Where's my degree

state.

How's that clamp transistor work?. I can see the need for one due the big -0.5V step but can't get my head round how it's actually clamping.

Nice drawing. Can you run some numbers for the lazy reader? We see a very small change in power dissipation, from 5.0 to 5.5mW, which we imagine won't cause much temperature rise in a device that can handle 500mW or more (is it 0.1 or 0.2C?). And we see a current sink with 5V across the CS-setting resistor(s), and a by-comparison microscopic change in Vbe from the aforesaid small change in temperature (about 0.1m/5V = 25ppm?). Just how much current change do you expect (about 0.1mV across the load)?

There's gotta be a better way!

Thanks for the diversion, hav'ta get back to the taxes!

Snip diagram.

I could not find a data sheet for the BC458 (npn) with a transient thermal resistance graph, so is this why you used the BC558 (matching pnp) data sheet?

Looking at the Semtech data sheet for the BC558, with a tp of 50mS and a tp/T of 50/250mS, rth-A looks to be about 180 K/W.

So it looks more like a 5mW step and 180 K/W.

Would it be better to step the dissipation by stepping the 25.7V on Q2's base (by 2V or so)?

Tony Williams a écrit :

Arghh. You're right, and 250uA.20V is 5mW, so it's even worse than I thought.

Yep.

I used:

The way they phrase it isn't very clear, but I think the abscisse time is the pulse duration, not the repetition rate (which wouldn't make sense for single pulse BTW).

Also, here we are not interested in peak temperature, rather in temperature excursion, so we have to use the single pulse curve, not the ones with duty ratio as parameter. This temperature excursion will ride on an elevated average temperature.

From this, I read the 'excursion Rth' to be Rth_ex = 0.25*Rth = 50K/W

I make it 5mW and 50K/W, so it's now 500uV that are missing.

Don't know why it should be better, but it'd be less simple.

Thanks.... might as well sing off the same songsheet.

Because you are currently looking for a small slope after a relatively large current step (10%), plus the change in Q1 current must produce a small change in Q1's Vbe,

Stepping the 25.7V can produce the same change in dissipation without having any Q1 current step, maybe a small front end transient only.

It's not that difficult. Put a 2222 ohm resistor in series with the base of Q2, and then move the top of the 20k to the base. That should step Vb(Q2) from 25.7 to 23.13, or by -10%.

But only during the pulse. The measurement, AIUI, is made afterwards. Stepping Vce would be cleaner then, since it doesn't directly affect Vbe, but the error from briefly stepping i(e) isn't great.

Since he's trying to measure Vbe(Q1), why not take the 'scope signal directly from Q1's emitter? That avoids the need for clamping (not knowing the clamp level, Q3 bugs me).

And, if he then steps Vce as you suggest, Q1's emitter gives the error signal directly.

Best regards, James Arthur

Buggerit. You've just trashed my next post. :)

Let's connect Q1's base to +0.7V (using a Vbe at 2.5mA) and the bottom end of the 2K to -5V.

Yeah, that's clean:

Vc | | | c +5---r1---b e | | | +------- scope | r2 | gnd

Step Vc from, say, +10 to +20. That will change power dissipation and the delta-Vbe waveform should say a lot about the junction thermal time constants. R1 is maybe 33 ohms, just to keep it from oscillating.

John

Right. The setup allows continuous real-time measurement of the thermal impulse-response, too.

James

Sounds good. Add a small potentiometer adjustment to that base voltage and he can d.c.-couple the 'scope, eliminating more potential gremlins (from 'scope bias currents, rectified RFI, blah blah blah).

Okay Fred, this one's whipped...next please!

James Arthur