440_hmk1_soln

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

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

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:

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

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online