Transcript DC-DC Converter

DC-DC PWM Converters
Lecture Note 5
DC Choppers. The converters that achieve the
voltage regulation by varying the on–off or time
duty ratio of the switching element using a control
technique called Pulse Width Modulation PWM.
Objective – to efficiently reduce DC voltage
!
The DC equivalent of an AC transformer
Iin
+
Vin
Iout
DC−DC
Converter
−
+
Vout
−
Lossless objective: Pin = Pout, which means that VinIin = VoutIout and
Vout
I in

Vin
I out
3
Linear Conversion
The load
R1
+
Vin
+
R2
−
Vout
−
R2
Vout  Vin 
R1  R2
Vout
R2


R1  R2 Vin
If Vin = 39V, and Vout = 13V, efficiency η is only 0.33
Unacceptable except in very low power applications 4
Linear Conversion
From what explained above, it is clear that a DC
conversion by a voltage divider presents some
drawbacks:
 A DC voltage higher than the input voltage cannot be
obtained;
 The output voltage depends on the load, in general;
 The efficiency is very poor.
Linear Conversion
the output voltage is given by:
• only a step down conversion is possible
• the efficiency remains low because all the power supplied by the source that it
is not utilized by the load have to be dissipated by the power BJT.
Switching Conversion
𝑉𝑜𝐴𝑉
𝑡𝑜𝑛
𝑡𝑜𝑛
= 𝑉𝑖𝑛
= 𝑉𝑖𝑛
= 𝑉𝑖𝑛 𝐷
𝑡𝑜𝑛 + 𝑡𝑜𝑓𝑓
𝑇
Duty Cycle
Switching Conversion
Transistor is operated in switched-mode:
 Switch closed: Fully on (saturated)
 Switch opened: Fully off (cut-off)
 When switch is open, no current flow in it
 When switch is closed no voltage drop across it.
Since P=V.I, no losses occurs in the switch.
 Power is 100% transferred from source to load.
 Power loss is zero (for ideal switch):
 Switching regulator is the basis of all DC-DC converters
Pulse Width Modulation

The output DC voltage of DC chopper can be varied by controlling
the width period (ton)

with constant switching/chopping frequency fs. This method is
called PWM method
Pulse Width Modulation
Choppers Types
Two of the most popular categories of DC-DC converters are:
Transformerless DC-DC Converters
Insulated DC-DC Converters.
Three basic types of non-isolated DC–DC converters are
Step-down converter
Step-up converter
Step-up-down converter
DC−DC Buck Converter
Step-Down Converter
Buck Chopper
12
On-State
Off-State
13
Switch is turned on (closed)
• Diode is reversed biased.
• Switch conducts inductor current
• This results in positive inductor
voltage, i.e:
• It causes linear
increase in the
inductor current
14
Switch turned off (opened)
• Because of inductive
energy storage, iL continues to
flow.
• Diode is forward biased
• Current now flows
(freewheeling) through the
diode.
• The inductor voltage can be
derived as:
15
Analysis
16
Analysis
17
Steady-state Operation
+ L
Realization using
iL(t)
power MOSFET
and diode
Vg +-
+
DTs Ts
VL(t)
D1
ic(t)
R
t
Unstable
Steady-state
18
Since the average voltage across L is zero
VLavg  D  Vin  Vout   1  D   Vout   0
DVin  D  Vout  Vout  D  Vout
The input/output equation becomes Vout  DVin
From power balance, Vin I in  Vout I out , so
I in
I out 
D
Note – even though iin is not constant
(i.e., iin has harmonics), the input power
is still simply Vin • Iin because Vin has no
harmonics
19
!
Output Voltage Ripple
20
Output Voltage
Ripple
21
Examine the inductor current
Examine the inductor current
Switch closed, vL  Vin  Vout ,
diL  Vout
vL  Vout ,

dt
L
Switch open,
 Vout
A / sec
L
iL
Imax
Iavg = Iout
Vin  Vout
A / sec
L
Imin
DT
diL Vin  Vout

dt
L
From geometry, Iavg = Iout is halfway
between Imax and Imin
ΔI
Periodic – finishes
a period where it
started
(1 − D)T
T
23
Examine the inductor current
 Vout
A / sec
L
iL
Imax
Vin  Vout
A / sec
L
Imin
DT
ΔI
Periodic – finishes
a period where it
started
(1 − D)T
Taking the derivative of above equation with respect to D and setting it to zero shows that ΔI is maximum
when D = 0.5
Examine the inductor current
The boundary of continuous conduction is when ΔiLmin = 0, as shown below:
vL  Vout ,
diL  Vout

