hw10sol - ECE 315 Homework 10 1(Frequency response of...

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ECE 315 Homework 10 1. (Frequency response of amplifiers) For an NMOSFET in an amplifier setup with C gs =0.5pF, C gd =0.1pF, C db =0.1pF, C L =1pF (including Cdb), g m =5mA/V, r o = 20k , and R sig = R L = 20k , (a) In the CS amplifier of Fig. 6.20, use the Miller theorem to find the midband gain A M and the 3- dB corner frequency f H . (4 pts) The Miller Theorem allows us to substitute the feedback capacitor, C gd , with two separate capacitors from the gate and the drain, both going to ground. The Miller Theorem estimates the equivalent capacitance on the gate as C in,miller = C gd [1 + g m (r o || R L )] = 5.1 pF The equivalent capacitance at the output at the output is given by C out,miller = C gd [1 + (g m (r o || R L )) -1 ] ≈ C gd Therefore, the total input capacitance is C in = C in,miller + C gs = 5.6 pF And we obtain the 3-dB corner frequency: f H = in sig C R π 2 1 = 1.42 MHz The flatband gain for this simple first-order (single pole) system is just A M = -g m (r o || R L ) A M = -50 (b) Repeat (a) with the open-circuit time constant method. Is the answer different? Give the percentage contribution to τ H by each of the three capacitances ( C gs , C gd and C L ). Give also the gain-bandwidth product. (6 pts) The open-circuit time constant method of finding the 3-dB frequency can be done by: 1) Find the equivalent resistance seen by each capacitor using Thevenin/Norton equivalent circuits (note that all other capacitors become open circuits, which is the origin of the name; don’t forget what you did when finding Thevenin/Norton equivalents either, namely shorting independent voltage sources to ground and opening independent current sources) 2) For each capacitor, find the product τ i = R i C i , where R i is the equivalent resistance seen by C i 3) Find the estimated 3-dB frequency by τ H = i i i C R Using this method we find: The equivalent resistance seen by C gs , R gs = R sig = 20 kΩ The equivalent resistance seen by C gd , R gd = R sig (1 + g m (r o || R L )) + (r o || R L ) = 1.03 MΩ The equivalent resistance seen by C L , R L,eq = (r o || R L ) = 10 kΩ 1
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Finding τ H = i i i C R = 0.123 μs f H = = H πτ 2 1 1.29 MHz Contribution from C gs R gs C gs = 10 ns = 8.1 % Contribution from C gd 103 ns = 83.8 % Contribution from C L = 10 ns = 8.1% Gain-Bandwidth product = 64.5 MHz∙V/V (c) Repeat (a) but use the exact analysis in Eq. (6.60). Find the exact values of f Z , f P1 and f P2 , and then estimate f H . How is this f H compared with the ones obtained in (a) and (b)? (6 pts) Looking at the numerator of Eq. 6.60, we see that a zero in the transfer function occurs when = = = = - gd m Z gd m Z m gd C g f C g s g C s π 2 0 1 7.96 GHz The two poles are located at roots of the polynomial in the denominator [ ] { } ( 29 [ ] ) || ( ) || )( ( )) || ( 1 ( 1 2 L o sig gd L gs gd L L o gd L sig L o m gd gs R r R C C C C C s R r C C R R r g C C s + + + + + + + + We can plug in values and solve this numerically:
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This note was uploaded on 09/15/2007 for the course ECE 3150 taught by Professor Spencer during the Spring '07 term at Cornell.

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hw10sol - ECE 315 Homework 10 1(Frequency response of...

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