Photovoltaic Systems
Ahmed G. Abo-Khalil
Power Electronics and Power Conversion , Assiut University
1
Components of PV Systems
Power Electronics and Power Conversion , Assiut University
2
Electrical Equivalent Circuit PV Cells
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3
I-V and P-V Characteristics
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4
Stand-alone PV System
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5
Introduction to
DC−DC Buck Converter
Power Electronics and Power Conversion , Assiut University
6
DC-DC Converter
The DC equivalent of an AC transformer
Iin
Iout
+
Vin
−
+
DC−DC
Buck
Converter
Vout
−
Lossless objective: Pin = Pout, which means that VinIin = VoutIout and
Vout
I
 in
Vin
I out
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Example of an inefficient DC−DC converter
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
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8
Another method – lossless conversion of 39Vdc
to average 13Vdc
Stereo
voltage
Switch closed
Switch open
39
+
39Vdc
–
Rstereo
0
Switch state
Closed, 39Vdc
DT
T
Open, 0Vdc
If the duty cycle D of the switch is 0.33, then the average voltag
e to the expensive car stereo is 39 ● 0.33 = 13Vdc. This is loss
less conversion, but is it acceptable?
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9
Convert 39Vdc to 13Vdc, cont.
+
39Vdc
–
Try adding a large C in parallel with the load to
control ripple. But if the C has 13Vdc, then
when the switch closes, the source current
spikes to a huge value and burns out the
switch.
Rstereo
C
L
+
39Vdc
–
C
Rstereo
Try adding an L to prevent the huge
current spike. But now, if the L has
current when the switch attempts to
open, the inductor’s current momentum
and resulting Ldi/dt burns out the switch.
lossless
L
+
39Vdc
–
C
Rstereo
By adding a “free wheeling” diode, the
switch can open and the inductor current
can continue to flow. With highfrequency switching, the load voltage
ripple can be reduced to a small value.
A DC-DC Buck Converter
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10
C’s and L’s operating in periodic steady-state
Examine the current passing through a capacitor that is operating in
periodic steady state. The governing equation is
dv ( t )
i(t )  C
dt
t
1 o t
which leads to v ( t )  v ( to )  C  i ( t )dt
to
Since the capacitor is in periodic steady state, then the voltage at ti
me to is the same as the voltage one period T later, so
v ( to  T )  v ( to ), or
The conclusion is that
t
1 o T
v ( to  T )  v ( to )  0 
i ( t )dt
C 
to T
 i ( t )dt  0
to
which means that
to
the average current through a capacitor operating in periodic stead
y state is zero
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Now, an inductor
Examine the voltage across an inductor that is operating in periodic
steady state. The governing equation is
di ( t )
v(t )  L
dt
which leads to
t
1 o t
i ( t )  i ( to ) 
v ( t )dt
L 
to
Since the inductor is in periodic steady state, then the current at tim
e to is the same as the current one period T later, so
i ( to  T )  i ( to ), or
The conclusion is that
t
1 o T
i ( to  T )  i ( to )  0 
v ( t )dt
L 
to T
 v( t )dt  0
to
which means that
to
the average voltage across an inductor operating in periodic steady
state is zero
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KVL and KCL in periodic steady-state
Since KVL and KCL apply at any instance, then they must also be valid
in averages. Consider KVL,
 v(t )
 0, v1 ( t )  v2 ( t )  v3 ( t )    v N ( t )  0
Around loop
t
t
t
t
t
1 o T
1 o T
1 o T
1 o T
1 o T
v1 ( t )dt 
v2 ( t )dt 
v3 ( t )dt   
v N ( t )dt 
(0)dt  0
T 
T 
T 
T 
T 
to
to
to
V1avg  V2avg  V3avg    VNavg  0
to
to
KVL applies in the average sense
The same reasoning applies to KCL
 i(t )
 0,
i1 ( t )  i2 ( t )  i3 ( t )    i N ( t )  0
Out of node
I1avg  I 2avg  I 3avg    I Navg  0
KCL applies in the average sense
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Buck converter
+ vL –
iL
iin
Iout
L
Vin
C
• Assume large C so that
Vout has very low ripple
iC
+
Vout
–
• Since Vout has very low
ripple, then assume Iout
has very low ripple
What do we learn from inductor voltage and capacitor
current in the average sense?
+0V–
iin
Iout
Iout
L
Vin
C
+
Vout
0A
–
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The input/output equation for DC-DC converters
usually comes by examining inductor voltages
+ (Vin – Vout) –
iin
Switch closed for
DT seconds
iL
Iout
+
V
(iL – Iout) out
–
L
Vin
C
Reverse biased, thus the
diode is open
vL  L
diL
,
dt
vL  Vin  Vout ,
Vin  Vout  L
diL
,
dt
diL Vin  Vout

