Question

ADM = \frac{\{v_a - v_b\}}{\{v_A - v_B\}} ACM = \frac{1/2\{v_a + v_b\}}{1/2\{v_A + v_B\}} 1.37. For the circuit shown in the above problem, the CMRR of the input buffer is 20 dB while the CMRR of the differential amplifier is 54 dB. What is the CMRR of the total amplifier (consisting of the buffer in series with the differential amplifier). Present theoretical validation of the results.

          ADM = \frac{\{v_a - v_b\}}{\{v_A - v_B\}}
ACM = \frac{1/2\{v_a + v_b\}}{1/2\{v_A + v_B\}}
1.37. For the circuit shown in the above problem, the CMRR of the input buffer is 20 dB while the CMRR of the differential amplifier is 54 dB. What is the CMRR of the total amplifier (consisting of the buffer in series with the differential amplifier). Present theoretical validation of the results.

ADM = ({va - vb})/({vA - vB})
ACM = (1/2{va + vb})/(1/2{vA + vB})
1.37. For the circuit shown in the above problem, the CMRR of the input buffer is 20 dB while the CMRR of the differential amplifier is 54 dB. What is the CMRR of the total amplifier (consisting of the buffer in series with the differential amplifier). Present theoretical validation of the results.

Added by Ruben H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vivek K

Step 1

In the given problem, the CMRR of the input buffer is 20 dB, which means that the differential mode gain is 20 dB higher than the common mode gain. Similarly, the CMRR of the differential amplifier is 54 dB, which means that the differential mode gain is 54 dB Show more…

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Text: {Ua-Ub} ADM= {UA-VB} 1/2{va+Ub} ACM= 1/2{0A+UB} RB WW WM R2 R R2 WM RA WW Rc Input Buffer U RI VA 1.37. For the circuit shown in the above problem, the CMRR of the input buffer is 20 dB while the CMRR of the differential amplifier is 54 dB. What is the CMRR of the total amplifier consisting of the buffer in series with the differential amplifier? Present theoretical validation of the results.