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
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