dt
L
The maximum required value of Lboundary occurs when D → 0. Therefore, the value of L
will guarantee
CCM for all D.
Effect of raising and lowering L while
holding Vin, Vout, Iout and f constant
iL
Lower L
Raise L
• Lowering L increases ΔI and moves the circuit toward
discontinuous operation
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Effect of raising and lowering f while
holding Vin, Vout, Iout, and L constant
iL
Lower f
Raise f
• Slopes of iL are unchanged
• Lowering f increases ΔI and moves the circuit toward
discontinuous operation
27
the rms value inductor current:
i (t )
 Imax  Imin 
the ripple
i (t )
Imax
0
I avg
=
Imin
+
the minimum value
I avg 
Imax  Imin 
2
Imin
0
the rms value inductor current:

2
I rms
 Avg i (t )  I min 2


2
2
I rms
 Avg i2 (t )  2i (t )  I min  I min

 
2
2
I rms
 Avg i2 (t )  2 I min  Avg i (t ) I min
2
I rms

I max  I min 2

 2I
3
I max  I min   I 2

min
min
2
Define I PP  I max  I min
2
I PP
2
2
I rms 
 I min I PP  I min
3
the rms value inductor current:
I
Recognize that I min  I avg  PP
2
2
I rms
2
I PP
I 
I 



  I avg  PP  I PP   I avg  PP 
3 
2 
2 

2
2
2
2
I PP
I PP
I PP
2
2
I rms 
 I avg I PP 
 I avg  I avg I PP 
3
2
2
2
I PP
I PP
2
2
I rms 

 I avg
3
i (t )
4
2
I PP
2
2
I rms  I avg 
12
Or
4
I avg
I avg 
I max  I min 
2
I PP  I max  I min
Component Ratings
Inductor current rating
Capacitor current rating
MOSFET and diode currents ratings
Voltage ratings
Inductor current rating
2
2
I Lrms
 I avg

 
1 2
1
2
I pp  I out

I 2
12
12
Max impact of ΔI on the rms current occurs at the boundary of
continuous/discontinuous conduction, where ΔI =2Iout
2Iout
iL
Iavg = Iout
ΔI
0
2
2
I Lrms
 I out

1
2
2I out 2  4 I out
12
3
2
I Lrms 
I out
3
Use max
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Capacitor current rating
iL
Iout
L
C
Iout
iC = (iL – Iout)
0
−Iout
(iL – Iout)
Note – raising f or L, which lowers
ΔI, reduces the capacitor current
ΔI
Max rms current occurs at the boundary of continuous/discontinuous
conduction, where ΔI =2Iout
Use max
2
2
I Crms
 I avg

1
2
2 I out 2  02  1 I out
12
3
I
I Crms  out
3
33
MOSFET and diode currents and current ratings
iL
iin
Iout
L
C
(iL – Iout)
2Iout
Iout
0
2Iout
Iout
0
Use max
Take worst case D for each
I rms 
2
I out
3
34
Voltage ratings
iL
iin
Iout
C sees Vout
Switch Closed
L
Vin
C
iC
+
Vout
–
Diode sees Vin
MOSFET sees Vin
iL
Switch Open
Iout
L
Vin
C
iC
+
Vout
–
• Diode and MOSFET, use 2Vin
• Capacitor, use 1.5Vout
35
!
There is a 3rd state – discontinuous
Iout
MOSFET
L
Vin
DIODE
36
C
Iout
+
Vout
–
• Occurs for light loads, or low operating frequencies,
where the inductor current eventually hits zero during
the switch-open state
• The diode opens to prevent backward current flow
• The small capacitances of the MOSFET and diode,
acting in parallel with each other as a net parasitic
capacitance, interact with L to produce an oscillation
vL = (Vin – Vout)
Switch
closed
vL = –Vout
Switch open
• The output C is in series with the net parasitic
capacitance, but C is so large that it can be ignored in
the oscillation phenomenon
 650kHz. With L = 100µH, this corresponds
to net parasitic C = 0.6nF
Impedance matching
Iout = Iin / D
Iin
+
+
Source
!
DC−DC Buck
Converter
Vin
Vout = DVin
−
−
V
Rload  out
I out
Iin
+
Vin
Equivalent from
source perspective
Requiv
−
Vout
Vin
Vout
Rload
D
Requiv 



