SMPS troubleshooting question

[Everton]

Hi,

I hqve several PC power supplies I am repairing (mostly for educational purposes).  I have found that the switching transistors blew (it wasn't hard to find since they almost caught fire!).  The part that has me scratching my head a little is the circuit on the base of the transistors.  There is a diode in series with a 39 ohm resistor which are shunted by a 1 uF cap.   The whole thing is in series with a 2.2 ohm resistor.  I'vre read somewhere that it is a pulse shaping circuit, but I can't understand what this would do.

I tried simulating it and as expected, the 1uF cap allows any pulses through essentially bypassing the diode.

                    1 uF
          ---------||-----------
          |     D      39R       |        2R2
       ------>|----/\/\/\------------/\/\/\/----- to base of transistor
                                     
Any help would be appreciated.

Thanks
                                 

[T3sl4co1l]

Did you simulate it just...naked?  Or did you have a PN junction to ground at the "base" end?

Like MOSFETs, BJTs require a lot of charge to switch on and off.  The coupling capacitor delivers this for sharp events (turning on and off), while the diode and relatively large resistor deliver the bias current required to keep the base forward-biased.

It's worth noting that the 1uF 50V electrolytic commonly used in that location has significant ESR (maybe 10 ohms), so it's not much better than the 39 ohm resistor.

BJTs are commonly referred to as current controlled devices, but this is a misnomer.  They are, in fact, voltage controlled; the current draw of the base is only an inconvenience that we deal with (or an added bonus, depending on how you design your circuit).

Tim

[Everton]

I did simulated it "naked".  I'll try it again tonight when I get home with a BJT to see the result.

If I understand correctly, the cap will give you the initial kick to turn the transistor on/off hard and the diode resistor are to keep the transistor biased correctly while it is supposed to be on?  That makes some sense to me, but then I am still confused about the 2R2 resistor. 

I am almost curious about the failure mode of this circuit.  I assume that a transistor must have gone short (possibly due to a AC line voltage surge) which took the other one with it on the following switching cycle.  But would this cause the base to short to the emitter and vaporize the 2R2?

[T3sl4co1l]

The drive is low voltage.  I'd be more concerned about getting a C-B short, or the fault current taking out the drive circuitry on the other side of the coupling transformer.

Tim

[Richard Head]

Everton
The 2R2 resistor is there to limit the pulse current from the capacitor. If it wasn't there the base current could momentarily exceed the forward (and reverse) bias safe operating area of the transistor.
Also, on turn off the capacitor provides a rapid negative bias to the base which is vital to get it to switch fast.
Luckily bipolars are very seldom used for high speed high current repetative switching nowadays. MOSFET's have almost totally replaced them in low to medium power applications.
Incidently, power bipolars also suffer from a failure mode called second breakdown. Even when the turn on and turn off trajectories were within the safe operating area of the datasheet failures occured. Bipolars were horrible things to use compared to MOSFETs.
The only reason you still see them used in converters is due to their very low cost compared to MOSFETs. That is because they use much less silicon for a given current rating compared to a MOSFET.

[T3sl4co1l]

Meh, bipolar isn't so bad.  Takes less voltage to drive and has more gain (pound for pound) than MOS.  Does tend to be slow and clunky though, hence the decreasing use these days.  They still dominate in CFLs and such; hard to beat two transistors and not much else.

Studying and playing with the old circuits is always illuminating.  In this case, literally.  (In fact, this circuit uses two old school properties: the self-driven BJT is turned off by the drive transformer saturating.  This isn't used for control, but it could be, in which case it would also use a magnetic amplifier.)

Tim

[Richard Head]

Tim

Yeh, I designed some electronic transformers for halogens years ago using a saturating base drive transformer (Jensen converter I think) and bipolars.
It's a very simple topology and low cost due to the bipolars, but tricky to design for various reasons. The short circuit protection relied on the leakage inductance of the transformer.
No control electronics available for current limit etc.
Also, the storage time of the transistors eroded the effective duty cycle which limited the operating frequency.
Bipolars, Backer clamps, proportional base drive, RBROA, Second breakdown, all a thing of the past, thank God!

Dick