Analog Circuits Test 6

In voltage series feedback mode,

Input impedance (R_in) = R_i (1 + Aβ)

Output impedance \(\left( {{R_{out}}} \right) = \frac{{{R_o}}}{{\left( {1 + A\beta \;} \right)}}\)

Where A is open loop gain

β is feedback factor

Given that, open loop gain (A) = 100 Ω

Feedback factor (β) = 0.99

Input impedance of amplifier (R_i) = 1 kΩ

Output impedance of amplifier (R_o) = 100 Ω

R_in = 1 × 10³ (1 + 100 × 0.99) = 100 kΩ

\({R_{out}} = \frac{{100}}{{\left( {1 + 100 \times 0.99} \right)}} = 1\;{\rm{\Omega }}\)

The closed-loop gain of a feedback amplifier is given as:

\(A_f(s)=\frac{A(s)}{1+A(s)β }\)   ---(1)

A = Open Loop gain

Af = Closed Loop Gain

β = Feedback/Transmission factor

The frequency-dependent open-loop gain is defined as:

\(A(s)=\frac{A}{1+s/ω_H}\)

A = mid-band gain or open loop gain

ω_H = Upper cut off frequency

Substituting A(s) to Equation (1) and after a little manipulation, we get:

\(A_f(s)=\frac{\frac{A}{1+Aβ }}{1+\frac{s}{ω_H(1+Aβ)}}\)

We observe that the upper cut off frequency is increased by a factor of 1 + Aβ.

Similarly, the lower cutoff frequency after feedback is given by:

\(ω_{Lf}=\frac{ω_L}{1+Aβ}\)

Application:

Given β = 0.99 and A = 100

Also, f_H = 100 kHz and f_L = 100 Hz

After feedback the new upper cut off frequency will be:

f_H(new) = f_H (1 + Aβ)

Putting on the respective values, we get:

fH(new) = 100k (1 + 99)

fH(new) = 10 MHz

And the lower cut off frequency will be:

\(f_{L(new)}=\frac{f_L}{1+Aβ}\)

\(f_{L(new)}=\frac{100}{100}=1 ~Hz\)