EE462L, Power Electronics, DC-DC Boost Converter Version ... - ECS

EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
Overview
Boost converters make it possible to efficiently convert a DC voltage from a lower level to a
higher level.
Theory of Operation
Relation Between V out and V in in Continuous Conduction
The idealized boost converter circuit is shown below in Figure 1. Under normal operation, the
circuit is in “continuous conduction” (i.e., i L is never zero)
i
I L i d I out
in
Variac
120/25Vac
Transformer
DBR
Remember – never connect a variac directly
toaDBR!
V in
Figure 1. DC-DC Boost Converter
L
i C
Leave the 10µF cap
across the input terminals
C
0.01Ω
+
V out
–
The circuit is assumed to be lossless so that P in = P out , or
V in • iLavg
= Vout
• Iout
, where Lavg Iin
i = . (1)
Assuming continuous conduction, the circuit has two topologies – switch closed, and switch
open. Both are shown in Figures 2a and 2b.
+ v L
−
+ v L
−
i L
I out
i L
I out
V in
L
C
i C
+
V out
–
V in
L
C
i C
+
V out
–
Figure 2a. Switch Closed for DT
Seconds
Figure 2b. Switch Open for (1-D)T
Seconds (Continuous Conduction)
When the switch is closed, the diode is reverse biased and open, and i L increases at the rate of
diL v V
= L =
in
, 0 ≤ t ≤ DT , (2)
dt L L
and the inductor is “charging.” When the switch is open, the diode is forward biased, and i L
decreases at the rate of
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EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
diL
dt
vL
Vin
−Vout
= = , DT < t < T , (3)
L L
and the inductor is “discharging.” The inductor voltage is shown in Figure 3.
V in
0
Vin − V out
Figure 3. Inductor Voltage in Continuous Conduction
Because of the steady-state inductor principle, the average voltage v L across L is zero. Since v L
has two states, both having constant voltage, the average value is
( V )
in DT + ( Vin
−Vout
T
)(1 − D)
T
= 0 ,
so that
Vin D + Vin
−Vout
−VinD
+ Vout
D = 0 .
Simplifying the above yields the final input-output voltage expression
Vout
Vin
= . (4)
1 − D
The graph of i L is shown in Figure 4.
Inductor Current in Continuous Conduction
Δ I
T
i
L max
i Lavg
i
L min
= i
= i
Lavg
Lavg
ΔI
+
2
ΔI
−
2
DT
(1–D)T
Figure 4. Inductor Current Waveform for Continuous Conduction
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EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
From (2),
diL
dt
Vin
ΔI
= = ,
L DT
so that
Vin
VinD
Δ I = • DT = , (5)
L Lf
where f is the switching frequency.
The boundary of continuous conduction is when
T
i L min = 0, as shown in Figure 5.
Δ I
iL max = 2I Lavg
i Lavg
i
L min = 0
DT
(1–D)T
Figure 5. Inductor Current at the Boundary of Continuous Conduction
Using Figure 5 and the “inductor discharging” slope from (3),
so that
( V −V
)( 1−
D) V −V
( 1−
D)
out in
in in
VinD
Δ I =
=
=
= 2I
Lavg , (6)
L f L f L f
boundary
boundary
boundary
L
boundary
VinD
= . (7)
2I
f
Lavg
From (1),
I Lavg = I in . (8)
Substituting into (8) into (7) yields
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EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
Lboundary
VinD
= . (9)
2Iin
f
Because the maximum value of D is 1, then
Vin
L > . (10)
2Iin
f
will guarantee continuous conduction for all D. Note in (9) and (10) that continuous conduction
can be achieved more easily when I out and f are large.
Discontinuous Conduction
At low load, the converter may slip into the discontinuous conduction mode. Referring back to
Figure 2b, this occurs when the inductor current coasts to zero. At that moment, the capacitor
attempts to reverse i L and “backfeed” the inductor, but the diode prevents current reversal. Thus,
the diode opens, and the circuit assumes the topology shown in Figure 6 until the switch closes
again. During this third state, all load power is provided by the capacitor.
+ 0 –
0
I out
V in
L
C
+
V out
–
Figure 6. Third State for Discontinuous Conduction
Once discontinuous, the voltage across the inductor is zero. The corresponding voltage
waveform is shown in Figure 7.
