(1) Design and build the differential amplifier shown in Fig. 1 such that the small signal gain (Vol Vs)=-100. The dc levels of Vo1 and Vo2 should be equal and about (3-5) volts. Draw the waveforms of Vol and Vo2 obtained on the dual-trace oscilloscope, and measure their ac vlaues more accurately using the digital multimeter. Compute the differential gain Ad, the common-mode gain Ac, and hence the CMRR (common-mode rejection ratio). Compare these experimental values with the theoretical ones. lov RC Re Fig.1. כלול RE 10K V02 -lov Q2 Hints: Ad (single-output) = Vol/Vs-gm- R/2 CMRR (theoretical)=1+2gm-RE For ac outputs, one output is (D+C) and the other is (-D+C), where D is the differential component and C is the common-mode component. By solving two equations, one can determine D and C, thus: Ad (experimental)=D/V Ac (experimental) = C/(V/2), CMRR (experimental-single output) = 20 log(A/Ac). Remember that the digital multimeter ac reading is an rms value, thus multiply by 2 to find the peak value.

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im using a 2N2222A transistor for Q1 and Q2, give me values of all components in the circuit for the experiments, Ad should equal to -100 and vo1 and vo2 should be about 3-5 volts

(1) Design and build the differential amplifier shown in Fig. 1 such that the small signal gain
(Vol Vs)=-100. The dc levels of Vo1 and Vo2 should be equal and about (3-5) volts.
Draw the waveforms of Vol and Vo2 obtained on the dual-trace oscilloscope, and measure
their ac vlaues more accurately using the digital multimeter. Compute the differential gain
Ad, the common-mode gain Ac, and hence the CMRR (common-mode rejection ratio).
Compare these experimental values with the theoretical ones.
lov
RC
Re
Fig.1.
כלול
RE 10K
V02
-lov
Q2
Hints:
Ad (single-output) = Vol/Vs-gm- R/2
CMRR (theoretical)=1+2gm-RE
For ac outputs, one output is (D+C) and the other is (-D+C), where D is the differential
component and C is the common-mode component. By solving two equations, one can
determine D and C, thus:
Ad (experimental)=D/V
Ac (experimental) = C/(V/2),
CMRR (experimental-single output) = 20 log(A/Ac).
Remember that the digital multimeter ac reading is an rms value, thus multiply by 2 to
find the peak value.
Transcribed Image Text:(1) Design and build the differential amplifier shown in Fig. 1 such that the small signal gain (Vol Vs)=-100. The dc levels of Vo1 and Vo2 should be equal and about (3-5) volts. Draw the waveforms of Vol and Vo2 obtained on the dual-trace oscilloscope, and measure their ac vlaues more accurately using the digital multimeter. Compute the differential gain Ad, the common-mode gain Ac, and hence the CMRR (common-mode rejection ratio). Compare these experimental values with the theoretical ones. lov RC Re Fig.1. כלול RE 10K V02 -lov Q2 Hints: Ad (single-output) = Vol/Vs-gm- R/2 CMRR (theoretical)=1+2gm-RE For ac outputs, one output is (D+C) and the other is (-D+C), where D is the differential component and C is the common-mode component. By solving two equations, one can determine D and C, thus: Ad (experimental)=D/V Ac (experimental) = C/(V/2), CMRR (experimental-single output) = 20 log(A/Ac). Remember that the digital multimeter ac reading is an rms value, thus multiply by 2 to find the peak value.
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