Engineering 6806: Design Project

9/10/02

Engineering 6806

Power and Heat

Michael Bruce-Lockhart

Power & Heat 2

Engineering 6806: Design Project

9/10/02

Modeling Heat Flow

• Use resistive equations• Power = Heat = Current• Temperature = Voltage• Heater is a current source• Fixed temperature is voltage source• Insulation is a resistance

Power & Heat 3

Engineering 6806: Design Project

9/10/02

Heating a Building

R

T o utT in

P

Heat flow through walls

Outside temperature

Tin = Tout + P x R

Inside temperature

Power dissipated in heaters =Heat generated by heaters

P = (Tin – Tout) / R

Power & Heat 4

Engineering 6806: Design Project

9/10/02

Piercing the Wall

RT o ut

T in

P

R

R

R

wall

w in

d o o r

gap • Piercing creates parallel heat paths

• Lowers resistance• More power to

maintain Tin

• Calculable: R = RSI/A

Power & Heat 5

Engineering 6806: Design Project

9/10/02

Heat in Semiconductors

T am bT j

P

jc c aTc

Thermal resistancejunction-to-case

θ conventionally used instead of R

Ambient temperature

Thermal resistancecase-to-air

Case temperature

Junctiontemperature

Average power

Dissipated In semiconductor

Power & Heat 6

Engineering 6806: Design Project

9/10/02

Temperatures

• Thus the case temperature is Tc = Ta + θca x Pav

• Notice it fluctuates up with ambient temp

• Junction temperatureTj = Tc + θjc x Pav = Ta + (θjc +θca) x Pav

• If Tj > 155 °C, junction melts!

Power & Heat 7

Engineering 6806: Design Project

9/10/02

Adding a Heatsink

T am bT j

P

jc c hTc hs

Thermal resistance rating

of heatsinkThermal gunk used to attachheatsink to case

If θch + θhs << θjc

ThenTc ≈ Tamb

&Tj ≈ θjc x Pav

Best you can do!

Power & Heat 8

Engineering 6806: Design Project

9/10/02

Supply Bypassing

• The supply pins on the H-Bridge must be bypassed

• Large inductive loads• V = L di/dt• Cutoff the current => high voltages• Need bypass caps to handle recirculating currents

• 100 µF for every amp of current

Power & Heat 9

Engineering 6806: Design Project

9/10/02

Power Calculations

Overheating is caused by power dissipation. A switching element dissipates power when it’s off, when it’s on and when it’s switching

sonoffD PPPP

The off power is the peak voltage times the leakage current (which should be very small) times the proportion of the time the switch is off–

where D is the duty cycle. The on power is the switched current (which we’ll call the peak current although it may vary depending upon what the motor is doing) times the switch drop times the proportion of the time it is on–

DVIP onpkon

D)(IVP offpkoff 1

Power & Heat 10

Engineering 6806: Design Project

9/10/02

Switching Power

The switching power takes a little more work. Suppose the voltage goes linearly from 0 to Vpk in ts seconds while the current starts from Ipk and drops linearly to 0 in the same time. Then the instantaneous power dissipated in the switching device is

)/(/)/1(/)( 2sspkpkspkspksi ttttIVttIttVtP

By integrating this over the switching time we can calculate the energy dissipated

6)3/2/(/)/(/ 22

0

2 spkpkssspkpk

t

sspkpkt

IVtttIVdtttttIVEs

ts

VpkIpk

Power & Heat 11

Engineering 6806: Design Project

9/10/02

Shape Factor

fttIVT

tIVP frpkpk

spkpks )(

6

1

3

Assuming the positive and negative going switching is symmetric, this occurs twice per switching cycle, so that the average power over one cycle is

The 1/6 represents a shape factor–that is it depends upon the exact shape of the currents and voltages during the rise and fall times. The details may differ (the waveforms might be more exponential, for example, and the rise and fall shapes might differ) but the only difference to the analysis is in the exact value of the shape factor. Thus the total power dissipated in the switching element is

ftkIVDVIftkIVDVIDIVP sshpkpkonpksshpkpkonpkoffpkD )1(

Ioff small

General shape factor