# obtaining thermal rise in diode

1. ## obtaining thermal rise in diode

This diode specifies Average Rectified Output Current @ TT = +75°C but referencing the If - Vf graph Figure 2 its Vf for 5A is ~1.15v power dissipated is 5.75W and Junction to lead thermal resistance is give as 10 C/W which would cause the junction to rise 57.5C not the stated 75C. https://www.diodes.com/assets/Datasheets/ds16007.pdf

Manufacturer specified 75C thermal rise for 5A is not accounted for even if its the sum of power dissipation and typical 25C ambient temperature which would be 82.5C. Isn't power dissipation for diodes the product of its forward voltage and average conducted current?  Reply With Quote

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2. ## Re: obtaining thermal rise in diode

Hi,

Junction to lead....temperature ruse is 57°C.
Maybe the lead temperature risees by 18°C
Making a total junction temperature rise of 75°C.

Klaus

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3. ## Re: obtaining thermal rise in diode Originally Posted by KlausST Hi,

Junction to lead....temperature ruse is 57°C.
Maybe the lead temperature risees by 18°C
Making a total junction temperature rise of 75°C.

Klaus
This diodes junction will be 75C + ambient temperature while averaging 5amps?  Reply With Quote

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4. ## Re: obtaining thermal rise in diode

We call this "sandbagging the spec". Conservative only costs
the customer money but tends to keep them safe; aggressive
makes field returns and ruins AOQ statistics.

There is also that diode I*V(I) is nonlinear and working with
averages embeds some assumptions about linearity and
superposition. Which may or may not be true / accurate.
Average of products being different than product of averages
in many cases.

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5. ## Re: obtaining thermal rise in diode

Vf for 5A is ~1.15v power dissipated is 5.75W and Junction to lead thermal resistance is give as 10 C/W which would cause the junction to rise 57.5C...
You are right, well almost.

The resistance has two ends; one end will be 57.5C if the other end is 0; if the exposed end is at 20C, the junction will be 77C. That is how I see it.

In addition, the heat transfer process is like diffusion and is highly non-linear. The thermal resistance is a function of temperature.

Fig 1 in the datasheet you have attached says that at 5A current, max terminal temp acceptable is 75C. The junction temp will be 75+57=132C! (good for frying an egg)

Max power dissipation can always be obtained as V.I; the exact power dissipation can always be obtained as the integral of V(t).I(t).dt over a complete cycle.

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6. ## Re: obtaining thermal rise in diode

@TT = +75C is the terminal temperature. It is in the "maximum" section, so the diode is rated for max 5A when the terminal temperature is +75C.
Vf @5A according to table 2 is about 1.05V @25C. Power dissipation = 1.05V * 5A = 5.25W
Temp rise = 10 * 5.25 = 52.5C
Junction temp = +75C + 52.5C = 127.5C but it will be less since the forward voltage decreases with increasing temp.
Approximate with -2mV/C so Vf @5A @+125C is about 0.85V.
Power dissipation = 0.85V * 5A = 4.25W
Temp rise = 10 * 4.25 = 42.5C
Junction temp = +75C + 42.5C = 117.5C but it will be a little higher since the temp is less than 125C, so Vf is a little higher than 0.85V.

Expect the junction temperature to be about +120C when the terminal temperature is +75C and the average current is 5A.

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7. ## Re: obtaining thermal rise in diode

but referencing the If - Vf graph Figure 2 its Vf for 5A is ~1.15v power dissipated is 5.75W...
You need to make a steady state approximation.

What I mean is that all the heat produced at the junction must be dissipated via the terminal ends.

About 5W of heat must be conducted away from the pads at 75C by radiation, conduction, etc etc...

Because the junction temp is not directly accessible to the end user, you need to use the terminal temp as the base for all your calculations.

If you are planning to use the diode at 5A for extended periods, you surely need some healthy heat sinking. Radiation will perhaps help about 1W/cm2 at 75C (guess value)

But if you increase the pad size, the terminal temp will perhaps decrease but the dissipation will remain the same. You have to get rid of the heat somewhere else somehow.

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