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

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

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?

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

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.

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

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

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

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

oops, Tja ^^^^^

John