Power Conversion Circuits And Diodes

6.002

CIRCUITS AND ELECTRONICS

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Power Conversion Circuits and Diodes

Power Conversion Circuits (PCC) PCC

+ – 5V DC

PCC

+ – 5V DC

110V 60Hz

solar cells, battery

3V DC

DC-to-DC UP converter Power efficiency of converter important, so use lots of devices: MOSFET switches, clock circuits, inductors, capacitors, op amps, diodes

R Reading: Chapter 16 and 4.4 of A & L. http://electrical.globalautomation.info

First, let’s look at the diode

⎛ VvD ⎞ T ⎜ iD = I S e − 1 ⎟ ⎜ ⎟ ⎝ ⎠ I S = 10 −12 A

iD

+ vD –

VT = 0.025V Boltzmann’s constant temperature in Kelvins charge of an electron

kT VT = q iD

iD

vD

− IS

vD V

mV

Can use this exponential model with analysis methods learned earlier „

analytical

„

graphical

„

incremental

(Our fake expodweeb was modeled after this device!) http://electrical.globalautomation.info

Another analysis method: piecewise–linear analysis P–L diode models: iD iD ≥ 0 Æ vD = 0

“short” or on

vD < 0 Æ i D = 0

0

vD

“open” or off

Ideal diode model

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Another analysis method: piecewise–linear analysis “Practical” diode model ideal with offset

+–

0.6V

iD Short segment Open segment

iD = 0

vD = 0

0.6V

vD

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Another analysis method: piecewise–linear analysis

Piecewise–linear analysis method „ „

Replace nonlinear characteristic with linear segments. Perform linear analysis within each segment.

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Example (We will build up towards an AC-to-DC converter)

0.6V +–

Consider + vI + –

R

vO –

vI

is a sine wave

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Example 0 .6 V +–

Equivalent circuit

+

vI + –

vO

R

– “Short segment”: iD = (vI − 0.6 ) / R vI ≥ 0.6

+ vI –

+–

+

0.6V

R

vO = vI − 0.6

– “Open segment”: iD = 0

vI < 0.6

+ vI –

+– 0.6V

+

R

vO = 0

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Example vI

vO 0.6 t

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Now consider — a half-wave rectifier 0.6V

+– vI + –

C

+ R

vO

–

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A half-wave rectifier

vI

diode on

diode off

vO

Demo

t

C current pulses charging capacitor

MIT’s supply shows “snipping” at the peaks (because current drawn at the peaks)

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se Do not u resistive s! el em ent

DC-to-DC UP Converter i

+ VI + DC –

vS

C vO

switch

S

load

–

vS S

S

closed

T

open

t Tp

The circuit has 3 states: I. II. III.

S is on, diode is off i increases linearly S turns off, diode turns on C charges up, vO increases S is off, diode turns off C holds vO (discharges into load)

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More detailed analysis I. Assume i(0) = 0, vO(0) > 0 S on at t = 0, diode off L

vO

i VI + –

VI T i (T ) = L

C

i

di VI = L dt

VI slope = L T

i is a ramp

t

1 ΔE = energy stored at t = T : Li( T )2 2 2

VI T 2 ΔE = 2L http://electrical.globalautomation.info

II. S turns off at t = T

diode turns on (ignore diode voltage drop) L

VI + –

VI T L

i

i

0 T T′

vO S

C

State III starts here

TP

t

1 ωO = LC

Diode turns off at T′ when i tries to go negative.

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II. S turns off at t = T, diode turns on Let’s look at the voltage profile

i

VI T L

0 T T′

ωO = 1

TP

t

LC

Capacitor voltage ignore diode drop

vO

III.

vO (T )

1 ωO = LC

0 T T′

ΔvO

TP

t

Diode turns off at T′ when I tries to go negative. http://electrical.globalautomation.info

II. S turns off at t = T, diode turns on Let’s look at the voltage profile

i

VI T L

0 T T′

ωO = 1

TP

t

LC

Capacitor voltage ignore diode drop

vO

III.

vO (T )

1 ωO = LC

0 T T′

ΔvO

TP

t

Diode turns off at T′ when I tries to go negative. http://electrical.globalautomation.info

III. S is off, diode turns off Eg, no load

+ VI + –

S

C vO

– C holds vO after T′ i is zero Capacitor voltage

vO

0

T′

TP

t

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III. S is off, diode turns off Eg, no load

+ VI + –

S

C vO

– C holds vO after T′ i is zero until S turns ON at TP, and cycle repeats I II III I II III … Thus, vO increases each cycle, if there is no load.

vO vO (n)

TP 2TP 3TP

t

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What is vO after n cycles Æ vO(n) ? Use energy argument … (KVL tedious!) Each cycle deposits ∆E in capacitor. 1 2 Δ = = L i ( t T ) E 2 2 1 VI T 2 ΔE = 2 1 ⎛ VI T ⎞ 2 L = L⎜ ⎟ 2 ⎝ L ⎠ After n cycles, energy on capacitor 2

nVI T 2 nΔE = 2L 1 This energy must equal CvO ( n )2 2 so, or

2

2 nV T 1 2 CvO ( n ) = I 2L 2 2

nVI T 2 vO ( n ) = LC

ωO =

1 LC

vO ( n ) = VI T ωO n

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How to maintain vO at a given value? + VI

+ –

vO

load

–

vO

pwm

control change T

T

Tp

compare + vref –

2

VI T 2 recall ΔE = 2L Another example of negative feedback: if if

(v (v

O O

− vref ) ↑

− vref ) ↓

then T ↓ then T ↑ http://electrical.globalautomation.info