# What does (PNP) BJT with shorted C-E do?

I was browsing through the TI datasheet for an LM78L05 and noticed this application schematic:

Note how Q2 has its collector and emitter shorted. I can't say I've ever seen that before and search didn't turn up anything.

What role would Q2 play in that configuration?

I kind of suspect a diode, but can't figure out why a plain old diode wouldn't work better and be a lot cheaper. The 2N4033 datasheet describes it as a General Purpose PNP Silicon Planar RF transistor.

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Taking a wild guess, it provides short circuit protection? Q1 seems to be a typical pass transistor, so Q2 would need to be that. How it works that way is the question A quick google shows a different layout for the same though, 2.bp.blogspot.com/-PKnSJB0ZxGw/T_aA_TxF2VI/AAAAAAAAAq4/… Good question. – Passerby Feb 19 at 4:47
It would act like a diode. If the LM78L05 stops sinking current, Q1 will turn off. – mkeith Feb 19 at 4:50
Funny, everyone else has the second layout, fairchild, st, onsemi. Only TI has that layout, and only on the lm78l05 data sheet? Their LM340 data sheet has the second layout. Figure 31 Page 15 ti.com/lit/ds/symlink/lm340-n.pdf Maybe that's just an uncaught error? – Passerby Feb 19 at 5:02
This layout is as old as the LM78L05. The 1980 Nat Semi Voltage Regulator Handbook has it. So TI just incorporated it into their data sheet from their buyout of National. – Passerby Feb 19 at 5:35

I think they goofed. Collector shorted to base is more common, more logical, and probably more accurate and more reliable. If you disconnect their collector from emitter and connect it to base, you get a current mirror or current multiplier. Google "current mirror". (On this topic, ignore the Wikipedia article.) You will see schematics of variations using two BJTs: two NPNs on the 0V or -V rail, or two PNPs at the +V rail. (But not many give practical applications like this power booster.) The scaling factor is decided by the ratio of the two emitter resistors. But the accuracy of the scaling is controlled by the VBE match. For the best VBE match, the transistors should be the same type, and their temperatures should be kept close, by mounting them on the same heat sink (even though Q1 has very little dissipation). Of course a plain diode works, but the match is not as good. Putting the plain diode on the heat sink with the transistor might be an improvement.

Re-drawing their circuit makes it more obvious what is going on. Q2 & R2 reduce the input voltage to the regulator, in order to measure the current it is pulling (most of which goes to the load). Q1 & R1 route 4 times the Q2 current around the regulator to the load. The regulator still regulates +5V on the load, even though 80% of the current is delivered via Q1. (R3 is more subtle. It reduces Q1's share of the load current when load current is small. The regulator also sends some current to ground. Without R3, the current mirror multiplies that current too, which would cause output voltage to exceed +5V, a disaster. With this deliberate imbalance, one could argue that the precision of the VBE match is not as important, so a matching transistor at Q2 is not as important, so diode or wrongly connected transistor is not a problem.)

simulate this circuit – Schematic created using CircuitLab

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This is a great answer! – Passerby Feb 19 at 16:49
I renamed the transistors to reflect Op's circuit, to avoid confusion. – Passerby Feb 19 at 16:59
Collector shorted to base makes a diode with a very low reverse breakdown voltage, – ilkhd Feb 19 at 17:21
The numbering comes from the short circuit protection being an add on to the original external pass transistor circuit, instead from scratch. They did have it top to bottom numbering though. – Passerby Feb 20 at 5:56
Redrawing clarifies simple circuits. Voltage divides and current flows top-to-bottom. Left drives and right receives. Number things left-to-right and top-to-bottom. The numbering in the app note disappoints me, in that R2 & Q2 convert drawn current into a voltage (relative to + rail), which voltage drives Q1 & R1 to push a multiple of that current. So Q2 is really the "first" transistor in THIS circuit. – A876 Feb 20 at 6:08

From the 1980 National Semiconductor Linear Regulator Handbook, section 7.1.3 has a High Current Regulator with Short Circuit Limit During Output Shorts, in an identical layout, but with Q2 being a simple Diode D.

