11/18/21

1

Switching Supplies IIBoost Converters and Other Things

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1

• Takes an input voltage and produces a loweroutput voltage

Buck Converter (from Tuesday)

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𝑪 𝑹𝑳+-𝒗𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳

𝒗𝑺 wasn’t actually changing…it was a constant and there were some switches…

2

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2

• Phase 1: When 𝑆!is closed, 𝑆" is open

• Current through the inductor builds up!

𝑪 𝑹𝑳+-𝑽𝑯𝑰

+

-

𝒗𝑶𝑼𝑻

𝑳𝑺𝟏

𝑺𝟐

𝒊𝑳 𝒕 = *𝒗𝑰𝑵 − 𝒗𝑶𝑼𝑻

3

• Phase 2: When 𝑆!is open, 𝑆" is closed

• Current through the inductor starts to decay away (but stays positive since it started positive

𝑪 𝑹𝑳+-𝑽𝑯𝑰

+

-

𝒗𝑶𝑼𝑻

𝑳𝑺𝟏

𝑺𝟐

𝒊𝑳 𝒕 = *𝟎 − 𝒗𝑶𝑼𝑻 + 𝒊𝑳 𝒕𝒔𝒘𝟏

Initial condition from when switches flipped

4

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Motivation• Transferring power over grid is best done at high

voltages.• Need a way to efficiently down-convert for local

electronics (which work at low voltages)• Buck converters allow us to do that efficiently!

• Mobile devices rely on batteries to store energy• Batteries often produce their energy using lower

voltages (few volts usually, Lipo Battery is at 3.7V nominally)• Some electronics need voltages higher than 3.7V:• Backlight, EEPROM (Flash), many sensors (gyroscopes)

• Need a way to convert voltages up, and ideally do so efficiently• Boost converters are one way to do this

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5

Bug zapper

DC-DC boost converters• 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 100’s of Volts!

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Smart phone

~3.7V to 20V!

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𝑪 𝑹𝑳+-𝒗𝑰𝑵

+

-

𝒗𝑶𝑼𝑻

𝑳

LoadDC Supply

𝑺𝟐

𝑺𝟏

• 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 • Variable Duty Cycle!! (we’ve seen this before)

• In the Buck converter, your output voltage would be (approx): "#!"

$

• With a Boost Converter it is Different

𝑫𝑻 𝑻 𝒕𝟐𝑻𝑻 + 𝑫𝑻

𝒗𝒄𝒐𝒏𝒕𝒓𝒐𝒍

𝟐𝑻 + 𝑫𝑻

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5

𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳

LoadDC Supply

𝑺𝟐

𝑺𝟏

• 𝑆! closed, 𝑆" open.

Stage 1

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Stage 2• 𝑆! open, 𝑆" closed.

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𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳

LoadDC Supply

𝑺𝟐

𝑺𝟏

10

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Stage 3• 𝑆! open, 𝑆" open.

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𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳

LoadDC Supply

𝑺𝟐

𝑺𝟏

11

𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳 𝑺𝟐

𝑺𝟏

• 𝑆!is closed, 𝑆" is open!• Current builds up through the inductor:• Since 𝒗𝑰𝑵 is constant:

• Therefore energy is being built up and stored in the inductor!

𝑖3 𝑡 =1𝐿*𝑣45 𝑑𝑡

𝒊𝑳 𝒕

𝒊𝑳 𝒕 = 𝒗𝑰𝑵 9 𝒕

Stage 1

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During Stage 1• The Capacitor and Load Resistor are Isolated, forming

their own cozy RC circuit away from the whole world…voltage on cap just decay

• The voltage source and inductor are isolated, building up energy

𝐷𝑇0

𝑣6

𝑡𝐷𝑇0

𝑖3

𝑡

𝑖3 t = 𝑣45𝑡

𝑇𝑇

𝑣6 t = 𝑣6 0 𝑒78/:!6

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𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳 𝑺𝟐

𝑺𝟏

• 𝑆! opens and 𝑆" closes at some point in time t=DT• Right before switch1:• Inductors don’t want much in life, but they do want 𝑖% 𝑡 to be continuous

𝑖3 𝐷𝑇7 =1𝐿 𝑣45 9 𝐷𝑇

𝒊𝑳 𝒕

Stage 2

1 Assuming current started at 0…otherwise this is the delta current built up during period11/16/21 14

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Stage 2 Math:𝑖3 𝑡 → 𝑖6 𝑡 + 𝑖: 𝑡

Where does the current go?

𝑖3 𝑡 = 𝑖3 𝐷𝑇 +1𝐿*;<

<𝑣45 − 𝑣= 𝜏 𝑑𝜏

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 𝑖3 moves over during Stage 2???

