BC857 Destroyed

D1 prevents it.

Cheers, James Arthur

I don't detect anything wrong with the circuit offhand.

I'd double-check the layout to be sure the clamp diodes exist and are wired as drawn on the schematic.

I'm with Spehro in thinking a bigger, stronger transistor will fix your problem, but I wouldn't be happy. AFAICT the BC556 shouldn't break, and you really need to find what's breaking it, and make sure you've fixed _that_.

I'd try to break more transistors--e.g. beadboard a version where you o parallel two relays for the load, and / or o cycle it very rapidly o increase the supply voltage etc.

Find out how much TR8 can really take, and learn thereby its mode of destruction & margin of safety.

Fixing it will be easy once you know what's breaking it.

Cheers, James Arthur

I printed out the schematic, but still have trouble telling what are connections and what are crossover. Anyhow, I am talking about what happens when TR8 and RL2 turn off. The problem would occur if TR8 turns off before C10 has discharged. Speaking of C10, not sure if the polarity is correct. It would be nice to know how this is supposed to work.

Tam

It *is* hard to read.

To me, it looks like C10(+) drives a common bus that drives the "high" side of all the relay coils.

That bus is clamped to +10v by one of the diodes (from pin 1 to 3) inside dual diode D1. The remaining diodes in dual diodes D1-2 are connected as inverse clamp diodes across the relay coils.

The negative side of C10 is returned to GND via 4.7k resistor R3.

OPERATION: when the relays are inactive, C10 charges to roughly 10v via D1(1,3) and R3. To fire a relay, TR8 is turned on, which raises C10(-) to +10v, and C10(+) to about +20v. That's the relay supply voltage, which drives the "high" side of all the relays. The "low" side of whichever relay is being fired is then pulled down by its associated driver transistor, TR2,3, or 4.

HTH, James Arthur

You mentioned that you thought they failed during build. Have you considered what happens when the uC is unprogrammed (outputs probably floating)? Or *during* programming, in case programming lines are shared with IO?

--
Ben Jackson AD7GD

http://www.ben.com/

That's not a bad guess. An open solder joint on D1 pin 1 or 3 would remove any control on the relays' flyback. The resulting negative transient could easily zap TR8...

--James Arthur

I was more thinking about trauma to TR8 itself.

But I guess it's time to crack out the DSO and do some careful measurements. Mostly this stuff happens because some spikes are pushing a semiconductor too close to the cliff.

--
Regards, Joerg

http://www.analogconsultants.com

What is the current gain (beta) of that transistor ** AT 200mA **?

From the NXP datasheet, it looks like the beta is about 40 at 200mA. Your

4k7 resistor cannot supply 200mA/40 = 5mA, because this would require more than 20Volts across the 4k7 resistor which you do not have.

If the base current is not sufficient, then VCE will be large (perhaps 10 Volts) and so the power dissipation may be several watts.

I suggest you measure VCE during the switching process, to see if the transistor is coming out of saturation. Unfortunately, this is probably a once-only phenomenon, and so you may only see the VCE becoming large on those transistors that are in the process of being destroyed.

If this is the problem, then you may be able to find a Zetex transistor that is specified to have a high beta AT A HIGH CURRENT.

Chris

I think you are right about the connections. Still causes problems for me. The top ends of all relays are at near +10 volts through D1 (1-2); Now, RL1 will operate whenTR2 is on, RL2 will operate whenTR4 is on, and RL3 will operate when TR3 is on. What does TR8 do? Or is he claiming the relays will operate on 20V, but not 10V? I guess that is the big secret.

What he needs to do is to put a storage scope on the collector of whatever transistor blows, and observe the waveform. In fact, I would do it with a blown board, and simulate the operation of the transistor with a clip lead between collector and base.

Tam

TR8 yanks C10's cathode up to +10v. Since C10 is already charged to

10v, the anode of C10 is then at +20v. That's bussed to the relays; that's their supply voltage.

Best, James Arthur

The symptoms sound like secondary breakdown. Very localised overheating causing collector to emitter short circuit.

--
Jim Backus running OS/2 Warp 3 & 4, Debian Linux and Win98SE
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