V in
Discontinuous
Vin − V out
0
Figure 7. Inductor Voltage in Discontinuous Conduction
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EE462L, Power Electronics, DC-DC Boost Converter
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Current Waveforms in Continuous Conduction
Continuous current waveforms are shown below.
Δ I
T
Inductor
i
= i
L max Lavg
i Lavg = I in
i
L min
= i
Lavg
ΔI
+
2
ΔI
−
2
DT
(1–D)T
Δ I
ΔI
MOSFET
iL
max = iLavg
+
2
i Lavg = I in
ΔI
iL
min = iLavg
−
2
Δ I
0
ΔI
i
Diode
L max = iLavg
+
2
i Lavg = I in
ΔI
iL
min = iLavg
−
2
0
Capacitor ( i
C
= i − I )
L
out
ΔI
iLavg − I out
+ 2
out
Δ I
0
iLavg
− Iout
= Iin
− I
ΔI
iLavg − − I out 2
− I out
(Note – compared to the other waveforms shown above, Δ I is exaggerated in the figure to
illustrate how the capacitor current can be negative in both DT and (1–D)T regions)
Figure 8. Current Waveforms for Continuous Conduction
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EE462L, Power Electronics, DC-DC Boost Converter
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Current Ratings in Continuous Conduction
Conservative current ratings for the inductor in continuous conduction correspond to the
situation where
Δ Imax = 2I in , (11)
which, as explained in the Buck Converter experiment, yields
1 2
2
⎛ ⎞
=
2 +
2 1
I Lrms, max Iin
( 2Iin
) = Iin
⎜1
+ ⎟ , (12)
12
⎝ 3 ⎠
so that
2
I Lrms , max = I in . (13)
3
Conservative current ratings for the MOSFET and diode are when D is large, so that (13) applies
for them also.
To determine the rms current through C, consider the capacitor current in Figure 8, and the
worst-case scenario in Figure 9.
2 I
−
in I out
0
Iin − I out
DT
(1–D)T
Figure 9. Maximum Ripple Current Case for Capacitor Current
When the switch is closed, the capacitor current is − Iout
. When the switch is open, the
capacitor current is iL
− Iout
. If the “switch closed” interval lasted for the entire T, the squared
2
rms value would be I out . If the “switch open” interval lasted for the entire T, the rms value
would be, for the maximum ripple case, 2 1
( I in − Iout
) + ( 2Iin
) 2 . The time-weighted average
12
of the two gives the squared rms current
−
I out
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EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
2 2
⎛
2 1 2 ⎞
I Crms = DIout
+ ( 1−
D)
• ⎜(
Iin
− Iout
) + ( 2Iin
) ⎟ . (14)
⎝
12 ⎠
Now, substituting in I = I ( 1−
D)
yields
out
in
2 2 2 ⎛
2 1 2 ⎞
I Crms = DIin
( 1−
D)
+ (1 − D)
• ⎜( Iin
− Iin
(1 − D)
) + ( 2Iin
) ⎟ .
⎝
12 ⎠
Simplifying yields
2 2 2 ⎛ 2 2 1 2 ⎞
I Crms = DIin
( 1−
D)
+ (1 − D)
• ⎜ IinD
+ Iin
⎟ ,
⎝ 3 ⎠
I
I
⎛
⎜ D(1
− D)
⎝
(1 −
+
3
) ⎞
⎟
⎠
2 2
2
2 D
Crms = Iin
+ (1 − D)
D
,
⎛ ⎞
⎜ − D + 2 + 1⎟
⎜
⎟
⎝
3
⎠
2 2 3 2 D
Crms = Iin
,
Setting the partial derivative with respect to D shows that the maximum occurs at
yields
2
I Crms , max =
3 I in .
1
D = , which
3
Since D = , then substituting for I in yields
2 I
I =
3 ⎞
⎜
⎛ 1
1 − ⎟
⎝ 3 ⎠
out
Crms, max = • Iout
. (15)
Voltage Ratings for Continuous Conduction
Referring to Figure 2b, when the MOSFET is open, it is subjected to Vout. Because of the usual
double-voltage switching transients, the MOSFET should therefore be rated 2Vout.
Referring to Figure 2a, when the MOSFET is closed, the diode is subjected to Vout . The diode
should be conservatively rated 2Vout .
