Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just read some first pages of "The art of electronics - Paul Horowitz". In chapter 2 transistor it says there are four properties of an NPN transistor (for PNP, it is reversed).

The 2nd property say:

The base-emitter and base-collector circuits behave like diodes. Normally the base-emitter diode is conducting and the base-collector diode is reverse-biased.

Then it says:

Note particularly the effect of property 2. This means you can't go sticking a voltage across the base-emitter terminals, because an enormous current will flow if the base is more positive than the emitter by more than about 0.6 to 0.8 volt.

I don't understand why? Current flow from base to emitter because the base-emitter is conducting-diode so why can't I stick a voltage on those two terminals. If I don't apply a voltage, how can there be a current flowing?


because an enormous current will flow if the base is more positive than the emitter by more than about 0.6 to 0.8 volt

What does this explanation mean? Why is the explanation that a voltage can't be applied to the base-emitter terminal?

share|improve this question
It just says, that without limiting the current (with a resistor for example) flowing through base-emitter junction the junction effectively creates a short circuit to ground. Because it behaves as a "normal" diode. – Golaž Aug 14 '14 at 12:48
Of course you can do whatever you like with your own transistor, but if you put 3V from base to emitter of a small NPN transistor it will be destroyed very quickly because several amperes will flow and excessive heating will cause irreversible damage. If you put 1K in series, then a few mA will flow and the transistor will be happy. – Spehro Pefhany Aug 14 '14 at 14:15
The book means, "You can't go sticking any RANDOM voltage across BE..." The same applies to any diode: connect BE directly to a 12V supply, it will pop like a fuse. – wbeaty Jan 23 at 8:05

So as you mentioned it says that the transistor is essentially two diodes.

You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows:

enter image description here

As the resistance across a diode is very low when this voltage is applied we can work out that the current would be extremely high: I = V/R, simple to see that the lower R is the higher the current, and this can be very damaging to the base terminal, I believe a datasheet of the particular transistor will give you more information on what size current it can take.

What this is saying is you need to have a current limiting resistor in front of the base terminal on the transistor which does exactly what its name describes, limits the current. As the voltage drop across the transistor will remain at 0.6-0.8V we can work out the size resistor we would need quite easily. R = (Vin - Vdrop)/I, 'I' being the base current that it can take, Vdrop being the voltage drop from the base to the emitter and Vin being the supply that is going into base, you also need to look at the hfe of the transistor so see if it will be able to give you the amount of current you need, which coincidentally can be limited, or 'tailored' with a resistor at the emitter pin so the transistor is less reliant on the hfe, but I am sure you will get on to that in the future!

share|improve this answer

Well, you are indeed applying voltage across B-E, and you also must limit current with a base resistor. You can find the max base current of a transistor in its datasheet.

Same story for diodes. If you want to power a LED you must include a current limiting resistor in your circuit.

share|improve this answer

The quote is poorly worded in my opinion. Of course, there must be a forward voltage across the base-emitter junction for there to be a current through.

However, once 'on', the current through can change drastically for a relatively small change in base-emitter voltage.

Thus, one must have some series resistance such that the current cannot exceed a safe amount.

Mathematically, the base current is approximately

$$i_B = \frac{I_S}{\beta}e^{\frac{v_{BE}}{V_T}}$$

In other words, the current increases exponentially with increasing voltage. A quick bit of algebra yields the following result:

  • the current doubles for a voltage increases of just about \$0.05V\$

So, controlling the base-current with a voltage source \$v_S\$ requires extremely precise voltage control.

Now, if a resistor \$R\$ is in series with the base-emitter junction, the equation for the base current becomes

$$i_B = \frac{v_S - v_{BE}}{R}$$

For a typical transistor and typical base currents, we have

$$0.6V \le v_{BE} \le 0.8V$$

Thus, the base current must be in the range

$$\frac{v_S - 0.8V}{R} \le i_B \le \frac{v_S - 0.6V}{R}$$

Moreover, when we look at the change in base current for a change in source voltage \$v_S\$, we find that, rather than the exponential relationship we had without the resistor, the relationship with the resistor is approximately linear. In fact, we have

$$\Delta i_B \approx \frac{\Delta v_S}{R}$$

for typical values of \$v_S\$ and \$R\$.

share|improve this answer

The transistor is a current controlled device. The emitter current is related to the base current as

I_e = (B+1) * I_b       ( B = beta )

In the forward bias mode for a diode ( using exponential characteristics), as soon as the voltage crosses a threshold ( about 0.7 V or so for silicon), the current value shoots up dramatically.

So if you directly plug a voltage source between the base and emitter terminals without a limiting resistor, a huge amount of current will begin to flow through the base, and since B ( beta) is typically 100 or greater for transistors in active mode, the emitter current will be even larger ( using above equation) which could damage the device.

share|improve this answer
Your description here regarding the beta of the transistor could be misleading. It the base emitter is connected up to a voltage source alone (collector is open) the base current would be equal to the emitter current. – Michael Karas Aug 14 '14 at 13:13
Actually, emitter current will drop to zero almost immediately when the base-emitter junction burns out and explodes the transistor. – JRE Aug 14 '14 at 14:45
@Plutonium smuggler: I see a kind of contradiction in your response. At first, you state that the transistor would be a current-controlled device (which is not true!) and according to the next sentence it is the B-E voltage which causes a dramatic increase (beyond 0.7 V). Can you clarify? – LvW Aug 15 '14 at 7:06
@LvW. What I mean is that although in a diode, the current is controlled by a voltage. But seeing a bigger picture (transistor as a whole), the emitter current is controlled by the base current. If at any point, i am incorrect, feel free to modify the answer. – Plutonium smuggler Aug 15 '14 at 19:19
No - I don´t think I should modify the answer from somebody else. However, it would be interesting to know how you can justify your assertion (BJT current-controlled). As far as I know there is absolutely no indication for that. In contrary, it is no problem to show that and why the BJT is voltage controlled. There are many people (witnesses with high reputation) supporting the voltage-control approach. Even from the energy point of view it is not possible that a large quantity is controlled by a smaller quantity of the same kind. – LvW Aug 16 '14 at 12:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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