Electronic – Why can’t you go sticking a voltage across the base-emitter terminals in a transistor


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?

Best Answer

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:

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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!