# chicken and the egg, or, diode V/I & Tj curves

Discussion in 'Electronic Design' started by [email protected], Mar 20, 2007.

1. ### Guest

I'm trying to figure out how hot the junction of a diode will get. I
know that 1.75A will be passing through it. I know the ambient
temperature willbe 25C. I look at the V/I table in the datasheet. It
has several curves depending on junction temperature. Looking at the
curves, you can see that the drop across the diode for a given current
changes depending on junction temperature. But junction temperature is
dependent on the power dissipated inside the diode, which is
calculated using the drop across the diode, which is affected by the
junction temperature....AHHHHH!! (brain exploding sound).

So it seems to be chicken or the egg. How do I find the "equilibrium"
point?

Its the MBRD650 (data sheet at www.irf.com)

2. ### EeyoreGuest

With or without forced cooling ? Will it *really* be 25C btw. Are there no other
hot components near it ?

Ok, it's D-pak so it's designed to be heatsunk to the pcb. Look for an
application note.

It's a 6A device so 1.75A simply isn't going to make it sweat.

Graham

3. ### John LarkinGuest

You could do one of those old classic load-line sorts of graphical
solutions, but that would take a lot of setup.

So iterate:

Calculate Theta_ja, based on the diode T_jc and the external
heatsinking.

Guess a temperature.

Look up the voltage drop.

Multiply by 1.75 to get the power

Multiply by T_ja to get temp rise

Add ambient to get junction temp.

Compare results to your guess and make a better guess and do it all
again, until you stop changing or straddle the solution point.

The hardest part will be finding T_ja.

John

4. ### Guest

I like load lines. How exactly would I do a load line for this kind of
problem? What is the line? Power dissipated versus Tja?

5. ### IanGuest

Tj max =150 deg C
Max Vf at 25 deg C at 3A 0.7V
Max Theta j-a no heatsink 80 deg C/W
Max dissipation 1.75 *0.7 = 1.225W
Tj < 123 deg C

So you're safe at room temp even without a heatsink.

Putting copper in the board round the part will have far more
effect than tolerances of Vf or variations with temperature.

Regards
Ian

6. ### EeyoreGuest

You need to know the thermal impedance of your pcb pattern.

Which you don't or apparently don't realise the importance of. Hence the
question is moot.

Graham

7. ### Noway2Guest

You make a good point here. Rather than trying to calculate how hot a
component will get running in the application, you are better off
calculating the worst case and seeing if you are safe. If you can
handle the worst case, which is often times easier to calculate, then
you have reasonable assurance that things will work in practice.

8. ### Jim ThompsonGuest

The way we "big boys" do it...

Run 1mA thru diode
Place in oven and measure (and record) forward drop versus temperature

In system, run load current until forward drop reaches equilibrium
Abruptly switch to 1mA and measure forward drop
Consult table you made from above

For chips I often add a diode that is used for nothing but temperature
recording.

...Jim Thompson

9. ### Richard HenryGuest

Do you connect the diode to I/O pins, or use some internal trick to

10. ### Jim ThompsonGuest

For the prototypes it's usually connected to a package pin, though it
may only be a probe pad.

It's normally not used in the final product.

The latest game in town uses a MUX to look at a variety of internal
points thru a single test pin.

...Jim Thompson

11. ### John LarkinGuest

Tja is constant, and you need to figure that out for any solution.

The graph paper axies will be power dissipation (Y) versus temperature
(X). There will be two curves to plot, call them "diode" and
"heatsink."

The heatsink curve is easy. It is 0 power at 25 deg C (or whatever you
call ambient) and rises straight-line with slope 1/Tjc.

For the diode curve, get a table of diode voltage drop vs temp at 1.75
amps and multiply by 1.75 to get some points of power dissipation vs
temperature. Plot those points and eyeball-interpolate the curve.

(Or use Excel, curve fit, and spoil all the fun.)

The intersection of the curves is the equilibrium operating point. I
think.

|
|
|
|
|
|
|
|
|
| /
| / ------------ diode
| / ------/
| ------+------/
| -------/ /
| /------/ /
|---- /
| /
| /
| / heatsink
| /
| /
| /
| /
| /
-------------------------------------------------------------
25c

The diode curve may actually slope up or down, depending on things. It
could even be flat or sorta parabolic. For most diodes operated at
moderate current, it will slope down. I ain't gonna redraw it.

John

12. ### John LarkinGuest

oops, Tja ^^^^^

John