0
\$\begingroup\$

If a BJT transistor has a voltage Vb at its base, its nominal emitter voltage Ve will be Ve = Vb - 0.7. If we know Re, we know the current through the emitter Ie = (Vb - 0.7)/Re.

If beta is high, nearly all of that current comes through the collector, and only about 1/beta comes through the base.

For example, in the circuit below, only 40uA comes through the base:

enter image description here

What I'm confused is:

  • When, if ever, will the transistor draw more current through the base?
  • Doesn't the BE junction behave like a diode? A diode will allow current through - shouldn't the transistor do so as well.
  • What happens if the current at the collector is limited - will the transistor draw the full Ie through the base?
  • How does the input impedance on the voltage to the base affect this?

Update

To further support my question, these lecture notes show

enter image description here

Shouldn't iB be the full current we'd get between base and emitter if we simply had a diode? If it's not, then how do we predict what iB is?

\$\endgroup\$
4

3 Answers 3

2
\$\begingroup\$

Shouldn't \$i_B\$ be the full current we'd get between base and emitter if we simply had a diode? If it's not, then how do we predict what \$i_B\$ is?

It is.

But you need to remember that \$V_{BE}\$ depends on both the base voltage and the emitter voltage.

If you tie the emitter to ground, then the base current will depend on the base voltage just like any diode.

But if you have a resistance in series with the emitter, then the emitter voltage will rise with the increase in current, and in fact, it will rise to the point where \$V_{BE}\$ is the correct value to produce a base current that is \$1/\beta\$ times the collector current. This is a form of negative feedback — an increase in collector(emitter) current causes a decrease in \$V_{BE}\$. The circuit reaches equilibrium when \$i_B = \frac{i_C}{\beta}\$.

If you externally constrain the collector current, as in your previous question, then the emitter voltage can't rise due to that current, and then the base must supply the remainder.

\$\endgroup\$
2
  • \$\begingroup\$ "If you externally constrain the collector current, as in your previous question, then the emitter voltage can't rise due to that current," - Yes, this is what I'm struggling with. Can you explain this further? It seems as if the transistor can somehow raise an emitter to the base, but only when it gets current from the collector to do so? As this is where I'm stuck, can you elaborate on this? \$\endgroup\$ Commented Jun 30 at 20:13
  • \$\begingroup\$ Is this correct: 1. It's just Ohm's law, if Ie goes up, and Re is fixed, then Ve goes up, obviously. 2. The I-V graph through the BE junction (or any diode, for that matter) is extremely high slope, so even a small increase in voltage at Ve will quickly lower current dramatically. These two combine so that just a little bit of ib current is enough to allow raise ic by beta * ib, which raises voltage ve, and therefore quickly stops Ib from growing further - the equilibrium you described. \$\endgroup\$ Commented Jun 30 at 20:34
1
\$\begingroup\$

There are two primary viewpoints for the BJT: the DC operating point and the AC behavior around that DC operating point. You must keep these ideas totally separate in your mind. The connection between them is simply that the DC operating point sets the value of the tangent line around which the BJT operates when a tiny AC signal is applied. But the AC behavior has nothing to do with the DC operating point. Nothing! So get that in mind.

Let's look at your schematic this way:

enter image description here

The first thing I can say about this is that the BJT cannot saturate. It will always be in active mode. The reason is that there is no way for the BC-junction to become forward-biased in this circuit. The base voltage cannot be higher than \$9\:\text{V}\$. Just not possible here. So the BJT is in active mode. And this means that the concept of \$\beta\$ applies well.

I don't even have to ask the question. The BJT is active, not saturated. Said and done.

This makes it very easy to work out, using KVL, all the details for the DC operating point of the circuit -- in theory. Of course, any specific physical BJT used in a real circuit will have varying details -- the \$\beta\$ can vary fairly widely, for example. (So can the saturation current.) But at least, in theory, we can write the following:

$$V_{_\text{TH}} -I_{_\text{B}}\cdot R_{_\text{TH}} -V_{_\text{BE}}-I_{_\text{E}}\cdot R_1=0\:\text{V}$$

where \$V_{_\text{TH}}=9\:\text{V}\cdot\frac{R_3}{R_2+R_3}\$ and \$R_{_\text{TH}}=\frac{R_2\,\cdot\,R_3}{R_2+R_3}\$.

Since you know the BJT is in active mode -- and I cannot emphasize it enough that you should be able to strongly agree with me on this point -- then as I said \$\beta\$ applies and we can say that \$I_{_\text{E}}=I_{_\text{B}}\cdot \left(\beta+1\right)\$. That allows us to substitute into the above equation, re-arrange, and find:

$$I_{_\text{B}}=\frac{V_{_\text{TH}}-V_{_\text{BE}}}{R_{_\text{TH}}+\left(\beta+1\right)\cdot\,R_1}$$

You will find that Spice programs will reach a value close to that, assuming you look at the BJT model details being applied by the program.

From the above, you can find all of the node voltages for the DC operating point.

Now, let's get to your updated portion of your question, given at the bottom.

enter image description here

None of that disagrees with the KVL I applied earlier. So it is consistent.

\$\endgroup\$
0
\$\begingroup\$
  • The transistor will draw more current from the base if you disconnect its collector.
  • Yes, the BE junction behaves like a diode but here its cathode follows the anode, so the current is too small.
  • You cannot limit the collector current (e.g., by inserting a collector resistor) since the collector behaves as a current source.
  • If you mean the internal resistance of the input voltage source, it does not noticeably affect the circuit operation because of the high input resistance (virtually increased by the following emitter voltage).
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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