DC Amplifier MCQ [Free PDF] – Objective Question Answer for DC Amplifier Quiz

The resonant frequency is also called peak frequency, which is determined by the combination of L and C.

f_p = 1/(2π√LC).

Frequency,

f_p= 1/[2π×√(LC.]

=1/[2π√(0.1µF×8mH)]

=1/1.776×10_-4= 5.63kHz.

=> X_L = 2πf_pL

= 2π×5.63kHz×8mH = 282.85.

The figure of merit of coil

Q_coil= X_L/R₁= 282.85/100Ω = 2.8285.

∴ R_p = (Q_coil)² ×R₁

= (2.8285²)×100Ω= 800Ω.

The gain of the amplifier at resonance is maximum and given by

A_F =-(R_F||R_p)/R₁

= -(10kΩ||800)/100Ω

=-740.740/100 = -7.407.

The gain of the amplifier at resonance is the maximum and given by,

A_F =-(R_F||R_p)/ R₁

=-[(R_p×R_F)/ (R_F+R_p)] /R₁

= -[ (10kΩ × 35kΩ)/ (10kΩ+35kΩ)] /1kΩ

=> A_F =- 7.78kΩ/1kΩ= -7.78.

The bandwidth of the peaking amplifier,

BW = f_p/Q_p,

where Q_p – figure of merit of the parallel resonant circuit

= (R_f||R_p)/X_l = (R_f×R_p)/[(R_f+R_p)× X_l]

=> BW = [f_p×(R_f+R_p)× X_l] / (R_f×R_p).

Given L=10mH and the internal resistance of the inductor R=30Ω. Assume R₁=100Ω. The gain times peak frequency=

10 × 32kHz = 320kHz

f_p= 1/2π√LC

=> C = 1/[(2π)²× (f_p)²×L]

= 1/ [(2π)²×(320)²×10mH]

= 1/252405.76 = 3.96µF ≅4µF.

Q_coil = x_L/R

=(2πf_pL)/R

=(2π×320kHz×10mH)/30

= 20096/30 =669.87

=> R_p= (Q_coil)²×R

= (669.88)²×30 = 13.5MΩ

To find R_f,

A_F= (R_F×R_p)/[R₁×(R_F+R_p)]

=>R_F = (A_f ×R_p ×R₁)/ (R_p -A_F ×R₁)

R_F = (-10×13.5×10⁶×100) / (13.5×10⁶-(10×100))=1000Ω

=> R_F = 1kΩ.

Thus the component values are R₁=100Ω, R_F= 1kΩ, L=10mH at R=30Ω and C = 4µF.

In the summing amplifier, the output voltage is equal to the sum of all input. Since the total input is a sum of negative input, the amplifier is an inverting summing amplifier.

If the internal resistors of the circuit are the same i.e

R_a=R_b=R_c=R_d=R (since there are four internal resistors)

Then, the output voltage for inverting amplifier is given as

V_o= -(R_f/R)×(V_a +V_b+V_c+V_d).

When the gain of the inverting summing amplifier gain is 1 then, the internal resistors and feedback resistors have the same value. So, the output is equal to the negative sum of all input voltages.

V_O= -(V_a+V_b+V_c)

=-(1.2+3.2+4.2)= -8.6v.

The weighted amplifier is also called a scaling amplifier. Here each input voltage is amplified by a different factor i.e. R_a, R_b and R_c are different in values ( which are the input resistors at each input voltage).

Since three input voltages are given assume the input resistors to be R_a, R_b, and R_c. In a scaling amplifier, the input voltages are amplified by a different factor

=> ∴ R_F/R_a ≠ R_F/R_b ≠ R_F/R_c

Therefore, the output voltage

V_o = -{[(R_F/R_a) V_a] +[(R_F/R_b) V_b]+[(R_F/R_c) V_c]}.