PNP BJT: Max. Collector-Emitter Voltage

[radian]

Hello all. I'm using a PNP BJT as a high-side switch. So normally Ve > Vc.

I'm trying to determine what would happen if Vc > Ve. For example, if the output is shorted to a higher voltage source. The datasheet only indicates a maximum value in one direction. Can I assume the same maximum value in the reverse direction? Same question for base-collector junction. See below for snippet from datasheet.



Thanks.

[Ian.M]

No.  Vc must *NEVER* significantly exceed Ve.  That would forward bias the collector-base junction and possibly damage your base drive circuit.  It would also reverse bias the base-emitter junction and if you Zener it by significantly exceeding the -4V Vebo abs.max. rating, depending on the current you pass through it either degrade or destroy the transistor.

[radian]

Ah, ok thanks Ian. What do you think about placing a series diode at the collector as a means to protect against this condition? Not worried about the voltage drop and current is around 40mA.

[Ian.M]

As long as you've got a resistor from base to emitter to handle the diode's leakage current, that should be fine up to the voltage rating of the diode, provided the reverse voltage doesn't have a very fast risetime.  If it does, you may also need a schottky diode from collector to emitter to shunt transient reverse current due to junction capacitance and reverse recovery of the series blocking diode.

[T3sl4co1l]

Um?  It's not a foregone conclusion that C-B forward bias causes problems.  There are situations where it might, and others (I suspect more common) where it doesn't matter.  And still others where it can't happen at all (but OP asking about this condition probably implies we're not considering that case).

In and of itself, the transistor's limit, without reference to the surrounding circuit: it's right there in the screenshot, Vebo.  The E-B junction has a lower breakdown voltage, typically acting something like a 7V zener diode.  (Vebo is the voltage from emitter to base, the remaining terminal (collector) being "open".)  Or in this case, potentially as low as 4V (note the limit is for guaranteed freedom from breakdown; the actual zener breakdown voltage might be 5-6V, I would guess).

The exact thing you asked for, Veco, isn't usually specified, but yes, the C-B junction will be forward biased under that condition.  More importantly, transistor action will amplify the breakdown current, effectively lowering the breakdown voltage; by how much, isn't specified here.  The forward version is (Vceo vs. Vcbo); and this is the explanation* why one is lower -- the C-B leakage forward-biases the B-E junction, multiplying the current, effectively causing breakdown to occur earlier, i.e. at a lower voltage.

*Probably there's a more refined explanation; alas, it's been long enough that I forget the intimate details of BJT operation, and whether this mode of operation has a weirder explanation, or if this really is the underlying mechanism.  Anyway, as a hand-waving explanation, it's fine.

Tim

[radian]

I appreciate the discussion. Good info. Let's consider the circuit below (ignore it's lack of usefulness):



Let's say Vout = 75V and Vin = 0V. The reverse leakage current for the 1N4148W under this condition is 50uA at Tj = 150C. Let's go with that.

Will the 50uA actually find it's way to R38 and generate a base voltage? Isn't there some sort of minimum voltage required to forward-bias the collector-base junction?

Assuming the current generates a base voltage, as long as it's below the Vebo I should be good, right? With the numbers for this example we get Vb = (50uA)(24K) = 1.2 which is less than 4V.

[magic]

This resistor will turn the transistor on. It would be better connected between base and emitter. A diode can also be used; it has the advantage of not conducting when the transistor is operated normally and providing reliable clamping to less than 1V for reasonable reverse currents.

[amyk]

Isn't there some sort of minimum voltage required to forward-bias the collector-base junction?

No. Forward bias just refers to the polarity of the voltage difference across the junction.

[T3sl4co1l]

The resistor is a B-E resistor because Vin = 0 in the example...

BJT breakdown is typically fine.  This condition can be handled up to the thermal limit, with some degradation of characteristics (hFE) proportional to charge passed in this way (at this current, it will take a long time to notice).  I'm not sure offhand if C-B breakdown causes degradation, but E-B breakdown is well known to.

Anyway, draw the equivalent circuit.  The transistor is in inverted operation, and depending on how much current diverts to the base resistor and inverted-collector current, versus breakdown, that will be your answer.



Inverted hFE is typically quite low, and rarely specified.  (It's rarely representative in SPICE models either -- can go looking for it, but it's probably not much help.)  To avoid breakdown, there needs to be enough Ie + Ib to sink the diode's leakage at Vb = Veco - Vcb, about 3.3V.

At 3.3V, 24k sinks 137uA, so even at hFE = 0, it's unlikely this causes breakdown.

Tim