General > General Technical Chat

BJT NPN: Apparently Ohm's Law Does Not Apply

(1/3) > >>

mbennett555:
Hello to everyone.  FYI, this is my first post to this forum. 

I believe I know a bit about electronics, and specifically bipolar junction transistors.  But I'm confused about something.  It's something that's confused me since I was at university years ago (I have a BS degree in Electrical Engineering).  It has to do with the voltages and currents observed when a BJT NPN is in saturation, with its emitter connected to ground.

There are 2 images in this post, and they're described below.
•   Image 1 shows the circuit diagram.
•   Image 2 shows:  the measurements I made, in a table (see the bottom-most row); and a zoomed-in view of a portion the circuit, namely the NPN transistor.

I do not understand why, as shown in Image 2, VCB= -572 mV.  According to Ohm's Law, that means there must be a current that flows from B to C, over some resistance.  But we can see from the measurements that there is a current IC= IE= 5 mA that flows from C to E.  How can there be a current that flows from B to C?  (Also as shown on Image 2, there is a current IB= 50 uA that flows into the base.  I believe this current can be neglected, since its value is 1/100 of IC.)

 Circuit_BJT NPN_1-NPN_202108_B1.pdf (68.09 kB - downloaded 98 times.)
 Circuit_BJT NPN_1-NPN_Measurements_202108_B1.pdf (1909.23 kB - downloaded 88 times.)
 Circuit_BJT NPN_1-NPN_P01= CircuitDiag_P02= Measurements_202108_B1.pdf (1973.39 kB - downloaded 96 times.)

bob91343:
I didn't download the files but when you draw reverse current through the BE junction there will be some laser operation and thus the collector voltage goes negative.  Or something like that.  Bob Pease explained it.

amyk:
The one-sentence answer is "because a BJT is not ohmic".

magic:

--- Quote from: mbennett555 on August 22, 2021, 07:18:38 pm ---According to Ohm's Law, that means there must be a current that flows from B to C, over some resistance.
--- End quote ---
Precisely according to Ohm's law, if there is a straight metal connection between B and C then it has a resistance and a voltage across those point causes a proportional current to flow.

In practice that's a PN diode so Ohm's law doesn't quite apply. But you are right, the BC junction becomes forward biased in saturation and steals base current away from the BE junction so a transistor can't saturate down to zero or negative Vce.

As for how current flows from 50mV collector to the emitter despite the 600mV base in between, that's because BE junction voltage attracts charge carriers (here electrons) from the emitter to the base layer which is thin enough that 99% of electrons ram into the collector before they get a chance to exit through the base terminal. A BJT with the sort of proportions that you have drawn wouldn't work very well (like, maybe β=1 if you are lucky).

antenna:
I don't think there is a current flowing from the collector to the base lead, aside from the minor current your volt meter is using to read the potential difference across the junction.  If you put your voltmeter across a battery that has nothing else connected to it, you still can read the voltage, because that voltage is now flowing through the 1MΩ resistor inside the voltmeter (ohms law satisfied). When the transistor is saturated, the B-E junction becomes forward biased and becomes conductive, and in the process, the holes in the p-type material are filled with electrons. Now, the p-type material at the base has an excess amount of electrons filling the holes in the B-C depletion region and the voltage across C-E can pull them through. There is a current between B and C, but the electrons are not coming from the base lead, they are coming from the emitter and going straight through the forward-biased junction to the P material at the base connection and filling the depletion region holes.  I've read somewhere, that the reason you cannot make a transistor with discrete diodes is because the electrons that are being attracted to the p-type material by the voltage at the base lead will not have the momentum to make it down the wire into the next diode junction to fill that junctions holes, which is why they need to be sandwiched together with a very thin p layer.

This video has a great explanation of the inner workings of a transistor if you are interested in how the semiconductor material behaves. https://youtu.be/J4oO7PT_nzQ

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod