-Refer to the figure above. If gm = 4000 µS and a signal of 75 mV rms is applied to the gate, the p- p output voltage is
A)1.13 Vp- p.
B)990 Vp- p.
C)2.8 Vp- p.
D)990 mV.

Question

-Refer to the figure above. If gm = 4000 µS and a signal of 75 mV rms is applied to the gate, the p- p output voltage is
A)1.13 Vp- p. 
B)990 Vp- p. 
C)2.8 Vp- p. 
D)990 mV.

Quizplus · Accepted Answer

The peak-to-peak output voltage can be calculated using the transconductance (gm) and the input signal. Given gm = 4000 µS (or 4 mS) and an input signal of 75 mV rms, first, we convert the rms voltage to peak voltage (Vp) since the question asks for peak-to-peak output voltage (Vpp). The peak voltage (Vp) is Vrms * sqrt(2), which is 75 mV * sqrt(2) = 106.07 mV approximately. The peak-to-peak voltage (Vpp) is twice the peak voltage, so for the output, we calculate it as Vpp = 2 * Vp * gm. Substituting the values, we get Vpp = 2 * 106.07 mV * 4 mS = 848.56 mV, which is approximately 0.85 V peak-to-peak. However, given the options, it seems there might be a misunderstanding in the calculation or the interpretation of the options. The correct approach is to use the transconductance value directly with the rms value for a more straightforward calculation: Vout(pp) = 2 * Vrms * gm = 2 * 75mV * 4 = 600mV peak, which translates to 1.2V peak-to-peak, but this exact value isn't an option. The closest provided option based on the correct calculation method is 2.8 Vp-p, which suggests there might be a mistake in the interpretation of the question or a typo in the options. The explanation provided is based on standard calculations, but the discrepancy suggests a reevaluation of the given options or a clarification in the question's details.

-Refer to the Figure Above