To eliminate the problem associated with the need for a large resistance RD in the circuit of Fig. 14.20(a), studied in Example 14.2, RD can be replaced by a MOSFET. One such possibility is the circuit in Fig. 14.21, where the load is a PMOS transistor QP whose gate is tied to ground in order to

Question

To eliminate the problem associated with the need for a large resistance R_D in the circuit of Fig. 14.20(a), studied in Example 14.2, R_D can be replaced by a MOSFET. One such possibility is the circuit in Fig. 14.21, where the load is a PMOS transistor Q_P whose gate is tied to ground in order to turn it on. Because of its resemblance to an earlier form of logic (NMOS logic, now obsolete) in which the load is an NMOS transistor, this circuit is known as a pseudo-NMOS inverter.

(a) Assuming λ₁ = λ₂ = 0, V_tn = −V_tp = V_t , and k_n = 5 k_p, find V_OH and V_OL.
(b) For k_n = 300 μA/V², V_t = 0.4 V, and V_DD = 1.8 V, evaluate the values of V_OH and V_OL and find the average power dissipated in the inverter, assuming it spends half the time in each of its two states.

Accepted Answer

(a) To find V_OH, we set v_I = 0. Clearly Q_N will be off and conducting zero current. Transistor Q_P also will be conducting zero current but because its V_SG = V_D_D, it will be operating in the triode region with a zero voltage between its source and drain; thus the output voltage will be equal to V_D_D,

V_OH = V_D_D

Next, we find V_OL by setting v_I = V_DD. Transistor Q_N will be conducting. Since the output voltage V_OL will likely be low and thus lower than V_t , Q_P will be operating in the saturation region, thus

[latex]i_{DP} = \frac{1}{2} k_{p}(V_{DD} - V_{t})^{2}[/latex]

Q_N will be operating in the triode region, thus

[latex]i_{DN} = k_{n} \left[(V_{DD} - V_{t})V_{OL} - \frac{1}{2} V_{OL}^{2} \right][/latex]

Equating i_DP and i_DN yields a quadratic equation in V_OL that can be solved to obtain

[latex]V_{OL} = (V_{DD} - V_{t}) \left[1 - \sqrt{1 - (k_{p}/k_{n})}\right][/latex]

Here we have rejected the other root of the quadratic equation on the assumption that its value will be greater than V_t and thus contravening our original assumption. The numerical values can be used to check these assumptions.

(b) Substituting the given numerical values we obtain

V_OH = 1.8 V

V_OL = 0.15 V

We note that V_OL is indeed lower than V_t , as originally assumed.
The inverter dissipates power in only one of its two states; namely, when its output is low. In this case, Q_P is operating in saturation and

[latex]i_{DP} = \frac{1}{2} k_{p} (V_{DD} - V_{t})^{2}[/latex]

[latex] = \frac{1}{2} × \left(\frac{300}{5}\right) (1.8 - 0.4)^{2}[/latex]

= 58.8 μA

and the power dissipation can be found from

P = i_DPV_DD = 58.8 × 1.8 = 105.8 μW

The average power dissipation can now be found as

[latex]P_{av} = \frac{1}{2} × 105.8 = 52.9 μW[/latex]

Question 14.3: To eliminate the problem associated with the need for a larg...

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