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BJT bipolar, MOSFET unipolar; BJT diff amp

Short answer

BJTs are bipolar because both electrons and holes participate in conduction. MOSFETs are unipolar because current involves only one type of carrier (electrons in n-channel, holes in p-channel). A differential amplifier uses two matched BJTs with common emitter resistor or current source to −VEE, collectors to RC to VCC, inputs at bases, outputs at collectors.

Why BJT is a bipolar and MOSFET is an unipolar device? And draw the circuit diagram of differential amplifier using BJT.

Why BJT is bipolar

In a BJT, current is carried by both types of charge carriers, electrons and holes. In an NPN transistor, the emitter injects electrons (majority carriers in the emitter) into the p-type base, where they are minority carriers. A small base current of holes replaces the holes lost to recombination in the base. Because both majority and minority carriers take part in conduction, the device is called bipolar.

Why MOSFET is unipolar

In a MOSFET, current flows through a channel carried by only one type of carrier, the majority carriers: electrons in an n-channel device or holes in a p-channel device. There is no minority carrier injection, and the gate is insulated, so no gate current flows. Conduction is controlled by the gate voltage, which creates or modifies the channel. Therefore it is a unipolar device.

Differential amplifier using BJT

                 +VCC
          +-------+-------+
          |               |
        [RC]            [RC]
          |               |
 vo1 o----+               +----o vo2
          |               |
         C|               |C
 v1 o----B| Q1        Q2  |B----o v2
         E|               |E
          |               |
          +-------+-------+
                  |
                [RE]  (or constant current source)
                  |
                -VEE

Two matched transistors Q1 and Q2 share the same emitter resistor or current source. The inputs v1 and v2 are applied at the bases, and the outputs vo1 and vo2 are taken from the collectors. The output depends on the difference v1 − v2 (differential gain Ad); a signal common to both inputs is rejected (low common-mode gain Ac), which gives a high CMRR = Ad/Ac.

✓ Answer checked by Claude Opus 5.5 (high effort)

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