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Just as an experiment, I decided to put together the following "circuit":

enter image description here

I then went ahead and measured Vbe, Vbc and Vec and this yielded the following results:

  • Vbe = 0.602 V
  • Vbc = 0.589 V
  • Vec = 0.013 V

Here is the question: As you can see from the circuit above, the Collector is not connected to anything so shouldn't Vbc be an absolute zero? Why am I measuring 0.589 V?

Thanks.

EDIT: I wanted to be more specific regarding what is really confusing me, I am not sure if this will make a difference or not but here it goes:

First let’s take the circuit mentioned above and replace the transistor with the diode equivalent analogy. I realize that replacing a transistor with two diodes is not truly equivalent but I just want to make a point.

enter image description here

Looking at the circuit above, it is clear that the diode representing the collector pin is doing absolutely nothing. As a matter of fact, it should be perfectly fine to just remove that diode and the circuit should behave exactly the same right?

enter image description here

What this reveals is that the collector plays no part whatsoever in this circuit (it should be like it is not there at all) and yet, somehow, I am getting a Vbc voltage. Is that crazy or what?

Of course, the fact is that the collector IS playing a role here, this is the only logical explanation for the reason there is a voltage across Vbc. But why is this happening? Is the diode analogy of a transistor not accurate?

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  • \$\begingroup\$ There's something wrong there. Are you measuring the 'base' voltage from the source side of the resistor? \$\endgroup\$ – Spehro Pefhany Mar 13 '15 at 3:48
  • \$\begingroup\$ @Spehro Pefhany: I am measuring the voltage directly on the pins of the transistor. I have updated the original question to reflect that fact. \$\endgroup\$ – T555 Mar 13 '15 at 3:50
  • \$\begingroup\$ Well- if you have Vbe = 5.89V you've either got the wrong kind of transistor (PNP or MOSFET or JFET or whatever rather than NPN BJT) , the transistor is damaged, or it is connected to the wrong pins! There are no other options I can think of. Vbe should be about 0.7V for reasonable base current. \$\endgroup\$ – Spehro Pefhany Mar 13 '15 at 4:28
  • \$\begingroup\$ @Spehro Pefhany: Sorry, my decimal points were on the wrong place (I fixed it now) and did't notice until you mentioned voltage across base / emitter. Still, that does't explain why the voltage across the Base and Collector is 0.589 V instead of 0 right? Or am I still missing something? \$\endgroup\$ – T555 Mar 13 '15 at 4:40
  • \$\begingroup\$ When the transistor is saturated (fully 'on') the C-E voltage will be very low (13mV in your case, with no current flowing). Since the voltages all have to add up (to zero around a loop taking direction into account- that's fundamental principle named after Gustav Kirchhoff), you must have almost the same voltage from base to collector as from base to emitter. That make sense? \$\endgroup\$ – Spehro Pefhany Mar 13 '15 at 4:52
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Okay, here's how the voltages add up:

enter image description here

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  • \$\begingroup\$ Thank you very much for being so patient trying to explain this to me. I am an electronics newbie that probably needs to get the basics first and move up to transistors later. The math you show on your drawing makes sense and it all adds up nicely. I still don't see how there is a 0.013 volts drop between the emitter and collector given that there is no voltage or current source of any kind that is connected between those two terminals (how can the transistor be saturated if there is nothing to saturate it with?)... Perhaps after a good night sleep I will see the connection tomorrow! Thanks! \$\endgroup\$ – T555 Mar 13 '15 at 6:11
  • \$\begingroup\$ The 13mV is a side effect of the current going into the base. The important thing is that if you connected a load like an LED with series resistor from the collector to some positive voltage (say 5V), the LED will turn on, because there will be only a small voltage from collector to emitter. It might be 0.1 or 0.2V with the LED rather than 13mV, but the point is that it's very small. \$\endgroup\$ – Spehro Pefhany Mar 13 '15 at 12:09
  • \$\begingroup\$ At the risk of making you go crazy trying to explain this concept to me, would you mind elaborating a little on what you mean by "The 13mV is a side effect of the current going into the base"? I am just unable to see how a current going through the base would affect the collector voltage. I also edited the original question and added some more information regarding why I find this behavior very confusing. I hope this helps clarify my question a little further. \$\endgroup\$ – T555 Mar 15 '15 at 3:38
  • \$\begingroup\$ At the risk of simplifying this a bit too much, if you look at the two-diode model and imagine a high value resistor (your meter) between collector and emitter. There's a bunch of current flowing into the base, and the C-B diode is forward biased so a bit of current flows out the collector through the meter resulting in your measuring a bit of voltage at the collector. In practice the collector is "almost" shorted to the emitter (13mV isn't much) and for most purposes you can treat 13mV ~= 0. Since collector voltage ~= emitter voltage, base-collector voltage ~= base-emitter voltage. \$\endgroup\$ – Spehro Pefhany Mar 15 '15 at 3:48
  • \$\begingroup\$ Ok, so when I plug the multimeter between Vce I am basically completing the circuit by adding a 10 megaohms resistor across Vce and because of that, current will flow through Vbe => Vec => Vcb. Ok, I got that no problem… But what about when I plug the meter between Vbc? In that case I am not closing any loops or completing any circuits am I? The voltage (in theory) should be zero right? But I think that what you are trying to tell me is that the internal construction of the transistor allows for some leakage current to flow between Vce even though ideally this should not happen? \$\endgroup\$ – T555 Mar 15 '15 at 4:09

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