Question

Assuming an ideal op-amp, design an inverting amplifier with a gain of 26 dB having the largest possible R1 value under the constraint of having to use resistors no larger than 1 MΩ. If the input voltage source of this amplifier has an internal resistance r0 = 50 kΩ, how will this affect the gain? Design a solution that will eliminate the effect of the internal source resistance without having to change the values of R1 and R2.

Answer 1

P3. R₂ RI M V + VC ya Vio -vec Fig: Inverting Amplitics. Gain, G= von - R₂ R, RL Magnitude of gain, Iol= Gain in dB, (Gl dp = 2o log Given Ioldo = 26 dB 26 dB = zolag (P.) – 20log R, loy Pi = - 1.3 R should be log R₂ - In order to get maximum value of R., maximum. . Let choose R₂ = AMIR log Ri log (106)-13 4.7 R = 50.12 kHz R 10b 8o=50kn R Vo 50.1alen vin ③ Sounce

Gain, - 106 ha = -9.988 (50+ 50.12) kn (Gl dB coloy 1-9.998) E 20dB sonya with 80-50kr 8 gain if There is 6 dB reduction in introdued in the ams plities OR Gain reduces to half of its original value. That is, -R. ~ - 20 h 50.ok - 106 -10 R2 R Ulv 50+56.1dka Somu & amplitics, this reduction if 가 in a butte introdued between be eliminated. can gain B2= 1 the thu 80=50kn V. Vo 50.12km -VUA Vu Butter SOWY has on out put resistanu, Since butter with Ideal op-amp output voltage, vol= vir the V. - R. vol hi - Ir vin 50.12 km - 20 Whe Vio