Contents
What are the operating regions of a transistor?
INTRODUCTION. Bipolar Junction Transistors have three semiconductor regions. The three regions are the emitter region (E), base region (B), and the collector region (c) and these regions are differently doped depending on the type of bipolar transistor it is.
When transistor is used as a switch it operates in which region?
saturation region
Transistor acts as a switch in the saturation region and cutoff region. The emitter-base junction and the collector-base junction is reverse biased in the cutoff region. Both the junctions are forward biased in the saturation region.
What is the difference between FET and BJT?
The major difference between BJT and FET is that in a field-effect transistor only majority charge carries flows, whereas in BJT both majority and minority charge carriers flow….Difference between BJT and FET.
| BJT | FET |
|---|---|
| BJT gain is more | FET gain is less |
| Its output impedance is high due to high gain | Its output impedance is low due to low gain |
How PNP transistor works as a switch?
For a PNP transistor, the Emitter potential must be positive with respect to the Base. Then the transistor operates as a “single-pole single-throw” (SPST) solid state switch. With a zero signal applied to the Base of the transistor it turns “OFF” acting like an open switch and zero collector current flows.
What kind of transistors are used for current mirror?
The basic current mirror can also be implemented using MOSFET transistors, as shown in Figure 2. Transistor M1 is operating in the saturation or active mode, and so is M2. In this setup, the output current IOUT is directly related to IREF, as discussed next.
How does a bipolar mirror convert voltage to current?
The simplest bipolar current mirror (shown in Figure 1) implements this idea. It consists of two cascaded transistor stages acting accordingly as a reversed and direct voltage-to-current converters. The emitter of transistor Q1 is connected to ground. Its collector-base voltage is zero as shown.
Why is the current of a mirror greater than zero?
Parameter β 0 is the transistor β-value for VCB = 0 V. If V BC is greater than zero in output transistor Q 2, the collector current in Q 2 will be somewhat larger than for Q 1 due to the Early effect. In other words, the mirror has a finite output (or Norton) resistance given by the ro of the output transistor, namely:
How is the drain current of a mirror determined?
The drain current of a MOSFET ID is a function of both the gate-source voltage and the drain-to-gate voltage of the MOSFET given by ID = f ( VGS, VDG ), a relationship derived from the functionality of the MOSFET device. In the case of transistor M1 of the mirror, ID = IREF.