dt
L
for DT seconds
Note – if the switch stays closed, then Vout = Vin
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Switch open for (1 − D)T seconds
– Vout +
iL
Iout
L
Vin
C
+
Vout
(iL – Iout)
–
iL continues to flow, thus the diode is closed. This
is the assumption of “continuous conduction” in the
inductor which is the normal operating condition.
vL  L
diL
,
dt
vL  Vout ,
 Vout  L
diL
,
dt
diL  Vout

dt
L
for (1−D)T seconds
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Since the average voltage across L is zero
1
T
DT
1
0 vL (t )dt  T
(1 D )T
 v (t )dt  0
VLavg  D  Vin  Vout   1  D   Vout   0
L
DT
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
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Examine the inductor current
Switch closed, vL  Vin  Vout ,
Switch open, v L  Vout ,
 Vout
A / sec
L
iL
diL Vin  Vout

dt
L
diL  Vout

dt
L
From geometry, Iavg = Iout is halfway
between Imax and Imin
Imax
Iavg = Iout
Vin  Vout
A / sec
L
Imin
DT
ΔI
Periodic – finishes
a period where it
started
(1 − D)T
T
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Effect of raising and lowering Iout while
holding Vin, Vout, f, and L constant
iL
ΔI
Raise Iout
ΔI
Lower Iout
ΔI
• ΔI is unchanged
• Lowering Iout (and, therefore, Pout ) 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
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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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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
I
2
Lrms
I
2
out
1
4 2
2
 2 I out   I out
12
3
2
I Lrms 
I out
3
Use max
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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
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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 larger
to the source
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Example of drawing maximum power from solar
panel
PV Station 13, Bright Sun, Dec. 6, 2002
6
Isc
Pmax is approx. 130W
(occurs at 29V, 4.5A)
5
I - amps
4
For max power from
panels at this solar
intensity level, attach
3
2
Rload 
1
29V
 6.44
4.5 A
0
0
5
10
15
20
25
V(panel) - volts
30
35
40
Voc
I-V characteristic of 6.44Ω resistor
45
But as the sun conditions
change, the “max power
resistance” must also
change
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Connect a 2Ω resistor directly, extract only 55W
PV Station 13, Bright Sun, Dec. 6, 2002
55W
6
130W
5
I - amps
4
3
2
1
0
0
5
10
15
20
25
30
35
40
45
V(panel) - volts
To draw maximum power (130W), connect a buck converter between the
panel and the load resistor, and use D to modify the equivalent load
resistance seen by the source so that maximum power is transferred
R
Requiv  load , D 
D2
Rload
2

 0.56
Requiv
6.44
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Buck converter for solar applications
The panel needs a ripple-free current to stay on the max power point.
Wiring inductance reacts to the current switching with large voltage spikes.
ipanel
+ vL –
iL
Iout
L
Vpanel
C
iC
+
Vout
–
Put a capacitor here to provide the
ripple current required by the
opening and closing of the MOSFET
In that way, the panel current can be ripple
free and the voltage spikes can be controlled
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DC−DC Boost Converter
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Buck converter
+ vL –
iL
iin
Iout
+
Vout
–
L
Vin
Boost converter
C
iin
+ vL –
iL
iC
Iout
L
Vin
C
iC
+
Vout
–
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Boost converter
iin
+ vL –
iL
iD
Iout
L
Vin
C
iC
+
Vout
–
This is a much more unforgiving circuit than the buck converter
• If the MOSFET gate driver sticks in the “on” position, then there
is a short circuit through the MOSFET – blow MOSFET!
• If the load is disconnected during operation, so that Iout = 0, then
L continues to push power to the right and very quickly charges
C up to a high value (250V) – blow diode and MOSFET!
• Before applying power, make sure that your D is at the
minimum, and that a load is solidly connected
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Boost converter
iin
+ vL –
iL
iD
Iout
L
Vin
C
iC
+
Vout
–
• Modify your MOSFET firing circuit for Boost Converter
operation (see the MOSFET Firing Circuit document)
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Boost converter
iin
+ vL –
iL
iD
Iout
L
Vin
C
iC
+
Vout
–
Using KVL and KCL in the average sense, the average
values are
Iin
+0V–
Iout
L
Vin
C
Iout
+
Vout
0A
–
Find the input/output equation by examining the voltage
across the inductor
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Switch closed for DT seconds
iin
+ Vin −
iL
Iout
L
Vin
C
diL Vin

