Question 12.DA.11: A MOSFET FEEDBACK CIRCUIT Objective: Redesign a BJT feedback...

(DC Design): For v_{O} = 0, the current in M_{3} is I_{D3} = 2  mA. Then
I_{D} = K_{n}(V_{GS3}  −  V_{T N})^{2}
or
2 = (1)(V_{GS3}  −  1)^{2}
which yields
V_{GS3} = 2.414  V

The voltage at the gate of M_{3} is then to be V_{G3} = 2.414  V. The current in M_{2} is 0.5 mA, so the resistance R_{D} is
R_{D} = \frac{12  −  2.414}{0.5} = 19.2  k \Omega
(AC Analysis): We can find the small-signal parameters as
g_{m1} = g_{m2} ≡ g_{m} = 2 \sqrt{K_{n} I_{D1}} = 2 \sqrt{(1)(0.5)} = 1.414  mA/V
and
g_{m3} = 2 \sqrt{K_{n} I_{D3}} = 2 \sqrt{(1)(2)} = 2.828  mA/V
The small-signal equivalent circuit is shown in Figure 12.63. Summing currents at the V_{1} node, we have
g_{m}  V_{gs1} + g_{m}  V_{gs2} = 0 ⇒ V_{gs2} = −V_{gs1}

Writing a KVL equation from the input, we find
V_{i} = V_{gs1}  −  V_{gs2} + V_{2} = −2V_{gs2} + V_{2}
or
V_{gs2} = \frac{V_{2}  −  V_{i}}{2}
We see that
V_{gs3} = −g_{m} V_{gs2} R_{D}  −  V_{o} = − \frac{1}{2} (V_{2}  −  V_{i}) R_{D}  −  V_{o}
Also
V_{2} = \left(\frac{R_{1}}{R_{1}  +  R_{2}}\right) V_{o} = \left(\frac{10}{10  +  40}\right) V_{o} = 0.2  V_{o}
so that
V_{gs3} = − \frac{1}{2} g_{m}[(0.2)V_{o}  −  V_{i}] R_{D}  −  V_{o}

Summing currents at the output node, we obtain
g_{m3} V_{gs3} = \frac{V_{o}}{R_{L}} + \frac{V_{o}}{R_{1}  +  R_{2}}
or
g_{m3} \left\{− \frac{1}{2} g_{m} [(0.2)V_{o}  −  V_{i}] R_{D}  − V_{o} \right\} = \frac{V_{o}}{R_{L}} + \frac{V_{o}}{R_{1}  +  R_{2}}
Combining terms, we obtain
\frac{1}{2} g_{m3} g_{m} R_{D} V_{i} = V_{o} \left[g_{m3} \left(1 + \frac{1}{2} g_{m}(0.2)R_{D} \right) + \frac{1}{R_{L}} + \frac{1}{R_{1}  +  R_{2}} \right]       (12.127)
Substituting parameters, we find
\frac{1}{2} (2.828)(1.414)(19.2)V_{i}

= V_{o} \left[(2.828) \left(1 + \frac{1}{2} (1.414)(0.2)(19.2) \right) + \frac{1}{4} + \frac{1}{10  +  40} \right]
The closed-loop voltage gain is then
A_{v} = \frac{V_{o}}{V_{i}} = 3.56
(Gain Variations): One of the advantages of feedback is that the closed-loop gain is relatively insensitive to changes in the individual transistor parameters. Determine the closed-loop gain if the conduction parameters decrease by 10 percent.
The new values of the small-signal parameters are
g_{m1} = g_{m2} ≡ g_{m} = 2 \sqrt{K_{n} I_{D1}} = 2 \sqrt{(0.9)(0.5)} = 1.342  mA/V
and
g_{m3} = 2 \sqrt{K_{n} I_{D3}} = 2 \sqrt{(0.9)(2)} = 2.683  mA/V
Substituting these values into Equation (12.127), we obtain
\frac{1}{2} (2.683)(1.342)(19.2) V_{i}

= V_{o} \left[(2.683) \left(1 + \frac{1}{2} (1.342)(0.2)19.2 \right) + \frac{1}{4} + \frac{1}{10  +  40} \right]
The closed-loop gain is then
A_{v} = \frac{V_{o}}{V_{i}} = 3.50
Comment: With a decrease of 10 percent in the transistor conduction parameters, the closed-loop gain has decreased by less than 2 percent. Even though we are considering a relatively simple feedback circuit with only three transistors, the advantage of feedback is observed.