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Engineering 6806
Engineering 6806: Design
Project
Power and Heat
Michael Bruce-Lockhart
9/10/02
Modeling Heat Flow
Engineering 6806: Design
Project
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Use resistive equations
Power = Heat = Current
Temperature = Voltage
Heater is a current source
Fixed temperature is voltage source
Insulation is a resistance
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Heating a Building
Engineering 6806: Design
Project
Inside temperature
Heat flow through walls
Outside temperature
Tin = Tout + P x R
P = (Tin – Tout) / R
Tin
Tout
R
Power dissipated in heaters =
Heat generated by heaters
9/10/02
P
Power & Heat
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Piercing the Wall
Engineering 6806: Design
Project
Rgap
Rdoor
Rwin
Tin
Tout
Rwall
• Piercing creates
parallel heat paths
• Lowers resistance
• More power to
maintain Tin
• Calculable: R = RSI/A
P
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Power & Heat
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Heat in Semiconductors
Engineering 6806: Design
Project
θ conventionally used instead of R
Thermal resistance
junction-to-case
Junction
temperature
Tj
P
Thermal resistance
case-to-air
Ambient temperature
 jc
Tc
 ca
Tamb
Case
temperature
Average power
Dissipated In semiconductor
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Temperatures
Engineering 6806: Design
Project
• Thus the case temperature is
Tc = Ta + θca x Pav
• Notice it fluctuates up with ambient temp
• Junction temperature
Tj = Tc + θjc x Pav = Ta + (θjc +θca) x Pav
• If Tj > 155 °C, junction melts!
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Adding a Heatsink
Engineering 6806: Design
Project
Thermal resistance rating
Thermal gunk used to attach
heatsink to case
 jc
Tj
Tc
 ch
of heatsink
 hs
Tamb
P
If θch + θhs << θjc
Then
Tc ≈ Tamb
&
Tj ≈ θjc x Pav
Best you can do!
9/10/02
Power & Heat
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Supply Bypassing
Engineering 6806: Design
Project
• 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
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Power Calculations
Engineering 6806: Design
Project
Overheating is caused by power dissipation. A switching element dissipates power when it’s off,
when it’s on and when it’s switching
PD  Poff  Pon  Ps
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–
Poff  V pk I off ( 1 D)
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–
Pon  I pkVon D
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Switching Power
Engineering 6806: Design
Project
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
Vpk
Ipk
Psi (t )  V pk  t / t s  I pk (1  t / t s )  V pk I pk / t s (t  t 2 / t s )
ts
By integrating this over the switching time we can calculate the energy dissipated
ts
t
E  V pk I pk / t s (t  t 2 / t s )dt  V pk I pk / t s (t s2 / 2  t s2 / 3)  V pk I pk s
6

0
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Power & Heat
10
Shape Factor
Engineering 6806: Design
Project
Assuming the positive and negative going switching is symmetric, this occurs twice per
switching cycle, so that the average power over one cycle is
ts
1
Ps  V pk I pk  V pk I pk (t r  t f ) f
3T
6
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
General shape factor
PD  V pk I off (1  D)  I pkVon D  V pk I pk k sht s f  I pkVon D  V pk I pk k sht s f
Ioff small
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