dt
L
Iout
+
Vout
–
Reverse biased, thus the
diode is open
for DT
seconds
Note – if the switch stays closed, the input is short circuited!
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Switch open for (1 − D)T seconds
+ (Vin − Vout ) −
iL
iin
Iout
L
Vin
C
diL Vin  Vout

dt
L
+
Vout
(iL – Iout)
–
Diode closed. Assume
continuous conduction.
for (1−D)T seconds
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Since the average voltage across L is zero
VLavg  D Vin  1  D  Vin  Vout   0
Vout  (1  D)  Vin  D  Vin  D  Vin
The input/output equation becomes
Vin
Vout 
1 D
A realistic upper limit on boost is 5 times
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Examine the inductor current
Switch closed,
diL Vin
v L  Vin ,

dt
L
Switch open,
vL  Vin  Vout ,
Vin  Vout
A / sec
L
iL
Imax
Iavg = Iin
Vin
A / sec
L
Imin
DT
diL Vin  Vout

dt
L
Iavg = Iin is half way between
Imax and Imin
ΔI
(1 − D)T
T
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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I in 2  4 I in2
12
3
2
I in
3
Use max
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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
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Worst-case load ripple voltage
iC = (iD – Iout)
0
−Iout
The worst case is where C provides Iout for most of the period. Then,
Q I out  T I out
V 


C
C
Cf
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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
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Continuous current in L
Vin  Vout
A / sec
L
iL
2Iin
Iavg = Iin
0
(1 − D)T
 1

Vin
V

1

1  D 
 Vin
in
Vout  Vin
1

D


2 I 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
use min
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Impedance matching
I out  1  D Iin
Iin
+
+
Source
DC−DC Boost V  Vin
out
1 D
Converter
−
Vin
−
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
1 D
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Example of drawing maximum power from solar
panel
PV Station 13, Bright Sun, Dec. 6, 2002
6
Isc
Pmax is approx. 130W
(occurs at 29V, 4.5A)
5
I - amps
4
For max power from
panels, attach
3
Rload 
2
1
0
0
5
10
15
20
25
V(panel) - volts
30
35
40
Voc
I-V characteristic of 6.44Ω resistor
45
29V
 6.44
4.5 A
But as the sun conditions
change, the “max power
resistance” must also
change
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Connect a 100Ω resistor directly, extract only 14W
PV Station 13, Bright Sun, Dec. 6, 2002
6
130W
5
4
I - amps
So, the boost converter
reflects a high load
resistance to a low
resistance on the
source side
3
2
14W
1
0
0
5
10
15
20
25
30
35
40
45
V(panel) - volts
To extract maximum power (130W), connect a boost converter between the
panel and the load resistor, and use D to modify the equivalent load
resistance seen by the source so that maximum power is transferred
Requiv  1  D  Rload , D  1 
2
Requiv
R
 1
6.44
 0.75
100
load , Assiut University
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44
BOOST DESIGN
Worst-Case Component Ratings Comparisons
for DC-DC Converters
Converter
Type
Boost
Input Inductor
Current
(Arms)
2
I in
3
10A
Output
Capacitor
Voltage
Output Capacitor
Current (Arms)
1.5 Vout
I out
Diode and
MOSFET
Voltage
2 Vout
120V
5A
120V
Likely worst-case boost situation
Diode and
MOSFET
Current
(Arms)
2
I in
3
10A
L. 100µH, 9A
C. 1500µF, 250V, 5.66A p-p
Diode. 200V, 16A
MOSFET. 250V, 20A
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BOOST DESIGN
Minimum Inductance Values Needed to
Guarantee Continuous Current
Converter Type
Boost
For Continuous
Current in the Input
Inductor
V
40V
L  in
2 I in f
200µH
2A
50kHz
L. 100µH, 9A
C. 1500µF, 250V, 5.66A p-p
Diode. 200V, 16A
MOSFET. 250V, 20A
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Questions?
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