Transistors - what is the collector/emitter voltage drop?

KeithRB: Don't get me started about Arduino books. I have half-a-dozen of them, and only -one- mentions millis().

Nick Gammon: That is a subpage of a university professor, sadly. I see other more subtle misleading things on a quick look at his website. The information he has on a JFET is specious, at best.

Arduino's native libraries is not what i'd call huge, there should not be an excuse to miss out any commonly used functions.

I think this comes from over thinking the transistor's equivalent circuit of looking like two diodes. Yes it looks like two diodes when you probe it with a resistance meter but it is not two diodes. You can not make a transistor from two diodes it simply would not work. This is because the only reason a transistor works is that the whole structure is build out of a single crystal.
The fact that there is a voltage between the collector and emitter even when the transistor is as hard on as it can get is an inevitable consequence of how the transistor works. Unlike a FET that does not have any saturation voltage but a saturation resistance.

It is all well and good pontificating about "errors" in web sites or books but Vce is of secondary importance and to a first approximation can be omitted from the discussion. This does not make these discussions wrong, just simplified. This happens all the time in every field of science and engineering. We still teach the idea that electrons "orbit" the nuclease in an atom where it has been known for more than half a century that this is not the case. However the idea of electron shells being in different orbits at different distances is so useful that it is still taught. It is the basis of the whole of chemistry but it is wrong.

The bottom line is don't go saying something is wrong just because it is a simplification because that is how people learn. Give them too much small scale second order irrelevant for what they are doing stuff and they will become swamped with "facts" and never learn anything. Ever thought why the miller effect in transistors is hardly ever mentioned on this forum? It is because most of the time it is not relevant but any explanation of a transistor action without mention of it would be wrong according to some.

I wasn't pontificating, Mike, I was trying to understand something.

Possibly I was wrong (it wouldn't be the first time), or I misunderstood, or the web page was wrong, or maybe misleading. I was just trying to work out which it was, and then move on to learning the next thing.

I wasn't pontificating, Mike,

I know, that was aimed at some others. :slight_smile:

So I'm a pedant. I don't like things so oversimplified that they aren't even wrong. :stuck_out_tongue:

Over simplification is sometimes the only way you can grasp something. In fact the explination of 99.99% of all physics is oversimplified to the extent of it being wrong. Take ohms law, it is still very useful but fundamentally it is wrong. But it is wrong to a degree that practically does not matter.
The same is true of Newton's laws of motion, it is wrong and an over simpilfaccition but it is good enough to fire a rocket from Earth and hit a spot within a foot or so on Mars.

If I interpret this entire discussion correctly, it got started because Nick Gammon was rather completely misled by an incorrect oversimplification, based on a stated observation that is in fact true only under some circumstances. But hey, whoever posted the original "explanation" of Vce(sat) was at least trying!

Nick,
Both the you tube and the educatior's home assignment are correct. The difference is that the utube guy demonstrates transistor acting as a switch, the professor(?) probably gives an assignment related to introduction to amplifiers. I am attaching a power point page to illustrate what I said. You can run two experiments now and post the results.
L

Arduino forum.pptx (77.8 KB)

x50505, sure, Vce(Sat) is different than Vce in linear mode .
Your ppt is OK, but, for the switch case, I think you are at the high limit for Rbase value. When I need to switch on a transistor, I look at the datasheet, there is a line which indicates the lowest Vce(SAT) and the corresponding Ic and Ib .
If these values are not available, I make the calculations, starting with the Ic I need, divide it by Hfe, which gives me the minimum Ib . Multiply that minimum Ib by 2 (or even 3 :wink: ) , and you're sure the transistor is fully saturated
Am I wrong here ?

You are right. It was pointed by the you tube man who computed 0.56 mA current and than used 1 mA to make sure the transistor is fully saturated. Also it should be noted that the Hfe can vary even between transistors selected from the same batch.

Grumpy_Mike, point taken about simplification. In this case, however, as jremington points out, oversimplification apparently was a problem in this case.

For sure, the models I use in my head to simulate circuits are extreme simplifications of reality and bear further analysis. In my head, I picture electrons being over-accelerated by the base current, requiring less pressure (voltage) to drive across the CBE junctions. Of course, that is not reality.

Of course, that is not reality.

