ECE2061 tutorial 1 - Solutions PDF

Preview

Summary

Solution for tutorial 1 in ECE2061, covering how electronics work in this course...

Description

MONASH UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER SYSTEMS ENGINEERING

ECE2061 ANALOGUE ELECTRONICS

PRACTICE CLASS #1: SOLUTIONS

FUNDAMENTAL DC CIRCUIT THEORY 1. Without directly using Ohm’s Law, determine the ratio Vo/Vi for the following voltage divider circuits.

ANS: (a)

(c)

Vo R2 3k = 0.3 = = + Vi R1 R2 3k + 7k

(b)

Vo 2k 2 = 0.24 = V i 6k 8 + 2k 2

Vo 4k 7 = 0.59 = Vi 3k 3 + 4k 7

(d)

Vo 22k = 0.13 = V i 150k + 22 k

2. It is required to produce a nominal 5V output from each three different DC voltage sources with respective magnitudes 12V, 48V, 200V. Calculate values for a two resistor voltage divider string using standard resistor values to achieve (as closely as possible) this output, with the restriction that the maximum power dissipation cannot exceed 0.5W for each of the two resistors in the resistor string. ANS: (a) 12V input: R2 >

72 52 > 98Ω . Make R1 = 100 Ω Æ R2 = 68 Ω, 3% error > 50Ω , R1 > 0.5 0.5

(b) 48V input:

R2 >

43 2 52 > 50Ω , R1 > > 3k 7 . Make R1 = 3k9 Æ R2 = 470 Ω, 3% error 0.5 0.5

(c) 200V input:

R2 >

1952 52 > > 76 k . Make R1 = 82k Æ R2 = 2k2, 5% error , R > 50Ω 1 0.5 0.5

3. Determine the current flowing through resistor R2 for the following circuits.

ANS: (a)

I R2 =

R1 3k Is = 3 = 1A (b) R1 + R2 3k + 6k

(c)

I R2 =

10 k 15 10 k 15 ∗ = 3.19mA ∗ = + + 10k 4k 7 10k 4k 7 10k 4k 7 3.2k

I R2 =

2 k2 10 ∗ = 4.55mA 2 k 2 + 2 k 2 1k1 (d) I R2 = 3.57 mA

4. Determine Thevenin and Norton equivalents for the following circuit.

ANS: Rth = 50 + 100 1k 0 = 140.9 Ω

Vth = 12

1k 0 = 10.9V 1k 0 + 100

IN =

10.9 V th = = 77.4mA Rth 140.9

CIRCUITS WITH DEPENDENT INTERNAL SOURCES

5. Determine the Thevenin equivalent circuit for the following circuit if gm = 0.002S and R1 = 22k. R1 v vs

gmv

ANS: vo / c = vs − R1( − g m ( vs − vo /c )) ⇒ vo / c (1 + g m R1) = vs (1 + g m R1) ⇒ vo / c = vs = vth i s / c = g mvs +

vs ⎛ v 1 ⎞ R1 ⎟⎟ v s = th ⇒ Rth = = ⎜⎜ g m + 1 + gm R1 R1 ⎝ R1 ⎠ Rth

6. Determine the Norton equivalent circuit for the circuit of Question 5 if gm = 0.025S and R1 = 4k7. ANS: R th =

4700 = 39.66Ω 1 + 4700 ∗ 0.025

i N = is/ c =

vs v = s A R th 39.66

7. Determine the Thevenin equivalent circuit for the following circuit if gm = 0.0025S, R1 = 4k7 and R2 = 1Mohm.

ECE2061 – Practice class #1

3

+ is

R1

v _

R2 gmv

ANS: vo / c = − R 2 g m v = − R2 g m R1is = vth i s / c = − g m v = − g m R1i s ⇒ Rth = vth = −10 6 ∗ 0.0025 ∗ 4700i s = 11.75 ∗ 10 6 ∗ i s

Rth = 1Mohm

v o/ c R2 g m R1 = = R2 is / c g m R1...