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EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
Capacitor Voltage Ripple
Re-examining the capacitor current in Figure 9, and re-illustrated in Figure 10, it can be seen that
the amount of charge taken from C when the switch is closed is represented by the dotted area.
2 I
−
in I out
0
Iin − I out
−
I out
DT (1–D)T ΔQ
Figure 10. Capacitor Charge Given Up While Switch is Closed
As D → 1, the width of the dotted area increases to fill almost the entire cycle, and the
maximum peak-to-peak ripple becomes
Iout
• T Iout
ΔV
max = = . (16)
C Cf
The Experiment (Important - to avoid high output voltages, always keep a load attached to
the boost converter output when input power is applied. Use a conventional 120V, 150W
light bulb as your load. Do not exceed 120V on the output.
1. Convert a buck converter to a boost converter, using the circuit shown in Figure 1 of this
document.
2. Double-check that the polarity of your converter’s output capacitor is correct.
3. Locate one of the 150W light bulb test load assemblies. Check the light bulb with an
ohmmeter to make sure it is not burned out.
4. Connect the light bulb test load to your circuit.
5. Connect an oscilloscope Channel #1 to view V GS , and Channel #2 to view V DS . The
ground clip of the Channel #2 probe should not be attached to the circuit, but instead it
should be clipped back onto its own lead in-cable so that it does not dangle.
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EE462L, Power Electronics, DC-DC Boost Converter
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6. Do not connect a DBR yet. Connect a MOSFET firing circuit to your converter, using
short wires, and then power-up your MOSFET firing circuit. Set the oscilloscope to
trigger on Channel #1. Observe your oscilloscope to confirm that the controls are
working properly.
7. Raise the D potentiometer to the max, and then set the duty cycle limiter so D cannot
exceed 0.8.
8. Set D to the minimum setting, and F ≈ 100kHz.
Important Note: the first time you energize your boost converter, feed the 120/25V
transformer through a variac to the DBR, so that you can SLOWLY increase the voltage
from zero and read the variac ammeter to detect short circuits before they become serious.
A common problem is to have the MOSFET in backward, so that its internal antiparallel diode
creates a short circuit. The ammeter on the variac is an excellent diagnostic tool. Once you are
convinced that your circuit is working correctly, the variac is then optional. Remember – your
boost converter requires DC input power from a DBR.
Does your circuit have a short If so, do the following:
1. Make sure that your MOSFET is not connected backwards.
2. Observe VGS on the MOSFET as you vary D. Does the waveform look correct
3. Unplug the wall wart. Does the short circuit go away If not, your MOSFET may be
shorted – so, disconnect the MOSFET from the converter, and perform the voltagecontrolled
resistance test on the MOSFET, or use the MOSFET tester.
9. Connect a variac to a 25Vac transformer, and 25Vac transformer to a DBR. Connect the
DBR to your boost converter, keeping the wires short. Then, energize the 25Vac
transformer and DBR. Slowly raise the variac, checking for a short circuit, so that Vac of
the transformer is approximately 27-28V.
10. With F ≈ 100kHz, slowly raise D so that Vout increases in steps of about 10V up to the
maximum obtained with D = 0.8 or 120V, whichever is less. Measure D, Vin, Iin, Vout,
and Iout as you go. Save a screen snapshot of VDS for Vout = 100V that shows the peak
value of VDS. Let the circuit run at the 100Vout condition for 1 or 2 more minutes, and
then turn off your circuit.
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EE462L, Power Electronics, DC-DC Boost Converter
Version Oct. 3, 2011
Save screen
snapshot #1
VDS for 90kHz, 120V
11. With your circuit turned off, quickly and carefully use your hand to check MOSFET
heat sink temperature and any other hot components. You can also use an infrared
thermometer for this step.
12. Compare your measured Vout/Vin to theory. Multiply voltages and currents to compute
input and output powers, and then compute the efficiency of your circuit for each Vout
measured. Plot your results in your written report.
13. Take can also take your solar lab data while you are doing the boost converter. Use a
150W light bulb as a load. Connect a DBR between a solar panel pair and your boost
converter to steady the panel current. Sweep D to find the maximum power point.
Compute the maximum power point value of Rload = Vout/Iout and the corresponding
equivalent max power resistance seen by the solar panels. Does the ratio of Rload
resistance to equivalent resistance seem by the solar panels check with D
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EE462L, Power Electronics, DC-DC Boost Converter
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