0
\$\begingroup\$

In this very interesting introductive video https://www.youtube.com/watch?v=CkX8SkTgB0g, I learned that a transistor is basically nothing more than a resistor whose value changes accordingly to what's applied at the Base.

So it should be possible to draw a characteristic curve of a NPN transistor with:

  • x-axis = voltage applied to Base
  • y-axis = value of the resistance between Collector and Emitter

Here is what I imagined (Am I totally wrong, only a little bit wrong, or nearly correct?) with my current understanding of a NPN transistor :

enter image description here

I haven't found such a curve, even in the datasheet of the transistor I have (2N3904) http://www.ece.rice.edu/~jdw/data_sheets/2N3904.pdf

Is it relevant to draw such a curve? Where can I find one for 2N3094?

\$\endgroup\$
  • 2
    \$\begingroup\$ That isn't how a transistor works. The article you read/saw is dumbing down things. \$\endgroup\$ – Andy aka Feb 13 '15 at 10:40
  • \$\begingroup\$ It's just transistor man at work \$\endgroup\$ – Spehro Pefhany Feb 13 '15 at 10:57
  • \$\begingroup\$ @SpehroPefhany So even if my understanding was wrong, it is more or less true, as a rule of thumb, that the higher the current or voltage in the base, the less resistance transistor man applies between collector and emitter? \$\endgroup\$ – Basj Feb 13 '15 at 11:11
  • \$\begingroup\$ @Basj No, the higher the current through the base, the more current 'transistor man' allows to flow between collector and emitter. Notice the second ammeter in the drawing; if he was just varying resistance in response to base current, he wouldn't have to observe C-E current at all! \$\endgroup\$ – Nick Johnson Feb 13 '15 at 12:44
  • \$\begingroup\$ @Basj What Nick said is correct. The resistance knob he's turning is just a way of getting a certain current. BTW, I hear Win Hill has eliminated transistor man in the new edition of AoE (on the shelves soon). \$\endgroup\$ – Spehro Pefhany Feb 13 '15 at 16:09
3
\$\begingroup\$

This is a more suitable diagram of a BJT's voltage and current characteristic: -

enter image description here

In the picture there are several curves; each for a different base current injected. So here's the first difference - base is normally depicted as being fed with a current not a voltage because this is more relevant - the BJT is likened to a device which amplifies current and has a current gain (Ic/Ib).

The next difference - imagine Ib was held at 150uA and you varied Vce and looked at Ic. The curve is right there in the picture above. If it were a resistive characteristic you'd see Ic rise linearly with applied Vce. As you can hopefully see, this is far from the case.

Typically once Vce has reached about 2 or 3 volts, Ic is very flat for quite significant changes in Vce - between 5 and 15 volts applied, Ic varies from about 16mA to about 17mA.

This is because a BJT acts like a current amplifier - a fixed base current of 150uA produces a largely fixed collector current of about 17mA i.e it has a current gain of about 113.

For a base current of 200uA, the collector current is about 23mA i.e. there is a current gain of about 115 - pretty much the same for the base current of 150uA.

Hopefully you can understand it's not a simple as the article seems to say.

\$\endgroup\$
  • \$\begingroup\$ "base is normally depicted as being fed with a current not a voltage because this is more relevant". Andy, what do you mean with this statement (more relevant)? Physically, the BJT is voltage-controlled. That´s a fact! \$\endgroup\$ – LvW Feb 13 '15 at 12:11
  • \$\begingroup\$ Andy,to avoid misunderstandings: With "control" I mean "physically determined" (cause and result). Of course, during circuit design we can proceed assuming that Ib would determine Ic (but I doubt if it helps at all!). Therefore, my desire to clarify the meaning of the term "more relevant". \$\endgroup\$ – LvW Feb 13 '15 at 14:23
  • \$\begingroup\$ @lvw yes it is base voltage that drives the bjt but it's more relevant to the op, at his level of understanding, to present the current gain scenario. I guess my answer is also dumbing down things a bit but I was hopefully careful enough with my words not to be downright incorrect! \$\endgroup\$ – Andy aka Feb 13 '15 at 14:49
  • \$\begingroup\$ @LvW if we started considering the drive to the base being a voltage we would have to recognize that temperature effects will lead to a drifting base current with time and then how could I easily explain the collector current varying slightly due to the base current has changing fractionally? If the op seeks understanding on this subtlety he will no doubt ask. \$\endgroup\$ – Andy aka Feb 13 '15 at 14:53
  • \$\begingroup\$ OK - I see your point. On the other hand: If somebody knows what happens when a voltage is applied across the pn junction of a simple diode - should it be a problem for him to assume that a very similar effect holds also for the B-E junction? \$\endgroup\$ – LvW Feb 13 '15 at 15:40
1
\$\begingroup\$

No - it is not true that a bipolar transistor (BJT) would be identical to "a resistor whose value changes accordingly to what's applied at the Base.". Therefore, the graph as shown has absolutely no meaning. If the BJT would work like such a simple resistor, the current Ic would be linearly depend on the applied voltage VCE. And that`s not the case. In contrary, the current Ic does depend only very little on VCE.

This over-simplified model totally ignores the difference between a static resistor (available as a discrete part) and the dynamic resistance of a non-linear device. A BJT is such a strong non-linear device, which acts as a base voltage-controlled current source, which - however- is non-ideal. An ideal current source would have an infinite source resistance. However, the emitter-collector path resembles such a current source having a finite dynamic source resistance (roughly 10...50 kOhms).

\$\endgroup\$
  • \$\begingroup\$ What do you mean by "the graph has no meaning"?. It is always posible to set Vbase = 0V, 0.1V, 0.2V, ..., 5V etc. and then for each to measure the resistance between collector and emitter, and then to draw a graph, no? So what do you mean by "no meaning"? \$\endgroup\$ – Basj Feb 13 '15 at 11:14
  • \$\begingroup\$ A BJT is a base current controlled current source. \$\endgroup\$ – Nick Johnson Feb 13 '15 at 11:42
  • \$\begingroup\$ @Basj How do you measure resistance? By putting a known voltage across it and measuring the current (or vice-versa). But a BJT's collector-emitter current will vary very little with applied voltage, so you'll get different values for resistance depending on the voltage you apply. \$\endgroup\$ – Nick Johnson Feb 13 '15 at 11:44
  • \$\begingroup\$ @NickJohnson, this question has been discussed extensively - also in this forum. Please, don`t confuse the questioner with such remarks (A BJT is a base current controlled current source.) Instead, read contributions from Berkeley, Stanford, MIT,... \$\endgroup\$ – LvW Feb 13 '15 at 12:07
  • 2
    \$\begingroup\$ @LvW First, I'd appreciate an end to the condescending tone. Saying "read contributions from Berkeley, Stanford, MIT" serves no purpose other than to make you feel superior; it certainly doesn't inform. While it's possible to look at a BJT as being voltage controlled, it's not terribly useful or informative - the relationship to voltage is exponential since the BE junction acts like a diode, while it's more or less linear with current, which is why a BJT is characterised by its gain in this fashion. \$\endgroup\$ – Nick Johnson Feb 13 '15 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.