# Transistor Question

1. Aug 3, 2011

### scaredcrow

What happens if a NPN transistor's collector voltage is below its emitter voltage?

Say we have a NPN transistor, and the base voltage is at 5V, the emitter voltage is 2V, and the collector voltage is at ground. Will current flow from the emitter to the collector, or is there an internal diode that blocks this? I ask because I never see this case in my book.

2. Aug 3, 2011

### sophiecentaur

I think that's because they are not designed to operate that way. You would, effectively have connected the transistor the 'wrong way up' and its e would be its c and its c would be its e, with an inappropriate be junction width.

3. Aug 3, 2011

### skeptic2

I have never heard of 3 volts between the base and emitter of a transistor unless the transistor is defective.

The construction of a transistor is such that there is the equivalent of a forward biased diode between the base and the emitter so it would be very difficult to get more than a volt between base and emitter without destroying the transistor.

There also is the equivalent a diode between base and collector. This diode is reversed biased under normal operation. if you were to bring the collector much more than about a 0.6 volts below the base voltage, you would probably destroy the transistor.

4. Aug 3, 2011

### Jiggy-Ninja

In principle, it can be operated that way, but it would have a really low hfe. They are made to optimize their operation in only one configuration.

5. Aug 19, 2011

### jsgruszynski

There's absolutely no reason you couldn't. The only thing that matters is the Vbe and Vbe relative voltages. As long as Vbe>0 and Vbc<0, then you are operation in linear mode and it works like a linear amplifier.

In your case, Vb-Ve = Vbe = 3V, and Vb-Vc = Vbc = -5V, ergo it's in linear. The reality though is that the Vbe voltage is set to 0.7V (in silicon) and the difference in voltage (5V - 0.7V) will be dropped across the Thevenin equivalent voltage of the base bias circuit. If it's a power supply then inside the power supply itself.

The absolute values of voltage never matter in any circuit - it's all only about relative voltages between nodes and across branches.

6. Aug 19, 2011

### yungman

You'll never get 5V on the base and 2V on emitter without burning the transistor!!! Let say if you forward bias the collector by about 0.7V and reverse biased the emitter, I think you'll make the NPN works somewhat like a normal NPN, but the beta is going to be low and not a good transistor. Also, the breakdown voltage of emitter base junction is quite low. We actually use the reverse biased B B junction as zener diode in integrated circuit design. NPN means exactly this......NPN and it should work as transistor if you forward-reverse biased it and keep the B E reverse voltage low enough to prevent it from becoming a zener diode. But the structure of the transistor is designed to work in the normal mode, and it won't be reliable and have good performance if you reverse it.

That's my understanding.