Transistor . Sensors Technology – MED4. Lecturer: Smilen Dimitrov. Introduction. The model that we introduced for ST. Introduction. We have discussed The units of voltage, current and resistance, from both a microscopic and macroscopic (electric circuits) perspective
Sensors Technology – MED4
We apply a moderate voltage between C and B -> C positive with respect to B
- polarity of voltage chosen to increase force pulling the N-type electrons and P-type holes apart….
…reverse-biased the B-C diode junction - widens the depletion zone between C and B and so no current will flow
In the absence of any externally applied electric field, we find that depletion zones form at both PN-Junctions, so no charge wants to move from one layer to another
We apply E-B voltage whose polarity is designed to forward-bias the E-B junction. This 'pushes' electrons from the E into the B region, ...
Most (but not all!) the electrons that get into B move straight on into the C,...
... and sets up a current flow across the E-B boundary. Once the electrons have managed to get into the B region they can respond to the attractive force from the positively-biased C region.
... provided the C voltage is positive enough to draw them out of the B region. That said, some of the electrons get 'lost' on the way across the Base.
Schematic symbols: The direction of the emitter arrow defines the type of transistor.
For proper polarization (of NPN), we say:
In a real transistor:
Regions of BJT operation
Rules for Bipolar Junction Transistors (BJTs):
For some transistors, the pin function can be identified from packaging:
But sometimes, we have to measure: Set a digital multimeter to diode test and an analogue multimeter to a low resistance range such as × 10, as described above for testing a diode.
We used a circuit like this to discuss transistor properties
– but there are no resistances in neither look A nor loop B – short circuit!
So, this is not a practical example....
Using a single resistor in the emitter is a practical example, but not interesting for us:
- since it cannot give any voltage amplification:
Vi - Vbe = Vo
If values are replaced: