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Catégorie :Category: mViewer GX Creator Lua TI-Nspire
Auteur Author: shlakene
Type : Classeur 3.6
Page(s) : 7
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Mis en ligne Uploaded: 09/10/2019 - 23:02:36
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EE4227 Fall 2019 NAME:_____________________________________




EE4227: Power Electronics

Homework 4 - Fall 2019

Q1. 20 pts
Problem
hidden)




The spec for a dc-dc converter has 12V input and a 3.3 V output at power levels between 10 W and 60 W at
the load. The switching frequency is 120 kHz.
(a) 10 pts. Draw a circuit that can perform this function.
(b) 10 pts. Select any inductor/capacitor values necessary to keep the output voltage ripple below ±1%.
Solution
hidden)




◼ Solution (a)
A buck converter is a good choice since
(1) Vout is less than Vin and is a large difference percentage from each other
(2) Large output voltage range




◼ Solution (b) Inductor only
The average voltage relationship
Vout = D Vin
Vout 3.3
D= Vin
=
12
= 27.5 %

Output resistance at minimum and full load
(Vout )2
RL = Pout
(Vout min load )2 3.32
RL,min = Pout,min
=
10
= 1.09 [Ω]
(Vout min load )2 3.32
RL,min = Pout,min
=
60
= 0.18 [Ω]

The easier approach would be to select the inductor large enough to keep the voltage ripple across
the load resistor within in specification with no capacitor and assuming ideal action and a large time
constant of the circuit
ⅆi
VL = L ⅆt
Δi
VL = L Δt

Considering switch 1 closed and switch 2 open. Using the smaller load power resistor value ⟶ more
restrictive Δi based on percentage




Rel. Date: 9/26/19 1 of 10
EE4227 Fall 2019 NAME:_____________________________________



Δi
VL = L Δt

L= VL ΔΔti
D1 Tsw
= (Vin - Vout ) Vout = 327 [ΜH]
0.02
RL

Intermediate calculation used in above equations
Vout 3.3
Δi = 0.02 iout = 0.02 = 0.02 = 61 [mA]
RL 1.09
1
Δt = D1 Tsw = 0.275 = 2.29 [Μs]
120 000
VL = 12 - 3.3 = 8.7 [V ]

◼ Solution (b) Alternative with inductor and capacitor
Select much smaller inductor and size based on capacitor; Choosing L = 20 [uH ]
Δi
VL = L Δt
Δt
Δi = VL = 0.996 [ A]
L
The inductor will impose a triangular current to the output and the capacitor current should carry the
triangular ripple such that iload will be fixed.
ⅆv Δv
ic = C ⅆt
=C
Δt
1
Δv = ∫ ic (t) ⅆ t = ta → area of shaded triangle below
C




Area of the shaded triangle
ta = 2  12 b h = 2 ⨯ 12 T Δi
2 2
= 4.16 10-6 [ A - s]
1 1 T Δi
Δv = < ⇒ C = > 31.5 [ΜF]
C 2 2 2




Rel. Date: 9/26/19 2 of 10
EE4227 Fall 2019 NAME:_____________________________________




Q2. 30 pts
Problem
hidden)




A power electronics circuit imposes a voltage pulse train from a equivalent source, as shown below as Vs (t),
on a R L series circuit. It is desired to serve a 10 Ω load with a 0.75 H inductor to filter the output. Assume the
peak to be at 50 V. Be sure to state any assumptions!




initialization
hidden)




(a) 10 pts. What is the average value of Vout ?
(b) 10 pts. Calculate the peak-to-peak inductor current variation?
(c) 10 pts. If a large capacitor is added in parallel to the resistor, briefly discuss how this would change the
peak-to-peak inductor current variation and output voltage values? Although, not required, you may want to
run a simulation to test your discussion.
Solution
hidden)




◼ Solution (a)
The average value is calculated as ...
1 0.005 0.01
〈Vin 〉 = ∫0 50 ⅆ t + ∫0.005 0 ⅆ t
0.01
= 25 V
If the inductor is large, the current/voltage ripple will be small, then the average value of the output
will be ...
〈Vout 〉 = 〈Vin 〉 + 〈VL 〉
=0
= 25 V

◼ Solution (b)
Assumptions:
(1) Steady State
(2) Equivalent Source pulse train, shown above, applied to a LR series circuit
Approach:
Solve differential equation for Inductor Voltage to find ripple current. Since the switching period (10
L 750 10-3
ms) than the time constant of the circuit, Τ = R
= 10
= 0.375 10-3 s, we can approximate with
Δi
Δt




Rel. Date: 9/26/19 3 of 10
EE4227 Fall 2019 NAME:_____________________________________



ⅆi Δi
VL = L ⅆt
=L
Δt
VL
Δi = Δt
L
(Vin -Vout D)
=
...

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