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Switcher Efficiency & Snubber Design

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Agenda• SMPS Basics• Control Methods• Losses• Example: Buck• Example: Boost• BJTs vs. MOSFETs• Snubber Design

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SMPS Basics

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Pin Pout

It ’s gettinghot in here!

The conversion mechanism generates heat...

The goal of a converter is to deliver powerpower

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η=Pout/Pin is called the efficiency

A 50% efficiency means Ploss = Poute.g. Pout = 100 W Ploss = 100 WPin = 150 W, Pout = 100 W η = 66%

⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅=−=−= 11

ηηPoutPoutPoutPoutPinPloss

Heat means that the energy transfer is notnotperfect

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The linearlinear approach:efficiency is poorgood noise performanceacceptable when (Vout-Vin) is smallcan only decrease the input level

The switchingswitching approach:efficiency is highnoise performance is poorworks with large (Vout-Vin)increase/decrease/invert the input level

R

ON or OFF

0 to ∞

Two different options exist to build a converter:

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50 V

Vout

100 V

10 A10 A

η=50%

Help!I am dissipating500 W!! ( )

The linearlinear approach:

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Hey!I feel much coolernow...

50 V

6 A10 A

Vout

Vin

0Vout

η=83%

The switchingswitching approach:

Vout

100 V

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ControllingControlling the power flow

VinDVinTstondttVout

TsVout

Ts

avg ⋅=⋅=⋅⋅= ∫ )(1

0

Ts TstonD =

ton

toff

PPulse WWidth MModulation (PWM) control

(Duty-cycle)

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Control Methods

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Regulation, keeping an outputoutput signal constant by…

• adjusting the duty-cycle via the PWM block• regulating the inductor peak current• regulating the inductor average current• adjusting the switching frequency• off time adjustment• …

Current-mode control…Voltage-mode control…

Two most popularmethods!

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The voltagevoltage--modemode method

I > Imax?reset

The error level sets the duty-cycle

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The dutyduty--cyclecycle factory…

+

-

-1.50

497m

2.50

4.50

6.50

verr,

ram

p in

vol

tspl

ot1

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10.0u 30.0u 50.0u 70.0u 90.0utime in seconds

-1.00

1.00

3.00

5.00

7.00

out i

n vo

ltspl

ot2

3

100%

0%

A ramp is compared to a DC level, the error voltage

PWMmodulator

Verror

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The currentcurrent--modemode method

The error level sets the current setpoint

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The peakpeak follows the error voltage

1 idrain 2 vfb 3 vfb 4 idrain

1.31m 1.65m 1.99m 2.33m 2.67mtime in seconds

-500m

500m

1.50

2.50

3.50

vfb

in v

olts

-1.00

0

1.00

2.00

3.00

idra

in in

am

pere

spl

ot1

12

2.50m 2.57m 2.65m 2.72m 2.80mtime in seconds

-1.00

0

1.00

2.00

3.00

idra

in in

am

pere

s

-500m

500m

1.50

2.50

3.50

vfb

in v

olts

Plo

t2

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S

R

Q

Q

Error voltage

Inductor peak current

MOSFET

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Losses

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Losses

ICConSWinout PPPPP −−−=PSW: Switching Losses

PCon: Conduction Losses

PIC: Power consumed by the chip

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IC Losses

qinIC IVP =Vin: IC input voltage

Iq: Quiescent current (read from the data sheet)

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Switching Losses• Losses which occur when the

power switch is turned on or off.• During this transition the voltage

and current on the FET are both high.

• Different for Buck and Boost configurations.

Y-A

xis

Time

Current(DS)

Voltage(DS)

ton

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Switching Losses (Buck)

SWoffonoutinSW FttIVP )(21

+=

Vin: Input Voltage

Iout: Average inductor current

ton: Turn on time of high side switch

toff: Turn off time of high side switch

FSW: Switching frequency

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Switching Losses (Boost)

SWoffonout

outSW FttD

IVP )(12

1+

−=

Vout: Output Voltage

D: Duty Cycle

Iout: Average inductor current

ton: Turn on time of high side switch

toff: Turn off time of high side switch

FSW: Switching frequency

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Reducing Switching Losses• Increase gate drive strength

– Increases cost (die area)– Increases EM emissions

• Decrease frequency– Requires a larger value inductor

• Use a smaller FET– Increases conduction losses

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Conduction Losses• Losses which occur when current flows through a resistive

path (I2*R), such as a FET, or a diode (V*I).• Major contributors include:

– Power Switch Rds,on

– Freewheeling Diode– Inductor DCR

• Different for synchronous and non-synchronous mode designs.

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Conduction Losses (Non-Synchronous)

DCRLdiodeLondsLCon RIDVIDRIP 2,

2 )1( +−+=

IL: RMS current through the inductor

Rds,on: On resistance of the power switch

D: Duty cycle

Vdiode: Forward voltage of the diode

RDCR: Winding resistance of the inductor

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Conduction Losses (Synchronous)

DCRLondsLondsLCon RIDRIDRIP 22,

21,

2 )1( +−+=

IL: RMS current through the inductor

Rds,on1: On resistance of the high side switch

D: Duty cycle

Rds,on2: On resistance of the low side switch

RDCR: Winding resistance of the inductor

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Example: Buck

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Example: NCV8851 Evaluation Board• Synchronous buck converter• Vin = 13.2 V• Vout = 5 V• Iout,max = 4 A• FSW = 170 kHz• Inductor: Wurth Electronics 7447709150 15 uH• Power switch (both): ON Semiconductor NTD5407N