• If 𝑣A is small/lower than 𝑣BC during Stage 2, 𝑖D will stay positive and grow!• Much of that current dumps into the capacitor, however, and because 𝑣A 𝑡 = E

A ∫ 𝑖F𝑑𝑡 that cap voltage will build up!

• If 𝑣A is larger than 𝑣BC during Stage 2, 𝑖D will potentially drop to 0 even if it leaves Stage 1 positive!

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Stage 3

𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳 𝑺𝟐

𝑺𝟏

• As long as 𝑖% 𝑡 is positive, 𝑆" stays closed letting current (energy) move to the right half of the circuit, but when 𝑖% 𝑡 gets low enough (negative), 𝑆" will open, locking that sweet energy in.• 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:𝑖3 𝑡 = 𝑖6 𝑡 + 𝑖: 𝑡

𝑉!𝐿 −

𝑣" 𝑡𝐿 = 𝐶

𝑑#𝑣"(𝑡)𝑑𝑡# +

1𝑅$𝑑𝑣"(𝑡)𝑑𝑡

How does the current go?

𝑉>𝐿𝐶 =

𝑑?𝑣=(𝑡)𝑑𝑡? +

1𝑅3𝐶

𝑑𝑣=(𝑡)𝑑𝑡 +

1𝐿𝐶 𝑣= 𝑡

2𝛼 𝜔@?Constant

𝑣= 𝑡 = 𝐴A𝑒7B8 cos 𝜔C𝑡 − 𝐴D

Homogenous response will therefore be one of three forms depending on what 𝛼and 𝜔@ are!

𝛼 < 𝜔@ 𝛼 = 𝜔@ 𝛼 > 𝜔@

𝑣= 𝑡 = 𝐴A𝑒7E"8 + 𝐴D𝑡𝑒7E#8 𝑣= 𝑡 = 𝐴A𝑒7E"8 + 𝐴D𝑒7E#8

Initial Conditions will then specify the constants in the solutions

𝑖$ 𝐷𝑇 +𝑉! 𝑡 − 𝐷𝑇

𝐿 −1𝐿3%&

'𝑣" 𝑠 𝑑𝑠 = 𝐶

𝑑𝑣"(𝑡)𝑑𝑡 +

𝑣"(𝑡)𝑅$

d/dt

𝜔$ = 𝜔%& − 𝛼&

𝑖3 𝐷𝑇 +1𝐿*;<

8𝑉> − 𝑣= 𝑠 𝑑𝑠 = 𝐶

𝑑𝑣=(𝑡)𝑑𝑡 +

𝑣=(𝑡)𝑅3

17

During Stage 2• The inductor dumps its

energy into the cap:• Inductor current drops• Capacitor voltage builds

• How do they change?• Depends on the circuit

coefficients 𝛼 and 𝜔&

𝐷𝑇0

𝑣6

𝑡

𝑣6 t = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐𝑠

𝐷𝑇0

𝑖3

𝑡

𝑖3 t = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐𝑠

𝑇

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If circuit is extremely underdamped (assume no load resistance)(𝛼 << 𝜔!)

𝒗𝒄(𝒕) = 𝑽𝑺 + 𝒗𝑪 𝟎 − 𝑽𝑺 𝟐 +𝑳𝑪 𝒊𝑳 𝟎

𝟐 𝐜𝐨𝐬 𝝎𝒐𝒕 − 𝐚𝐭𝐚𝐧𝑳𝑪

𝒊𝑳 𝟎𝒗𝑪 𝟎 − 𝑽𝑺

𝒊𝑳 𝒕 = −𝑪𝑳 𝒗𝑪 𝟎 − 𝑽𝑺 𝟐 +

𝑳𝑪 𝒊𝑳 𝟎

𝟐 𝐬𝐢𝐧 𝝎𝒐𝒕 − 𝐚𝐭𝐚𝐧𝑳𝑪

𝒊𝑳 𝟎𝒗𝑪 𝟎 − 𝑽𝑺

𝝓Cos + phase form

𝒗𝒄 𝒕 = 𝑨𝟎𝒆7𝜶𝒕 𝐜𝐨𝐬 𝝎𝒅𝒕 − 𝑨𝟏

𝒗𝒄 𝒕 = 𝑨𝟎 𝐜𝐨𝐬 𝝎𝒐𝒕 − 𝑨𝟏

𝛼 tiny so 𝒆7𝜶𝒕 disappears and 𝝎𝒅 ≈ 𝝎𝒐

Initial Conditions yield these terms!11/16/21 19

19

If 𝛼 < 𝜔! for example…• The inductor dumps its

energy into the cap:• Inductor current drops• Capacitor voltage builds

• How do they change?• Depends on the circuit

coefficients 𝛼 and 𝜔&

𝐷𝑇0

𝑣6

𝑡

𝑣6 t = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐𝑠 from the partial arc of a sinusoid

