Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_072

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Step 8.040P Refer the circuit diagram of current -source load of the common source (CS) amplifier in Figure 7.3(a) in the textbook. From the figure, it is clear that the PMOS device acts as the load to NMOS CS amplifier. Write the expression for overdrive voltage Here threshold the source to gate Rewrite the Rewrite the It is clear that, source voltage at PMOS device is equal to Supply voltage Substitute 2.5 for for V. and for =2.5-0.8 =1.7 V Thus, gate voltage is 1.7 V Step (b) There is single current (drain current) flows through the entire circuit. Drain current is, Write the expression for current Substitute for 100, for 0.5 for V. and 100 uA for Solve for 100x10 Thus, width to length ratio of PMOS transistor Step Recall the current equation. Substitute for for 0.5 for and 100 for Solve for = Thus, width to length ratio of NMOS transistor IS Step Refer to the circuit diagram of small-signal equivalent circuit of the amplifier in Figure 3(c) in the textbook Write the expression for gain of the circuit. Here 8ml is the trans conductance of the NMOS transistor is the ac output resistance of the NMOS is the output resistance of the PMOS Step of 6 Write the expression for early voltage. L the channel is the early voltage per length. In the fabricated process, for CS amplifier the channel length is 0.4 Substitute 0.4 for L and for 20 V/µm in the equation. (20 =8V Write the expression for I Here is the early voltage. Substitute 20 V/µm for and 100 for in the 8V 100 80 Thus, the ac output resistance of the NMOS transistor is 80 Similarly the ac output resistance of the PMOS transistor is 80 Step 6 Write the expression for trans-conductance of NMOS transistor 2/ (0.3) Recall the gain the circuit expression. Substitute for for and 80 for 2.66 V/V Therefore the voltage gain of circuit 2.66 V/V 66