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: ω p1 = 8.2 Mrad/s f p1 = 1.31 MHz ω p2 = 938 Mrad/s f p2 = 149.3 MHz f H ≈ f p1 = 1.31 MHz This is on the same order of frequency as found in parts (a) and (b). In fact, it is only 0.11 MHz off from the answer of part (a), which is less than 10 % and only 0.02 MHz off from the answer of part (b).
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