )}=\frac{V_{\bmod (\min )}-(-V)-1.4 \mathrm{~V}}{R_{\mathrm{BIAS}}}=\frac{1 \mathrm{~V}-(-9 \mathrm{~V})-1.4 \mathrm{~V}}{56 \mathrm{k} \Omega}=154 \mu \mathrm{A}
\\g_m=K I_{\mathrm{BIAS}(\min )}=(16 \mu \mathrm{S} / \mu \mathrm{A})(154 \mu \mathrm{A})=2.46 \mathrm{mS}
\\A_{v(\min )}=g_m R_L=(2.46 \mathrm{mS})(10 \mathrm{k} \Omega)=24.6
\\V_{\text {out }(\min )}=A_{v(\min )} V_{\text {in }}=(24.6)(50 \mathrm{mV})=1.23 \mathrm{~V}
The resulting output voltage is shown in Figure 12–23 .
P R A C T I C E EXERCISE
Repeat this example with the sinusoidal modulating signal replaced by a square wave with the same maximum and minimum levels and a bias resistor of 39 kΩ.
