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I'm unable to find a definitive answer to this (trust me, I've looked, unless it's right under my snout) so please do point out the correct answer if I have. I'll feel foolish but at least then I'll know!

This isn't about mirrors, etc. but more specifically something I must have missed although I'm sure it must have something to do with output impedance, I can't wrap my head around the idea that we always (seem) to use NPN transistors (or N-type mosfets) to sink current from a load and vice versa. Even writing this, I'm starting to question if I have that the right way around... My poor noodle is all a tizzy over this.

Assuming the device is saturated (BJT) then all the current flows through the circuit equally with a little bit extra coming via the base-emitter junction to drive it. With a beta of 100 that's probably a fairly insignificant quantity.

My best guess was this would apply with split rail voltages (referenced to ground) but I've seen NPN and PNPs used as sources and sinks in single rail applications without explanation and that's where I'm stumped.

What am I missing here? It must be so obvious that I can't see it.

Marc Draco
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    In normal conditions when the NPN transistor emitter is at GND and the load at the collector side. The collector current can only flow from "collector to emitter". The collector can only "sink" current. But if we put a load between the emitter and GND and collector to the power supply. Now notice that now the emitter can "source" the current and cannot "sink" current (current can only flow out of the emitter and never flow into the emitter. – G36 Aug 25 '21 at 20:44
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    I can use an NPN to make a current sourcing or a current sinking circuit. The same applies to a PNP. So your question's statement that PNPs source and NPNs sink is **wrong**. – Bimpelrekkie Aug 25 '21 at 20:44
  • And this statement is true "PNP transistors source current and NPN transistors sink it". In the case when NPN or PNP is connecter as a CE amplifier. Emitter at GND load at collector for NPN transistor. And for PNP when the emitter is at Vcc and the load is at the collector side. – G36 Aug 25 '21 at 20:48
  • @marco, draw a schematic with the tool (edit your question) or find a circuit that matches your description. – Voltage Spike Aug 25 '21 at 21:19
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    they are used as they are because one lead needs to be a lower voltage than the other, which lends itself to sink/source but doesn't mandate it per-se. – dandavis Aug 25 '21 at 21:35
  • Apologies to anyone who thought I was a complete novice. The accepted answer filled in the blank I was looking for. The devil is in the detail - and the question - rather like the way that all (AF) bets are off when designing PCBs carrying high-frequency signals near to sensitive components - which, is incidentally, what I'm learning at the moment: looking at fields rather than current and allowing for the skin effect. "Energy travels through the dielectric" Rick Hartley. Fascinating stuff. – Marc Draco Aug 26 '21 at 01:17
  • @dandavis pedantically, they can work in reverse active mode (and I am sure reverse saturation must be possible). – user253751 Aug 26 '21 at 09:04

3 Answers3

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Have you ever heard of an emitter follower?

An NPN transistor connected in the emitter follower configuration sources current:

This figure is from the Wikipedia page on the common collector (emitter follower) circuit:

enter image description here

It sources current just fine.

This emitter follower (also from the Wikipedia page) made with a PNP transistor sinks current:

enter image description here

Your assumption that an NPN always sinks current and a PNP always sources current is incorrect.

The transistors can do either. How they are connected makes the difference.

JRE
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    In fact, they always do both. All the current that goes in one place goes out another place. – user253751 Aug 25 '21 at 20:51
  • I'll have to find somewhere that shows this in more detail and I know most "basic" circuits. I do know how to Google and I do own a (rather worn) copy of Art of Electronics (2nd Edition). I guess this must be more to do with convention than anything else since Kirchhoff's current law has to apply - I learned that one >30 years ago. – Marc Draco Aug 26 '21 at 01:04
  • Such a shame this wasn't the accepted answer. – Simon Fitch Aug 26 '21 at 04:31
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    It would be worth pointing out the characteristics of the circuit are different! – user253751 Aug 26 '21 at 09:05
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    This doesn't really address OP's confusion, though. In control circuits OP's point is generally true - that NPN and PNP are used for sinking and sourcing, respectively, and for good reasons - you can't *generally* build circuits the other way around that work. I think this answer fails to appreciate the fields where OP's oversimplified claim is generally true - anyone working in controls would generally live by the same rule of thumb. These emitter followers are special cases with hard limitations that make them unsuitable in many (most?) applications where you need to sink or source current. – J... Aug 26 '21 at 12:07
  • @J...: Emitter followers are hardly rare or obscure. TTL logic gates are built entirely out of NPN transistors, but actively source current as well as sinking it, and a 741 op amp uses NPN transistors to source current and PNP transistors to sink it. – supercat Aug 26 '21 at 20:36
  • @supercat I didn't say they were rare or obsure - in fact I said the notion of "NPN=sink" and "PNP=source" was *oversimplified*. It is, nevertheless, a real and very present rule of thumb **in certain fields**, particularly in automation and logic controller I/O, where you will rarely, if ever, see it broken. For people whose electronic experience does not extend beyond the sphere of that field, this is the only rule they know, and it exists for a very simple and obvious reason (namely, that with the emitter pinned to either Vcc.(PNP) or ground (NPN) you produce a simple and reliable switch). – J... Aug 26 '21 at 20:55
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This is because it's more typical to connect the load to the collector. You can use the emitter as an output, but that has certain limitations...the voltage gain will always be 1 in an analog emitter follower configuration, and in either an analog or switching configuration, driving the load from the emitter requires either a compromise in output range due to B-E drop, or the addition of some kind of boost circuitry to produce base voltages outside the supply rails. Sometimes the higher mobility of NPN (or N-channel) devices makes it worth implementing a boost mechanism, but most designs use complementary collector driving schemes.

Cristobol Polychronopolis
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  • Thank you Christobol, that makes a lot more sense. Perhaps I gave the other folks here the idea that I was more of a noob than I am, which is entirely my fault of course. I just got this idea into my head that I was missing something (i.e. what you've pointed out) and I couldn't see the wood for the major deciduous forest. – Marc Draco Aug 26 '21 at 01:07
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I think the only correct way is to go back to the electrical equivalent of a BJT (or mosfet) in different configurations. Reading those (a bit hard... I agree..) formulas You can see that You can have different practical behaviours.

So engineers will choose formulas based on the scope: You will use a different equivalent circuit with wanted parameters, (gain / output voltage / current output / output impedance and so on..). Chosen the required parameters, you will have to adopt the corresponding circuit (common emitter/collector or even base)

to sum up: you use a circuit not reasoning how current flows ( up/down/whatever...) but once decided the application of your circuit that fits better.

ingconti
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