𝐷𝑇0

𝑖3

𝑡

𝑖3 t = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐𝑠 from the partial arc of a sinusoid

𝑇

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20

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Cyclic Operation

𝐷𝑇 𝑇+𝐷𝑇0

𝑖3

𝑡𝑇 2𝑇

𝐷𝑇 𝑇+𝐷𝑇0

𝑣6

𝑡𝑇 2𝑇

higherhigher

higher

Cap starts with no voltage across it and that voltage gradually builds up!

higherhigher

2𝑇+𝐷𝑇

3𝑇

Quasi-qualitative plots of operation:

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Eventually decay from load will eventually cancel gains from inductor current injection! And that’s 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

𝐷𝑇 𝑇+𝐷𝑇0

𝑖3

𝑡𝑇 2𝑇

𝐷𝑇 𝑇+𝐷𝑇0

𝑣6

𝑡𝑇 2𝑇

Build Up Energy

Consume Stored Energy

Dump stored Energy

Build upStored Energy

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Cyclic Operation

0

𝑖3

𝑡

𝐷𝑇

𝑇+𝐷𝑇0

𝑣6

𝑡𝑇 2𝑇

• Depending on 𝐷, 𝑇, 𝛼and 𝜔& (𝑅% 𝐿, 𝐶) you can get different behaviors!• 𝑇 − 𝐷𝑇 shorter?

Current might not drop all the way to 0?• Current will eventually

reach cycle-to-cycle steady state average value

𝐷𝑇 𝑇+𝐷𝑇

𝑇 2𝑇

𝑣45

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13

Cyclic Operation

𝐷𝑇 𝑇+𝐷𝑇0

𝑖3

𝑡𝑇 2𝑇

𝐷𝑇 𝑇+𝐷𝑇0

𝑣6

𝑡𝑇 2𝑇

• Depending on 𝐷, 𝑇, 𝛼and 𝜔& (𝑅% 𝐿, 𝐶) you can get different behaviors!• More damping?

(power-hungry load, aka 𝑅% is low)…• Current will drop quick

and may hit 0

𝑣45

S2 opens to prevent current

going negative

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25

Cyclic Operation𝐷𝑇 𝑇+𝐷𝑇0

𝑖3

𝑡𝑇 2𝑇

𝐷𝑇 𝑇+𝐷𝑇0

𝑣6

𝑡𝑇 2𝑇

• Depending on 𝐷, 𝑇, 𝛼and 𝜔& (𝑅% 𝐿, 𝐶) you can get different behaviors!• Waaay too much

damping? (power-hungry load, aka 𝑅% is low)…

𝑣45

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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!

11/16/21 27https://www.ti.com/download/trng/docs/seminar/Topic_3_Lynch.pdf

27

MIT Ion plane• aka flying boost converter• The key

• 200 V à 40 kV• Low mass: 1.2 kW/kg

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𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳

LoadDC Supply

𝑺𝟐

𝑺𝟏

• What were those switches?

Boost Converter

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29

𝑪 𝑹𝑳+-𝑽𝑺

+

-

𝒗𝑶𝑼𝑻

𝑳

LoadDC Supply

𝒗𝒄𝒐𝒏𝒕𝒓𝒐𝒍

• Top switch is a Diode• Control Switch is a Transistor

FET

diode

Boost Converter

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What about those “magic” parts?

The so-called “diode” and “transistor” that were doing all this switching? How are they working?

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Transistor• Go into them next week• Three terminal device:• Middle terminal enables/disables

connectivity between two outside ones• 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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17

Diode

• Historically it was the first Non-Linear device we as humans developed and worked with (whether it was known at the time or not)• What does it meant to be non-linear?

𝒗

𝒊

𝒊

𝒗+ −

𝑖 = 𝐼g 𝑒hijN − 1

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Diode Can Act Like a Switch• Voltage in one way: Conduct (act like a wire)• Voltage in the other way: Don’t conduct (act like an open)

https://www.electronics-tutorials.ws/diode/diode_5.html

𝒗

𝒊

Simpler model of the system:

𝒊

𝒗+ −

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18

Diode Can Act Like a Switch• From current perspective:• Current Positive? Let it through (be a short)• Current Negative? Block it! (be an open)

• From voltage perspective:• Voltage across Positive? Be a short• Voltage across negative? Be an open

https://www.electronics-tutorials.ws/diode/diode_5.html

𝒗

𝒊Simpler model of the system:

𝒊

𝒗+ −

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Diodes are One Way Electrical Valves• 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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