I think about it like this. There is a hole injected into the base P region, now holes are very attractive to electrons, just like a pretty girl is attractive to men. The hole attracts the attention of lots of electrons who move in across the potential barrier, then they all rush towards the hole. The winner combines with the hole and the attraction goes away. But all the other electrons are also running flat out and have enough energy to fall over the potential barrier between base and collector..

Hah! I like that.

If the transistor is on, the Collector voltage is 1.6 volts higher than the Emitter voltage.

I didn't see any theoretical backup for this statement on that page; he claims it's an innate property of the transistor.
I might even believe that that's what he measures, with the circuit shown (100k base resistor? THAT'S not going to saturate a 2222!)
So yeah, it's pretty wrong, the the broader context.

You know what I think he's done? He measured 1.6V drop over the 220 ohm resistor (which sounds about right) and then put that figure under the resistor on the diagram.

Then looking at it later he took that as the collector voltage level, and since the emitter was zero at ground, he concluded that the 1.6V must be dropped over the collector-emitter junction.

Nick,
That was my original thought. I posted my response few days ago and quickly realizing my mistake cancelled the post. It was past midnight, so I did not try to post correct reply.

The voltage drops through all components must add to zero. The upper end of the LED is at 5 V. Lower end at 3.1V which menas that voltage drop is 1.9V through the LED. Then the lower end of the resistor shows 1.6 V. so the resistor voltage drop is 1.5V (3.1V-1.6V). We are now left with transistor. Collector shows 1.6 V and the other end is at 0V. So the resistor drop has to be 1.6V.

By the way, the voltages shown on the schematics are voltages measured with respect to ground.

Now lets add the voltage drops and see if they equal of the battery voltage: 1.9+1.5+1.6 =5 V. It is equal. Does it make sense?

To make it clear, I attach another power point file (3pgs). This time I actually run LT SPICE program and compared hand calcs with the LT SPICE output. I could run the test. Didn't do so because I remember running similar tests, displaying results on the oscilloscope and verifying the stuff by hand analysis in my lab classes.
L

Junk.pptx (135 KB)

OK, so to help me get my head around this, if the transistor is not fully saturated then there will in fact be a voltage drop between collector and emitter, as he stated.

However in the other source (the video) presumably it was fully saturated.

in the other source (the video) presumably it was fully saturated.

Yes. Or "enough saturated", anyway.
The explanation about the two junctions "canceling each other out" was a simplification, though. After all, if it was really two junctions, one of them would be reverse-biased and it wouldn't conduct at all. I think I'd have to dig out the solid state physics books to remind myself what really happens, but IIRC it's something like the inherent current carriers in the base material, TOGETHER with the carriers injected as part of the base current you're adding, combine so that the C-E overall behavior is as if the junctions had been nullified.

Referring to your original post:
"I measured 12.06 V at the +12 V point, and 10.28 at point C, giving a voltage drop of 1.78 V in circuit."
Incorrect. You measured 1.78 V voltage drop across the LED. This leaves 12V – 1.78 V = 10.22 V to be accounted for.
Now, the current at the emitter is19.2 mA. Part of this current will flow through base, so let us subtract
that current 19.2 – 19.2/100 =19 mA.
This current will flow through transistor, resistor and LED to positive battery terminal (or if you prefer the other way). Note that the current has identical value at any point along this path. Knowing current and the value of resistor I can compute voltage drop through the resistor: V = I R = .019A560 Ohm=10.64V. Voltage balance: 10.64+1.78 = 12.42V Measured voltage drop is 12.06 V. Therefore 12.06V-12.42V = - 0.36V. I suspect the 560 Ohm is a nominal value of the resistor. Did you measure the real value? This leaves 0Vdrop for the transistor.

Latest post: "However in the other source (the video) presumably it was fully saturated."
Because it was fully saturated there is no voltage drop between collector and emitter. Are you referring to 0.60 mV measured between collector and emitter? Forget it. This is so small that it can safely be assumed to be zero. As a matter of fact it may be due to measuring instrument inaccuracy or the manufacturing process.
Also do not be concerned with voltage drops between the base and emitter (about 0.7V) and the base and collector (.3V or so?). They are always present no matter what you do. They do not come into play, at least not at the sophistication level of circuitry we are dealing with.
L