# A question on how the transistor works

Now I am not getting how the second equation is made. Because how can you measure VBC? Because then it will be like measuring the potential difference between two split wires... Something like this...

So is that possible?

• Just to be sure, you're not trying to interpret the transistor symbol as a wire connecting the collector and base are you? Commented Jan 29, 2014 at 12:23
• No!!I am not trying to think what you said... Commented Jan 29, 2014 at 12:30
• Why, then, do you think that measuring V_BC is like measuring "between two splitted wires..."? Commented Jan 29, 2014 at 12:33

You can measure the voltage between any two points. For example, I could measure $V_{kitchensink,tongue}$ like this:

1. place positive lead of voltmeter on kitchen sink
2. place negative lead of voltmeter on tongue

While this is probably not a meaningful measurement, it is defined anyway, because my tongue exists at some electric potential, and my kitchen sink exists at some other electric potential, and the difference between them is the "voltage", which is more accurately called the electric potential difference.

• Right but incomplete. It would be good to point out that therefore going around any circle will yield net 0 volts. That is exactly the problem the OP has. He is measuring the voltages between the three leads of a transistor in a circle an asking why it must always be 0. Commented Jan 29, 2014 at 13:22
• @OlinLathrop True. I think the OP is pretty confused on a number of points, so it's hard to be complete. I'm a little surprised he accepted my answer so quickly...certainly more complete answers are possible. Commented Jan 29, 2014 at 13:26

Here's a better picture I think (NPN transistor): -

It's fairly easy to see that: -

$V_{BE}+ V_{CB}=V_{CE}$ and therefore the 2nd equation in the question = zero.

They're all very easy to measure with a multimeter.

• No, you have the polarity of Vce shown backwards in the diagram and in your equation. The OP's second equation and the one you show disagree. Yours would be Vbe + Vcb - Vce = 0. Note the difference in sign on Vce. Commented Jan 29, 2014 at 13:02
• @OlinLathrop I too thought about the same thing... Commented Jan 29, 2014 at 13:09
• @OlinLathrop Andy's equation looks fine to me: If you take the OP's equation and note that Veb = -Vbe and that Vbc = - Vcb, then rearrange, you get Andy's equation which looks more intuitive to me. Note that the sign and letter reversal was also done in the diagram in the OP's question.
– Tut
Commented Jan 29, 2014 at 13:29
• Commented Jan 29, 2014 at 13:29
• @PhilFrost Yes, that's what I was taught. Vxy would be measured with the positive lead on x. Vyx would be measured with the positive lead on y.
– Tut
Commented Jan 29, 2014 at 13:34

Transistor is not like two splitted wires as Alfred mentioned in his comment. You should see trnsistor equivalent circuits for more clarification.

The second equation in your diagram is an example of one of Kirchoff's circuit laws, namely, Kirchoff's voltage law (KVL). Generally stated as, the sum of all voltages around a loop is zero.

Intuitively, imagine voltage as a ladder, each ladder step represents an increase in voltage. Start at one step on the ladder. This is your reference, let us call it point A. First, count up four steps to point B. Then from B, count another 3 steps up to point C. In total you will have counted up 7 steps. If you now measure from point C back to the original step (A), it will be -7 steps. Thus, if you add all these measurements together, you will get zero. This works no matter how many times you go up and down, or how many steps you take each time you move, when you loop back to the original point the sum of all measurements will equal zero.

In this instance, it is easy to make a loop. Think of the reference as the base, point B in the ladder as being the collector, and ladder point C as the emitter.

1. Start at the base, and measure the voltage to the emitter. V_EB.
2. You are now at the emitter, measure to the collector. V_CE.
3. You are now at the collector, close the loop by returning to the base. V_BC.
4. From KVL, V_EB + V_CE + V_BC = 0.

The measurement between the base and the emitter is not analagous with the measurement in your second diagram, because the base is not short circuited with the emitter. Because of this, a voltage difference may exist between these two points. However, it is possible to make the measurement shown in the second diagram.

