2
\$\begingroup\$

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.

enter image description here

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?

\$\endgroup\$
  • 2
    \$\begingroup\$ Of course current will flow if the voltage is high enough. But it will likely cause permanent damage to the transistor. I don't remember seeing a specification for maximum allowable Vec, but a lot of small transistors have a maximum Veb of around 5V. So I guess it would not take much reverse voltage to damage a BJT. \$\endgroup\$ – mkeith Mar 15 '18 at 16:15
  • \$\begingroup\$ That's why transistors have a max voltage..... BUt in your scenario current flows from the collector to the emitter, not the other way around. \$\endgroup\$ – Trevor_G Mar 15 '18 at 16:41
  • \$\begingroup\$ The subject line for the question specifically says from emitter to collector. That is why I answered the way I did. Maybe the question could be clarified. \$\endgroup\$ – mkeith Mar 15 '18 at 17:06
  • \$\begingroup\$ yes, if it is a photo transistor \$\endgroup\$ – jsotola Mar 16 '18 at 3:55
1
\$\begingroup\$

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.

schematic

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 :

schematic

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

\$\endgroup\$
0
\$\begingroup\$

There will always be some leakage current, even with the base tied to the emitter. However, that is "small" in practical applications to the point it is generally ignored.

If you apply enough voltage to anything, current will eventually flow. At about 1000 V per millimeter, air conducts, for example. Each transistor has a maximum C-E voltage spec for that reason. Up to that voltage, the transistor works like a transistor, and the C-E current with the base open is leakage that should also be specified in the datasheet. If you apply more than the maximum C-E voltage, there is no guarantee what the transistor may do. At some point enough current will flow to irreversibly damage the transistor.

The leakage will be higher with the base floating than with it tied to the emitter. One way to look at this is that the little bit of leakage across the reverse biased C-B junction ends up being base current. If you really want a BJT to be off, actively hold its base close to the emitter voltage.

A transistor in a circuit with its base floating can also pick up and amplify noise.

\$\endgroup\$
0
\$\begingroup\$

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?

There are three useful circuits that employ a transistor without any voltage supply to the base (through external wires, at least).

The first is a temperature-compensated Zener diode: base-emitter breakdown may be about 6V, and a forward-biased collector adds a single diode drop to the breakdown. The temperature coefficient of the Zener breakdown and the forward diode are opposite, and approximately equal, so that this makes a good voltage reference. This application note AN-71 recommends 2N3252.

The second is as a phototransistor; in a transparent package, light impinging on the collector-base junction causes leakage current and can bias the transistor into conduction. This is a very common kind of light sensor element.

The third is rather exotic; if, instead of Zener-breakdown in the base-emitter, one breaks down the collector-base junction with a high applied voltage, one may find (depending on the internal structure of the transistor) an avalanche-transistor capabiity to do very fast switching of high current. Many transistors do not do this usefully, but a few types are available ZTX-415 datasheet

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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