A MOSFET FEEDBACK CIRCUIT

Objective: Redesign a BJT feedback circuit using MOSFETs.

Specifications: The circuit in Figure P12.36 is to be redesigned using MOSFETs. The new circuit configuration is shown in Figure 12.62. The output voltage is to be zero for $v_{i} = 0$ .

Choices: Assume that NMOS devices are available with parameters $V_{T N} = 1 V, K_{n} = 1 mA/V^{2}$ , and λ = 0.

Question

A MOSFET FEEDBACK CIRCUIT

Objective: Redesign a BJT feedback circuit using MOSFETs.

Specifications: The circuit in Figure P12.36 is to be redesigned using MOSFETs. The new circuit configuration is shown in Figure 12.62. The output voltage is to be zero for [latex]v_{i} = 0[/latex].

Choices: Assume that NMOS devices are available with parameters [latex]V_{T N} = 1 V, K_{n} = 1 mA/V^{2} [/latex], and λ = 0.

Accepted Answer

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

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

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

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

[latex]= 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][/latex]
The closed-loop voltage gain is then
[latex]A_{v} = \frac{V_{o}}{V_{i}} = 3.56[/latex]
(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
[latex]g_{m1} = g_{m2} ≡ g_{m} = 2 \sqrt{K_{n} I_{D1}} = 2 \sqrt{(0.9)(0.5)} = 1.342 mA/V[/latex]
and
[latex]g_{m3} = 2 \sqrt{K_{n} I_{D3}} = 2 \sqrt{(0.9)(2)} = 2.683 mA/V[/latex]
Substituting these values into Equation (12.127), we obtain
[latex]\frac{1}{2} (2.683)(1.342)(19.2) V_{i}[/latex]

[latex]= 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][/latex]
The closed-loop gain is then
[latex]A_{v} = \frac{V_{o}}{V_{i}} = 3.50 [/latex]
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.

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

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