A voltmeter essentially tells you the difference in voltage between two points, one point with respect to an other. A measurement and a reference point. Often, voltage measurements are made with respect to ground, but this need not be the case. If you place both reference and measurement in the same place, the potential difference in between those two places, the voltage, will be 0.

Lets start at a very unsophisticated level. You ask, how can there be a VBC? Its because there is a diode junction between the Base and the Collector. Its not a split wire. Its a semiconductor diode, just like VEB

Remember that your basic bipolar transistor is a three layer object with TWO diode junctions: They're back to back, so NPN means, literally, N--|<|--P--|>|--N. or Cathode-Anode-Cathode, or Emitter Base Collector. E--|<|--B--|>|--C.

PNP means opposite doping, the base is the single cathode, same physical structure: P--|>|--N--|<|--P

You can verify this with a buzzlight, a 3V continuity tester with an incandescent lamp and two conventional alkaline cells. They're VERY useful tools for understanding small signal circuits and physics experiment. You can sort out random bipolar transistors by which directions they conduct and which they do not, too.

Back to the NPN transistor, there are two junctions: One, Base to Emitter, is forward biased to turn-on the transistor, less-than-forward biased to turn it off.

The other, Base to Collector, is reverse biased, dead off.

N--|<|--P--|>|--N = E--|<|--B--|>|--C

So if you hook up a 3V battery, a load (that buzzlight lamp will work) and a biplolar transistor to act as a switch, you'll get:

gnd---- E--|<|--B--|>|--C ---(-\/\/\-)--- ---[+ -][+ -]----gnd

VEB = Vbatt, 3V for talking purposes. You can measure the voltage at the Base, but it is the result of leakage currents and cosmic rays. Lets put a nice 360 ohm resistor on B and connect it to ground. What happens?

Nothing. VBC is 3.0V, hard-back-biasing the Base-Collector junction, and making both sides of the Emitter-Base junction the same voltage. No meaningful current is flowing. Lamp off.

Now move the resistor so it connects between the Base and the +3 battery output. Oh, now the lamp lights! Why?

Briefly: Voltage at the base rises until the Base-Emitter junction turns on. When the Base-Emitter junction turns on, current flows, just like any 360 ohm resistor and semiconductor diode. (lets say Silicon for the moment). You can work out the current, Vf across both diode junctions is 0.7v, approximately.

But here's the quantum mechanical thing: The Base is so physically small, so thin, that when electrons flow from ground to the Emitter and then into the Base, to the 360 Ohm resistor, and then the + terminal, why, some of the electrons pass all the way through the base and find themselves in the COLLECTOR! They don't slow down, they go on to the lamp, through it, and into the + terminal on the battery. In fact, MOST electrons flowing through the emitter do that.

Now the tricky part: What's VBC? Its the voltage at the collector, relative to the base. The base is biased up to +3 with a resistor, the collector is connected to +3 by the load (lamp). So there is no, first order, connection BC to measure. There IS a VBC, but its the product of

VEB, more or less a forward biased silicon diode, 0.7V with VOM black lead at Emitter and the VOM red lead at Base;

subtracted from

VEC, black lead at Emitter, red lead at Collector, which is the voltage at the middle of the divider formed by RLoad and ROn of the transistor, across the 0 (ground) to 3V.

Its NOT a forward diode drop, because there is NO diode junction between the Emitter and the Collector... They are VIRTUALLY CONNECTED. So potential at the Collector is closer to that at the Emitter (0V) than anything else. It is, no doubt, a lower value, relative to the Emitter, than the Base, so VBC is going to be a negative number, although VEB is a positive 0.7V. Or so.

Note that the 360 Ohm resistor isn't accidental. Try 360K, Probably won't light the lamp. Your typical, conversational, bipolar transistor, has a 'HFE' or 'beta' or forward current gain, of about 50. If the lamp is about 10 Ohms, the current is about 0.3A, and 1/50 of 0.3A is 0.006. 3V - .7V = 2.3V. 2.3V/0.006 = 383 Ohms. 360 Ohms is slightly less, less Ohms, more amps, make sure the transistor is ON.

Does that explain how VBE can be measured? Have you built a circuit and tried measuring VBE? That's probably the best way to study this sort of thing.