440_hmk1_soln

440_hmk1_soln - ECE 440 - Electronics I Solutions to...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 440 - Electronics I Solutions to Homework #1 1. DC Analysis: replacing the capacitors with open circuits: V G = 100 300 (8) = 2.67 = V GS + 1 K (.5 m ) V GS 1 ( ) 2 V GS = 2.08 I D = 0.587 mA V DS = 8 11 KI D = 1.54 V g m = 2(.5 m )(2.08 1) = 1.08 mA / V Setting the DC source to ground and replacing the capacitors with short circuits: Z i = 200 K ||100 K = 66.7 K Ω v o = (10 K ||20 K ) g m v GS ; v GS = Z i Z i + 10 K v s A v = (10 K ||20 K ) g m Z i Z i + 10 K = 6.26 Z o = 10 K Ω 2. Setting V DD to ground and replacing the capacitors with short circuits: Since the gate is open: Z i = 200 K || 300 K = 120 K Ω v o = 3 K ||3 K ( ) g m v GS ; v GS = v G 2 Kg m v GS v GS = v G 1 + 2 Kg m v G = Z i Z i + 10 K v s A v = 3 K ||3 K ( ) g m 1 + 2 Kg m Z i Z i + 10 K = 0.462 Z o = 3 K Ω 3. Setting V DD to ground and replacing the capacitors with short circuits: Since the gate is open: Z i = 100 K Ω v o = (2 K ||4 K ) g m v GS ; v GS = v G (2 K ||4 K ) g m v GS v GS = v G 1 + (2 K ||4 K ) g m v G = 100 K 110 K v s A v = (2 K ||4 K ) g m 1 + (2 K ||4 K ) g m 100 K 110 K = 0.79 Z o = 2 K || 1 g m = 182 Ω 4. Performing the DC analysis:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
V GS + 10 KI D 5 = 0 I D = 5 V GS 10 K = 3 m V GS 1 [ ] 2 V GS = 1.35 g m = 2 K N ( V GSQ V T ) = 2.1 mA / V , r o → ∞ Setting the DC sources to ground and replacing the capacitors with short circuits: v o = 5 K ||4 K ( ) g m v GS ; v GS = v i A v = 5 K ||4 K ( ) g m = 4.7 Z o = 5 K Ω 5. Setting the DC sources to ground and replacing the capacitors with short circuits: Z i = 1.8 K || 1 g m = 486 Ω v o = 2.5 K ||10 K ( ) g m v GS 2 ; v GS 2 = v G 2 2.5 K ||10 K ( ) g m v GS 2 ; v GS 2 = v G 2 1 + 2.5 K ||10 K ( ) g m v G2 = (15 K ||1 M ) g m v GS 1 ; v GS 1 = v i A v = 2.5 K ||10 K ( ) g m 1 + 2.5 K ||10 K ( ) g m (15 K ||1 M ) g m = 16.6 Z o = 2.5 K || 1 g m = 526 Ω 6. For the DC analysis assume that all the capacitors are open and refer to the bottom device as Q1 and the top device as Q2: V G 1 = 50 K 300 K (5) = 0.833 = V GS 1 + 10 KI D 1 5 = 0 I D 1 = 5.833 V GS 1 10 K = 0.8 m V GS 1 1.2 [ ] 2 V GS 1 = 1.9 I D 1 = 0.8 m 1.9 1.2 [ ] 2 = 0.39 mA = I D 2 V GS 2 = 1.9 g m = 2(0.8 m )(1.9 1.2) = 1.12 mA / V , r o → ∞ When we redraw the circuit at AC the capacitors are short and the DC sources are grounded.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 11

440_hmk1_soln - ECE 440 - Electronics I Solutions to...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online