I know Vbe needs to be +0.6V or so but for a NPN bipolar transistor does the base voltage have to be below the collector voltage for current to flow through the transistor. My lecturer said that a transistor will turn on if the collector is at GND and the base is positive, is this true?

 

Hi
For an NPN transistor Emitter = N, Base = P, Collector = N. As you must be knowing for Transistor to work in Active Region Base-Collector junction has to be reverse biased(or rather it should not be forward biased). So in this case Base has to be negative and Collector has to be positive for pure reverse bias. But as you know from diode working that if Base is positive and Collector is negative the Base-Collector junction wont be forward biased till some voltage around 0.4V(Normally for diode we consider 0.7V but in BJTs we consider 0.4v). Beyond this voltage Base-Collector Junction will be forward bias and the Transistor will be in cut off. So I would say your Lecturer is not entirely correct. Do speak to him for more clarification.

Thanks
Salil

 

Thanks alot

 

It is a simplification to say that the collector-base junction needs to be reverse biased for active transistor operation.

As a transistor is biased on harder and harder by increasing base current, it may be possible for the collector potential to fall below that of the base. This condition can sometimes occur at the extremity of the active region, just before true saturation occurs. In the case of fully saturated operation, Vbe>Vce is not at all uncommon.

This situation is often very perplexing to a student who meets it for the first time, bur it helps to emphasize the point that a transistor really does not behave like a pair of resistors in series.

Edit: To answer your original question, while it can sometimes be possible to have conduction with Vbe>Vce, this cannot continue to the extent that Vce falls to zero. In fact, if you short the emitter to ground,a small forward bias current may flow from base to collector. This isn't really a "transistor" action though, more a case of diode action.

 

The reason I ask this question is based on the circuit shown. I'm unsure whether a negative or positive voltage is required to cut off the transistor, keeping in mind the negative I/P is a virtual ground. In the notes handed out it apparently turns off when the comparator o/p is positive, but in the lecture he contradicted this and said it should be negative.

 

The reason I ask this question is based on the circuit shown. I'm unsure whether a negative or positive voltage is required to cut off the transistor, keeping in mind the negative I/P is a virtual ground. In the notes handed out it apparently turns off when the comparator o/p is positive, but in the lecture he contradicted this and said it should be negative.

The transistor shown is PNP, and therefore requires a negative base drive to turn it on. The collector is returned to a negative supply, and its emitter will be at or close to ground potential if the amplifier feedback condition is satisfied. We might therefore expect the transistor to turn on for any base voltage more negative than about -0.6V (for silicon).

There may of course be errors in the drawing, or in the notes, or in what your lecturer said, or in your hearing or recall of it. You would really need to ask him for clarification of that.

 

Thanks for the assistance