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Example: NCV8851 Evaluation Board• D = Vin / Vout = 5 V / 13.2 V = 0.379• ton = 5 ns (empirical)• toff = 7 ns (empirical)• Rds,on1 = Rds,on2 = 50 mΩ (including self heating temperature

effects)• RDCR = 26 mΩ• Iq = 15 mA• IL = Sqrt(Iout

2 + (0.609 A)2 / 3)

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Example: NCV8851 Evaluation Board• PIC = 13.2 V * 15 mA

= 0.198 W• PSW = (1/2) * 13.2 V * Iout * (5 ns

+ 7 ns) * 170 kHz = 0.01346 * Iout W

• PCON = (Iout2 + 0.12378 A2) * (50

mΩ) * (0.379) + (Iout2 + 0.12378

A2) * (50 mΩ) * (0.621) + (Iout2 +

0.12378 A2) * 26 mΩ = (0.076 * Iout

2 + 0.009407) W

Sources of Power Dissipation

0

0.2

0.4

0.60.8

1

1.2

1.4

0 1 2 3 4

Iout (A)Po

wer

(W)

PiqPswPcon

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Example: NCV8851 Results

8851 Efficiency

0.00%10.00%20.00%30.00%40.00%50.00%60.00%70.00%80.00%90.00%

100.00%

0.000 1.000 2.000 3.000 4.000

Iout (A)

Effic

ienc

y

ActualCalculated

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Example: Boost

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Example: NCV8871 Sample Application• Non-synchronous boost converter• Vin = 13.2 V• Vout = 18 V• Iout,max = 9 A• FSW = 170 kHz• Inductor: Vishay IHLP6767GZER330M11 33 uH• Power Switch: ON Semiconductor NTD5803 x 2• Diode: ON Semiconductor MBRB1645T4G

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Example: NCV8871 Sample Application• D = 1 – (Vin / Vout) = 0.267• ton = 30 ns• toff = 20 ns• Rds,on = (12 mΩ) / 2 = 6 mΩ (including temperature effects)• RDCR = 37 mΩ• Iq = 10 mA• IL = Sqrt ((Iout / (1 – D)) ^ 2 + (0.3137)^2 / 3)

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Example: NCV8871 Sample Application• PIC = 13.2 V * 10 mA = 0.132 W• PSW = (1/2) * 18 V * Iout / (1-0.267)

* (20 ns + 30 ns) * 170 kHz = 0.10436 * Iout W

• PCON = (Iout2 + 0.0328 A2) * (12

mΩ) * (0.267) + Sqrt(Iout2 +

0.0328 A2) * (0.5V) * (0.733) + (Iout

2 + 0.0328 A2) * 37 mΩ = (0.0402 * Iout

2 + Sqrt(Iout2 +

0.0328)*0.3665 + 0.00131869 W

Power Dissipation

0

1

2

3

4

5

6

0 3 6 9

Iout(A)Po

wer

(W)

PicPswPcon

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Example: NCV8871 Results

Efficency

0.00%

20.00%

40.00%

60.00%

80.00%

100.00%

0 3 6 9

Iout(A)

Effic

ienc

y

Calculated

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BJTs vs. MOSFETs

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Switches and converters…

The bipolar transistor is often used:1. In high voltage - high current applications 2. In low-cost converters

The bipolar transistor

=

Ic

Vce

When saturated…

sat avgPcond Vce Ic= ⋅

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Switches and converters…

The bipolar transistor

The bipolar transistor switching losses:1. Depend on temperature (storage time, current tail) 2. Watch-out for hot spots!3. Often needs proportional drive (shallow saturation)

Vce Ic

( )3

Vce Ic ton toffPswTsw

⋅ ⋅ +=

⋅

losses

ton toff

If ton = toff…

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Switches and converters…

The MOSFET transistor is the most popular:1. Ease of drive (capacitive input)2. Avalanche rugged3. BVdss of 600 V for SMPS, 500 V for PFCs…

)(² ONdRMS RdsIPcond ⋅==

d

s

g

d

s

g

Rds(ON)

body diode

The MOSFET transistor

Often used infreewheel function

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To enhanceenhance a MOSFET, bring it charge

VT

Plateau levelMiller effect

How many coulombs to turn on the MOSFET: Q = i x t…

d

s

g

Crss

Coss

Ciss

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Snubber Design

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When is it needed?• Parasitic inductances

and capacitances from the power devices form a RLC filter that resonates

• Excessive ringing can cause damage to the devices

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Snubber Design• Measure the frequency of the ringing (fc) at maximum input

voltage– Use a low capacitance probe

• Find out either the L or C of the circuit– L is dominated by the top power switch– C is dominated by the body diode of the bottom power switch or the

capacitance of the freewheeling diode

• Calculate the characteristic impedance of the circuit– If L is known: Z = 2πfc L– If C is known: Z = 1 / ( 2 πfc C )

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Snubber Design• Choose RSNUB = Z• Choose CSNUB = 1 / (2 πf R)• Power dissipation in RSNUB is CV2fs• Put RSNUB and CSNUB in series across the device causing

ringing.• Test in circuit. RSNUB can be fine turned further to reduce

ringing if it is found to be insufficient

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For More Information

• View the extensive portfolio of power management products from ON Semiconductor at www.onsemi.com

• View reference designs, design notes, and other material supporting automotive applications at www.onsemi.com/automotive