ECE_315_Homework_6_Solutions_07

ECE_315_Homework_6_Solutions_07 - ECE 315 Homework 6...

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ECE 315 Homework 6 Solutions October 18, 2006 1. (How small is small signal operations) Consider an nMOSFET having k n ’W/L =2mA/V 2 . Let the transistor be biased at V OV =1V. (a) For operations in saturation, what is I D ? (6 pts) I D = 0.5 k n ’(W/L)V OV 2 I D = 1 mA (b) If a +0.1V signal is superimposed on V GS , find the corresponding increment in I D in comparison with V OV =1V? (6 pts) I D ’ =0.5 k n ’(W/L)(V GS + 0.1V – V th ) 2 I D ’ = 0.5 k n ’(W/L)(V OV + 0.1V) 2 I D ’ = 1.21 mA ΔI D = I D ’ – I D ΔI D = 0.21 mA (c) If a -0.1V signal is superimposed on V GS , find the corresponding change in I D in comparison with V OV =1V? (6 pts) I D ’ = 0.5 k n ’(W/L)(V GS + 0.1V – V th ) 2 I D ’ = 0.5 k n ’(W/L)(V OV + 0.1V) 2 I D ’ = 0.81 mA ΔI D = I D ’ – I D ΔI D = -0.19 mA (d) Use the results in (b) and (c) to estimate g m , and compare with the small signal g m evaluated at V OV =1V. (6 pts) g m = ΔI D /ΔV GS = 2.1 mS (for ΔV GS = 0.1V) g m = -1.9 mS (for ΔV GS = -0.1V) we can estimate g m to be approximately the average of these two solutions: g m = 2.0 mS find small signal g m : g m = ∂I D /∂V GS = ∂/∂V GS [0.5 k n ’(W/L)(V GS – V th ) 2 ] g m = k n ’(W/L)(V GS – V th ) = k n ’(W/L)V OV g m = 2 mS The estimate g m and the calculated small signal g m values are in agreement. 2. (Current source biasing) For the circuits in Fig. 4.33(a) in the textbook with I = 1mA, R G =0, R D =5k , V DD =10V, and V SS = -10V, consider the behavior in each of the following two cases. In each case, find the voltages V S (potential at the source node), V D and V DS = V D – V S
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(a) V th = 1V and k n ’W/L =0.5mA/V 2 (6 pts) V D = V DD – IR D = 10V – 1mA×5kΩ V D = 5 V First, let’s assume saturation: I D = 0.5 k n ’(W/L) (V GS – V th ) 2 = 1 mA V G = 0 V, V th = 1 V, k n ’(W/L) = 0.5 mA/V 2 I D = 0.5 k n ’(W/L) (-V S – 1) 2 = 1 mA V S = 1 V or -3 V If V S = 1 V, then V GS = -1 V < V th then nMOSFET must be in subthreshold mode, which cannot support a current of 1 mA Therefore, V S = -3 V Then, V DS = 5 – (-3) V V DS = 8 V Now, we must check our assumption: Is V GS – V th < V DS for saturation? Yes, V GS – V th = 2 V, V DS = 8 V Therefore, we can conclude that: V D = 5 V, V S = -3 V, V DS = 8 V (b) V th = 2V and k n ’W/L =1.25mA/V 2 (6 pts) Again, V D = V DD – IR D = 5 V Let’s assume saturation again: I D = 0.5 k n ’(W/L) (V GS – V th ) 2 = 1 mA V G = 0 V, V th = 2 V, k n ’(W/L) = 1.25 mA/V 2 Solving for V S V S = -0.74 V or -3.26 V If V S = -0.74 V, then V GS = 0.74 V < V th then nMOSFET must be in subthreshold mode, which cannot support a current of 1 mA Therefore, V S = -3.26 V Then, V DS = 8.26 V Check assumptions: Is V GS – V th < V DS for saturation? Yes, V
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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 University (Engineering School).

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ECE_315_Homework_6_Solutions_07 - ECE 315 Homework 6...

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