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Power dissipation in a 5% duty cycle actuator

Posted on 2010-09-21
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Last Modified: 2013-11-10
If you have a 5% duty cyle switched load as follows:

Supply voltage 14 Volts
Duty Cycle 5% on
Impedance 2 Ohms

How Do I calculate the power dissipation in the load?

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Question by:AutoTest
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jon47 earned 84 total points
ID: 33733603

Pavg = (D*Pon)+((1-D)*Poff)
where
D = Ton/(Ton+Toff) = duty cycle
Pavg = average power dissipation over the cycle
Pon = Power when on
Poff = Power when off

P = I^2*R = (V^2)/R

However, the average power dissipation is often not useful.  If you're doing thermal design then you probably need to consider the peak power dissipation (i.e. Pon as defined above).  But it can get complicated, as the power coming out of the load is dissipated into a heatsink, and so when off the temperature drops slowly - and often not back to the level before you last turned on the load.  so over time the temperature of the load and heatsink increases.  Hence consider peak power instead.
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by:d-glitch
d-glitch earned 83 total points
ID: 33734649
At 100% duty:

    Power = V²/R  =  14²/2  =  196/2  =  98 watts

5% of this is 4.90 watts.

I would disagree with jon47:  Average power dissipation is often very useful.

You do have to consider the time scales.  
Are talking about 5% of a second or 5% of an hour?
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by:jon47
ID: 33735803
The usefulness or otherwise of the average power dissipation depends on the application.  In practice it depends on the specific heat capacity of the load and its thermal management systems and what the length of the on part of the cycle is.

I hadn't provided a specific answer because this looked like it might be some student work, and E.xv of the terms of use don't allow me to do the work, just help - "xv.  Violating the guidelines for academic honesty or other unethical behavior; helping a student with a project is allowable, but not doing it for them."  Though the sum is so trivial it doesn't make much difference.

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by:AutoTest
ID: 33739461
Thanks for your responses  btw i havent been a student for 30 odd years :)

Both of you have multiplied the 100% duty cycle power by the actual duty cycle.

What if

Calculate average voltage  = 14 * .05  = 0.7 Volts

Therefore average power = 0.7 * 0.7 / 2 ~ 0.25 W

This was my quandry as the 2 answers are wildly different.

Any thoughts?

Also I understand the thermal part but it is not an issue in this instance
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by:jon47
ID: 33739782
You have to calculate the peak power, and then average that over the cycle.  You can't average the voltage, it doesn't work.
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by:rwj04
rwj04 earned 83 total points
ID: 33760038
the formula for power dissipation does not vary with time.  Power dissipation in a purely resistive load is simply:

P = I * V       where I = V / R    

therefore,

P = V^2 / R

P = 14^2 / 2 = 196 / 2 = 98 Watts.


Power is Energy over Time. so you would use your time variance to calculate power with the integral

W (work energy) = closed integral of P(t) dt, from t0 to t1

since 98 Watts = 98 Joules / second

and if we could assume a consistent 5% use, this would average out to just slightly under 20 Joules / second.


the end result is that, regardless of your duty cycle, you would not want to use anything less than than is capable of safely dissipating 100W.   probably even more than that depending on other factors.   i

because it is very difficult to quantify exactly how much heat generation your resistor could tolerate, and depends on multiple variables related to thermal coupling such as board type, equipment case,  ambient temperature, air flow, solder joints, etc. etc.


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by:rwj04
ID: 33760082
d-glitch wrote:
>> At 100% duty ...  98 watts>> 5% of this is  ... 4.90 watts.


no, no, no...  you can not do this.   what i tried to say above, power dissipation does not change with time or duty cycle.

you can integrate the power over time, to determine the energy (work) in joules.   but you can not say that a 50 watt resistor is only 5 watts because you left it on 1 hour instead of 10.   or variations on that theme

in other words, say you have a 1000 Watt hair drier.   toggling it on for 1 second and off for 9 seconds, does not turn it into a 100 Watt hair drier.

you can determine how much energy is consumed (or generated) by multiplying the power by time.  

the 1000 Watt hair drier, if left on for one hour, will consume 1000 Watt-hours of energy, also known as 1 kilowatt-hour, which is 3,600,000 Joules.

in any event, that 1000 Watt hair drier has to be constructed to safely dissipate 1000 Watts, whether you plan to use it for 15 seconds, or to leave it on for an hour.


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by:rwj04
ID: 33760095
rwj04 wrote:
>> and if we could assume a consistent 5% use, this would average out to just slightly under 20 Joules / second.

oops!

sorry, i meant to say that since 98 Watts = 98 Joules/second, if the load were consistently exercised at 5% duty cycle over a time period, this would average out  to just slightly under 5 Joules / second.

dur..  it's late.  ive got to go to bed.


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by:rwj04
ID: 33760101
(i was thinking 1/5, instead of 5%)


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