# Choosing Load dump diode for truck power supply

Discussion in 'General Electronics Chat' started by vam3kor, Jul 26, 2011.

1. ### vam3kor Thread Starter New Member

Jul 18, 2011
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0
I am designing a power supply for Truck application of 28V battery system, I need to choose a load dump diode to place across battery line.

how to calculate its peak power, energy, etc.

Load dump clip voltage is 40V,

Vs= 176 V
Ub= 28V
Ri = 2 ohms
Td= 400ms ( pulse duration)
Tr= 10ms ( 10 to 90% rise time)

2. ### lokeycmos Active Member

Apr 3, 2009
432
7
could you give more info about your setup and what you want to accomplish? are you putting the diode across the terminals of the battery? if you goto digikey's website you can click on the specs you need and if they have it in stock check the prices.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Cat=4261062

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3. ### #12 Expert

Nov 30, 2010
17,818
9,137
The way I see this is that a 40 volt limit will be placed on a surge with the same polarity as the 28 volt battery. That is, the 28 volt battery shall not be allowed to have more than 40 volts at its terminals when a 176 volt generator with 2 ohms of impedance surges for 400 milliseconds.

176-28 = 148V
148/2 = 74 amps max
times .4 seconds = 29.6 amp seconds, and I think that's "joules"...or are joules watt-seconds?

Problems: I'm getting squat for a single device that can do that.
The battery should contribute quite a lot to quenching this surge, so the math here is very conservative.

Last edited: Jul 26, 2011
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Jul 18, 2011
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5. ### vam3kor Thread Starter New Member

Jul 18, 2011
5
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I need to choose a TVS diode to clip the transient above 40V, So the TVS diode should be able to consume the energy grnerated by the transient

As you told the voltage of the signal is 172-40= 132V
Resistance is 2
so current is 132/2= 66A

Now how to calculate the exact power of this signal ?
because its going to be a half sine wave with amplitude of 132V and Current 6A
is the below way is correct?
RMS value of the voltage is (Vm/√2)=132/1.414
RMS value of current is (Im/√2)= 66/1.414

I don't know how to calculate power from this information.
Is Sinθ info required?