Switching Supplies II
Transcript of Switching Supplies II
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Switching Supplies IIBoost Converters and Other Things
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β’ Takes an input voltage and produces a loweroutput voltage
Buck Converter (from Tuesday)
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πͺ πΉπ³+-ππΊ
+
-
ππΆπΌπ»
π³
ππΊ wasnβt actually changingβ¦it was a constant and there were some switchesβ¦
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β’ Phase 1: When π!is closed, π" is open
β’ Current through the inductor builds up!
πͺ πΉπ³+-π½π―π°
+
-
ππΆπΌπ»
π³πΊπ
πΊπ
ππ³ π = *ππ°π΅ β ππΆπΌπ»
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β’ Phase 2: When π!is open, π" is closed
β’ Current through the inductor starts to decay away (but stays positive since it started positive
πͺ πΉπ³+-π½π―π°
+
-
ππΆπΌπ»
π³πΊπ
πΊπ
ππ³ π = *π β ππΆπΌπ» + ππ³ ππππ
Initial condition from when switches flipped
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Motivationβ’ Transferring power over grid is best done at high
voltages.β’ Need a way to efficiently down-convert for local
electronics (which work at low voltages)β’ Buck converters allow us to do that efficiently!
β’ Mobile devices rely on batteries to store energyβ’ Batteries often produce their energy using lower
voltages (few volts usually, Lipo Battery is at 3.7V nominally)β’ Some electronics need voltages higher than 3.7V:β’ Backlight, EEPROM (Flash), many sensors (gyroscopes)
β’ Need a way to convert voltages up, and ideally do so efficientlyβ’ Boost converters are one way to do this
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Bug zapper
DC-DC boost convertersβ’ Sometimes you want to
create a large DC voltage from a small DC voltage
Camera flash (Xenon bulb)
Camera flash (Xenon bulb)
Stun gun
Defibrillator
~3V to 100βs of Volts!
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Smart phone
~3.7V to 20V!
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πͺ πΉπ³+-ππ°π΅
+
-
ππΆπΌπ»
π³
LoadDC Supply
πΊπ
πΊπ
β’ The diode and the FET act like switches just like before in the Buck Converter!
Boost Converter
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Drive with a Periodic Signal β’ Variable Duty Cycle!! (weβve seen this before)
β’ In the Buck converter, your output voltage would be (approx): "#!"
$
β’ With a Boost Converter it is Different
π«π» π» πππ»π» + π«π»
ππππππππ
ππ» + π«π»
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³
LoadDC Supply
πΊπ
πΊπ
β’ π! closed, π" open.
Stage 1
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Stage 2β’ π! open, π" closed.
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³
LoadDC Supply
πΊπ
πΊπ
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Stage 3β’ π! open, π" open.
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³
LoadDC Supply
πΊπ
πΊπ
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³ πΊπ
πΊπ
β’ π!is closed, π" is open!β’ Current builds up through the inductor:β’ Since ππ°π΅ is constant:
β’ Therefore energy is being built up and stored in the inductor!
π3 π‘ =1πΏ*π£45 ππ‘
ππ³ π
ππ³ π = ππ°π΅ 9 π
Stage 1
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During Stage 1β’ The Capacitor and Load Resistor are Isolated, forming
their own cozy RC circuit away from the whole worldβ¦voltage on cap just decay
β’ The voltage source and inductor are isolated, building up energy
π·π0
π£6
π‘π·π0
π3
π‘
π3 t = π£45π‘
ππ
π£6 t = π£6 0 π78/:!6
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³ πΊπ
πΊπ
β’ π! opens and π" closes at some point in time t=DTβ’ Right before switch1:β’ Inductors donβt want much in life, but they do want π% π‘ to be continuous
π3 π·π7 =1πΏ π£45 9 π·π
ππ³ π
Stage 2
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Stage 2 Math:π3 π‘ β π6 π‘ + π: π‘
Where does the current go?
π3 π‘ = π3 π·π +1πΏ*;<
<π£45 β π£= π ππ
Current built up at end of stage 1 Current change during stage 2
Either is fine. R consumes power, C stores it for consumption later by R. So how much π3 moves over during Stage 2???
β’ If π£A is small/lower than π£BC during Stage 2, πD will stay positive and grow!β’ Much of that current dumps into the capacitor, however, and because π£A π‘ = E
A β« πFππ‘ that cap voltage will build up!
β’ If π£A is larger than π£BC during Stage 2, πD will potentially drop to 0 even if it leaves Stage 1 positive!
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Stage 3
πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³ πΊπ
πΊπ
β’ As long as π% π‘ is positive, π" stays closed letting current (energy) move to the right half of the circuit, but when π% π‘ gets low enough (negative), π" will open, locking that sweet energy in.β’ Whatever voltage/energy is remaining on the capacitor
can now only be dissipated by the load resistance (nothing can flow back to the inductor)!!
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Stage 2 Math:π3 π‘ = π6 π‘ + π: π‘
π!πΏ β
π£" π‘πΏ = πΆ
π#π£"(π‘)ππ‘# +
1π $ππ£"(π‘)ππ‘
How does the current go?
π>πΏπΆ =
π?π£=(π‘)ππ‘? +
1π 3πΆ
ππ£=(π‘)ππ‘ +
1πΏπΆ π£= π‘
2πΌ π@?Constant
π£= π‘ = π΄Aπ7B8 cos πCπ‘ β π΄D
Homogenous response will therefore be one of three forms depending on what πΌand π@ are!
πΌ < π@ πΌ = π@ πΌ > π@
π£= π‘ = π΄Aπ7E"8 + π΄Dπ‘π7E#8 π£= π‘ = π΄Aπ7E"8 + π΄Dπ7E#8
Initial Conditions will then specify the constants in the solutions
π$ π·π +π! π‘ β π·π
πΏ β1πΏ3%&
'π£" π ππ = πΆ
ππ£"(π‘)ππ‘ +
π£"(π‘)π $
d/dt
π$ = π%& β πΌ&
π3 π·π +1πΏ*;<
8π> β π£= π ππ = πΆ
ππ£=(π‘)ππ‘ +
π£=(π‘)π 3
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During Stage 2β’ The inductor dumps its
energy into the cap:β’ Inductor current dropsβ’ Capacitor voltage builds
β’ How do they change?β’ Depends on the circuit
coefficients πΌ and π&
π·π0
π£6
π‘
π£6 t = ππ¦ππππππ
π·π0
π3
π‘
π3 t = ππ¦ππππππ
π
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If circuit is extremely underdamped (assume no load resistance)(πΌ << π!)
ππ(π) = π½πΊ + ππͺ π β π½πΊ π +π³πͺ ππ³ π
π ππ¨π¬ πππ β ππππ§π³πͺ
ππ³ πππͺ π β π½πΊ
ππ³ π = βπͺπ³ ππͺ π β π½πΊ π +
π³πͺ ππ³ π
π π¬π’π§ πππ β ππππ§π³πͺ
ππ³ πππͺ π β π½πΊ
πCos + phase form
ππ π = π¨ππ7πΆπ ππ¨π¬ ππ π β π¨π
ππ π = π¨π ππ¨π¬ πππ β π¨π
πΌ tiny so π7πΆπ disappears and ππ β ππ
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If πΌ < π! for exampleβ¦β’ The inductor dumps its
energy into the cap:β’ Inductor current dropsβ’ Capacitor voltage builds
β’ How do they change?β’ Depends on the circuit
coefficients πΌ and π&
π·π0
π£6
π‘
π£6 t = ππ¦ππππππ from the partial arc of a sinusoid
π·π0
π3
π‘
π3 t = ππ¦ππππππ from the partial arc of a sinusoid
π
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Cyclic Operation
π·π π+π·π0
π3
π‘π 2π
π·π π+π·π0
π£6
π‘π 2π
higherhigher
higher
Cap starts with no voltage across it and that voltage gradually builds up!
higherhigher
2π+π·π
3π
Quasi-qualitative plots of operation:
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Eventually decay from load will eventually cancel gains from inductor current injection! And thatβs ok
http://ctms.engin.umich.edu/CTMS/index.php?aux=Activities_BoostcircuitA
The system will reach a steady-state operation
Just like in the PWM circuits!
What will that steady-state voltage be?
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Let's assume that our Boost converter reaches some sort of steady-state, meaning that at the beginning and end of a period (T), the voltage across the capacitor and the current through the inductor are the same.
During Stage 1 (when the inductor is "charging" up), it is in a circuit that looks like this:
vIN is a constant, so at the end of the Stage 1 (at t = DT):
In Stage 2, the time-domain expression for iL is now based on what the capacitor voltage is!
If we want the finishing current to be the same as the starting current (assuming the system is in a stable, cyclic behavior), then iL(T) = iL(0)!
If we assume that the capacitor is large enough to not vary much we can treat vc(t) as a constant, making the integration easy.
O IT KE FDI
I I D ve Us I
nT
Left with something pretty simple:
D varies from 0 to 1...so this means our capacitor's average voltage (which is our output!) will vary from 0 to infinity!!!
In reality other things, non-idealities get in the way, but under certain circumstances, this expression holds true...by varying our duty cycle we can get higher output voltages than what we put in!!!
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Cyclic Operation
π·π π+π·π0
π3
π‘π 2π
π·π π+π·π0
π£6
π‘π 2π
Build Up Energy
Consume Stored Energy
Dump stored Energy
Build upStored Energy
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Cyclic Operation
0
π3
π‘
π·π
π+π·π0
π£6
π‘π 2π
β’ Depending on π·, π, πΌand π& (π % πΏ, πΆ) you can get different behaviors!β’ π β π·π shorter?
Current might not drop all the way to 0?β’ Current will eventually
reach cycle-to-cycle steady state average value
π·π π+π·π
π 2π
π£45
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Cyclic Operation
π·π π+π·π0
π3
π‘π 2π
π·π π+π·π0
π£6
π‘π 2π
β’ Depending on π·, π, πΌand π& (π % πΏ, πΆ) you can get different behaviors!β’ More damping?
(power-hungry load, aka π % is low)β¦β’ Current will drop quick
and may hit 0
π£45
S2 opens to prevent current
going negative
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Cyclic Operationπ·π π+π·π0
π3
π‘π 2π
π·π π+π·π0
π£6
π‘π 2π
β’ Depending on π·, π, πΌand π& (π % πΏ, πΆ) you can get different behaviors!β’ Waaay too much
damping? (power-hungry load, aka π % is low)β¦
π£45
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Booost converter
10k
190k
Under the Hood of a DC/DC Boost Converter by Brian T. Lynch included as supplemental fun reading
Costs a few dollars!
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MIT Ion planeβ’ aka flying boost converterβ’ The key
β’ 200 V Γ 40 kVβ’ Low mass: 1.2 kW/kg
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³
LoadDC Supply
πΊπ
πΊπ
β’ What were those switches?
Boost Converter
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πͺ πΉπ³+-π½πΊ
+
-
ππΆπΌπ»
π³
LoadDC Supply
ππππππππ
β’ Top switch is a Diodeβ’ Control Switch is a Transistor
FET
diode
Boost Converter
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What about those βmagicβ parts?
The so-called βdiodeβ and βtransistorβ that were doing all this switching? How are they working?
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Transistorβ’ Go into them next weekβ’ Three terminal device:β’ Middle terminal enables/disables
connectivity between two outside onesβ’ Like an electrical modulated resistor (can
use one signal to toggle between low and high resistance between other two terminals
FET
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Diode
β’ Historically it was the first Non-Linear device we as humans developed and worked with (whether it was known at the time or not)β’ What does it meant to be non-linear?
π
π
π
π+ β
π = πΌg πhijN β 1
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Diode Can Act Like a Switchβ’ Voltage in one way: Conduct (act like a wire)β’ Voltage in the other way: Donβt conduct (act like an open)
https://www.electronics-tutorials.ws/diode/diode_5.html
π
π
Simpler model of the system:
π
π+ β
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Diode Can Act Like a Switchβ’ From current perspective:β’ Current Positive? Let it through (be a short)β’ Current Negative? Block it! (be an open)
β’ From voltage perspective:β’ Voltage across Positive? Be a shortβ’ Voltage across negative? Be an open
https://www.electronics-tutorials.ws/diode/diode_5.html
π
πSimpler model of the system:
π
π+ β
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Diodes are One Way Electrical Valvesβ’ Use them all the time in Power Conversion, Signal
Processing, Logic, Computation!
https://www.electronics-tutorials.ws/diode/diode_5.html11/16/21 36
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