This current boost circuit takes advantage of the internal current limiting characteristics of the regulator to provide short-circuit current protection for the booster as well. The regulator and $Q1$ share load current in the ratio set between $R2$ and $R1$ if $V_d = V_{be}(Q1)$

$I1 = \dfrac{R2}{R1} \cdot I_{REG}$

During output shorts

$I1(sc) = \dfrac{R2}{R1} \cdot I_{REG}(sc)$

If the regulator and $Q1$ have the same thermal resistance $0jC$ and the pass transistor heat sink has $R2/R1$ times the capacity of the regulator heat sink, the thermal protection (shutdown) of the regulator will also be extended to $Q1$. Some suggested transistors are listed below.

The minimum input-to-output voltage differential of the regulator circuit is increased by a diode drop plus the Vr1 drop.

Considering the identical layout, and NatSemi being the source of the layouts, the shorted Q2 PNP C-E will act the same. As @Robherc suggests, it's likely used as a matched pair, to provided some performance gain compared to a random diode which would have a much different performance. Unmatched, I suspect the different IV curves can lead to over or under current conditions, or too much cycling/oscillation. Of course, given that the application note suggests a diode, that's probably not the case.

### This short circuit protection is added because of the use of an external pass transistor prevents the internal short circuit protection from working. It could just be omitted, if short circuit protection is not needed.

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I love that handbook you linked to Passerby! The applications section in there is terrific. Who knew you could build a switching converter with an LM317? :) – scanny Feb 19 at 7:00
@scanny, well, anyone who grabbed the LM117 (which is the LM317, too) datasheet when it was still a National Semiconductor publication and stashed it on their computer, even though not knowing that this was a great idea as TI was going to ruin all the datasheets when they ate Nat Semi... – Ecnerwal Feb 19 at 17:10

My guess is that they're using the C-E shorted transistor to compensate/balance the B-E offset voltage of Q1.

While a diode could technically accomplish the same function, using a matched transistor should give a more similar response.

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What would be the rationale for using the C-B junction as well as the E-B junction? Would that be likely to have the same voltage drop but allow more current or something? – scanny Feb 19 at 5:02
Still guessing, but to me shorting C-E together in that installation may be for protecting the transistor from damage if a 'floating lead' picks ul a harmful voltage, or damage that purportedly can be caused by running a transistkr with too much base current vs collector-emitter current (I remember back in my early EE days reading textbooks warning of premature death from oversignalling transistors without enough avail. C-E current to satisfy gain). Or possibly it's just to 'tie up loose ends' and not leave a floating lead. – Robherc KV5ROB Feb 19 at 5:11
@RobhercKV5ROB This is odd, because this is exactly how BJT switches work; they are deeply saturated, with the base current way higher than minimal current required for the required collector current. – ilkhd Feb 19 at 17:29
@scanny the rationale could be prevention of the breakdown of the E-B junction when biased wrong way. – ilkhd Feb 19 at 17:31

A transistor is kind of like two diodes paralleled when you short the C and E together. I have heard of using NPN's as diodes with just the NP (but why do that when you can get a diode? I think I remember trying this when I was a kid experimenting with electronics. I have never used them in the configuration of the question schematic.

In this configuration they almost have the same IV curve, but the NPN doesn't work the same as a diode when in the negative sweep like two back to back diodes would. Notice all the nodes have the same curve except 2 and 4. I can't speak for the real world configuration as I haven't actually used a transistor like this but it did almost what I thought it would.

simulate this circuit – Schematic created using CircuitLab

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Both transistors here are PNP. – Passerby Feb 19 at 6:04
You'll still end up with two diodes in parallel – laptop2d Feb 19 at 6:21
Sure, but proper documentation and symbols, and polarity is important. – Passerby Feb 19 at 6:26
In the interest of time I get lazy – laptop2d Feb 19 at 19:59

A transistor connected like this acts as a diode with SUPER fast on and off times, as well as super low forward resistance.

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Why do you say that? Have you any citations or explanation? I'd be inclined to believe such a configuration would have storage time on the same order as a regular diode, given the scale/geometry of the PN junctions, and that it's forward voltage would also be similar given a Silicon PN junction is still at work. – scanny Feb 24 at 19:34