# Electronic – Is it possible to make current flow through a transistor from the emitter to the collector without supplying voltage to the base

transistors

I've started studying bipolar junction transistors, and I'm trying to understand how they work.

When learning about how a transistor work, you always see it connecting 2 circuits, the smaller one involving only the emitter and the base, and the largest one with the collector too.

According to what I understood current flow in the largest, only if it flows in the smallest.

If I disconnect the base and the smaller cirucuit, or remove the voltage generator from that circuit, current stop flowing in the largest one too.

For this reason a transistor in a circuit if no voltage is supplied to the base (or the base not attached to anything) is said to be "off".

As my current understanding (it may be wrong), in this scenario, however you put the voltage generator the current can't flow due to the presence of an electric field inside the transistor ( depletion layer).

But what would happen if I increase the voltage?

Is there a trashold limit where the depletion layer is overcomed? Or is, more generally, possible to make electricity flow from the emitter to the collector without supplying voltage to the base?

The NPN transistor in reverse direction can be viewed as Emitter-base Zener diode with the breakdown voltage larger than 5V plus "ordinary" PN diode between base and collector.

simulate this circuit – Schematic created using CircuitLab

And the typical values for Veb is

BC337-40

Veb=8.2V at I=5.5mA

BC549B

Veb=8.3V at I=5.5mA

BD139-16

Veb=8.5V at I=5.5mA

BC639

Vbe=7.7V at I=500uA

BC337

Veb=7.9V at I=500uA

2SC945

Veb=8.1V at I=500uA

And the emitter-collector breakdown voltage (Vec) is:

BC337-40

Vec=6.7V at I=5.5mA

BC549B

Vec=7.2V; I=5.5mA

BD139-16

Vec=6.7V; I=5.5mA

BC639

Vec=6.3V; I=500uA

BC337

Vec=6.4V; I=500uA

2SC945

Vec=7.5V; I=500uA

And the "equivalent" circuit is :

simulate this circuit

And what is more interesting is that the emitter-collector "equivalent" diode act as a "tunnel diode" hence the negative-resistance region in emitter-collector avalanche breakdown.

Look at the exampel:

http://www.cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html

http://jlnlabs.online.fr/cnr/negosc.htm