2
I in I out  D I out  D
D2
So, the buck converter
makes the load
resistance look 37larger
to the source
Example 1: Step-Down DC-DC Converter supplied by 230V DC voltage. The load resistance
equal to10Ω. Voltage drop across the chopper when it is ON equal to 2V. For a duty cycle of
0.4, calculate:
a) Average and RMS values of output voltage
b) Power delivered to the load and
c) Chopper efficiency.
DC−DC Boost Converter
Step-Down Converter
Boost Chopper
39
Buck converter
+ vL –
iL
iin
Iout
L
Vin
Boost converter
C
iin
+ vL –
iL
iC
Iout
L
Vin
+
Vout
–
C
iC
+
Vout
–
Boost (step-up) converter
41
Boost Analysis: Switch Closed
42
Boost Analysis: Switch Opened
43
Average Output voltage Expression
The net energy in the inductor is should be equal to zero over T period
Average voltage across inductor is 0
VLavg  ton  Vs  toff  Vs  Vout   0
Vout
ton  DT
Vout
ton
toff
Vs  Vo
Vout  toff  Vs toff  Vs ton
T
 Vs
toff
Vs
T
 Vs
T  ton
Vout
T
 Vs
T  DT
Vout
Vs

1 D
Output Characteristics
Vout
Vo
Vs

1 D
Infinity
Vs
D
0
1
Output Characteristics
Io
Pout  Ps
Is
Vout  I out  Vs  I s
Vs
 I out  Vs  I s
(1  D)
I out  (1  D) I s
D
0
1
As 1 → D , the width of the ΔQ area increases to fill almost the entire
cycle, and the maximum peak-to-peak ripple becomes
V 
Q I out  T I out


C
C
Cf
47
Examine the inductor current
Switch closed,
diL Vin
vL  Vin ,

dt
L
Switch open,
diL Vin  Vout
vL  Vin  Vout ,

dt
L
Vin  Vout
A / sec
L
iL
Imax
Iavg = Iin
Vin
A / sec
L
Imin
DT
Iavg = Iin is half way between
Imax and Imin
ΔI
(1 − D)T
T
48
Continuous current in L
Vin  Vout
A / sec
L
iL
2Iin
Iavg = Iin
0
(1 − D)T
 1

Vin
Vin 
 11  D 
 Vin
V V
1 D 
2 I in  out in  1  D T  1  D
 1  D T 
Lboundary
Lboundary
Lboundary f
2 I in 
Vin D
Lboundary f
,
V D
Lboundary  in
2 I in f
Then, considering the worst case (i.e., D → 1),
V
L  in
2 I in f
use max
guarantees continuous conduction
49
use min
Inductor current rating
2
2
I Lrms
 I avg

 
1 2
1
2
I pp  I in

I 2
12
12
Max impact of ΔI on the rms current occurs at the boundary of
continuous/discontinuous conduction, where ΔI =2Iin
2Iin
iL
Iavg = Iin
ΔI
0
2
2
I Lrms
 I in

I Lrms 
1
2Iin 2  4 Iin2
12
3
2
I in
3
Use max
50
MOSFET and diode currents ratings
iin
+ vL –
iL
iD
Iout
L
Vin
C
iC
+
Vout
–
2Iin
0
2Iin
0
Use max
Take worst case D for each
I rms 
2
I in
3
51
Capacitor current and current rating
iin
iL
iD
Iout
L
Vin
C
iC
+
Vout
–
iC = (iD – Iout)
2Iin −Iout
0
−Iout
Max rms current occurs at the boundary of continuous/discontinuous
conduction, where ΔI =2Iout
Use max
I Crms  I out
52
Voltage ratings
Diode sees Vout
iin
iL
Iout
C sees Vout
+
Vout
–
L
Vin
C
iin
iL
Iout
L
Vin
C
+
Vout
–
MOSFET sees Vout
• Diode and MOSFET, use 2Vout
• Capacitor, use 1.5Vout
53
Impedance matching
I out  1  DIin
Iin
+
+
Source
DC−DC Boost
Converter
Vin
−
Vin
1 D
−
Vout 
V
Rload  out
I out
Iin
+
Vin
Equivalent from
source perspective
Requiv
−
1  D Vout  1  D 2 Vout  1  D 2 R
V
Requiv  in 
load
I out
I in
I out
54
1 D
Example 2: A boost chopper has input voltage of 20 V with switching frequency
equal to 1 kHz. Calculate:
• The required duty cycle that can be applied to the switch to boost the input
voltage to 60V.
• The ON and OFF period for the constant switching frequency operation.
• Output current if the resistance load equal to 10 Ω.
• Average input inductor current.
• The maximum and minimum currents via the input inductor if the inductance is
10mH.