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3 years ago

What’s a pnp transitor?

Engineering

2 answers:

Dafna1 [17]3 years ago

7 0

Answer:

A transistor is of two types: PNP and NPN

Explanation:

A transistor is current controlled device in which the current flow by the movement of the electrons and wholes.

The NPN and PNP transistor are the types of the BJT (Bipolar junction transistor) that have two uses i.e switching and amplification.

A PNP transistor have three junction i.e P, N and P where the base is connected to the N - junction. The P junction is high in number of wholes and N- junction have high numbers of electrons.

In the PNP transistor the voltage is provided to the emitter and base terminal with base connected to the negative terminal of the voltage source and emitter connected to its positive terminal. The direction of the flow of the current is from emitter to collector whenever a significant voltage is provided to the emitter base junction.

Damm [24]3 years ago

6 0

Answer:

PnP transistor is a current controlled device. It has two crystal diodes connected back to back. The left side of the diode is known as the emitter-base diode and the other side which is the right side of the diode is known as the collector-base diode.

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3 years ago

The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery

Lynna [10]

Answer:

capacity = 0.555 mAh

capacity = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C

capacity = 2 × ampere × second ...............1

put here value and we get

and 1 Ah = 3600 C

capacity = $\frac{2}{3600}$

capacity = 0.555 mAh

and

when it is in parallel than capacity will be

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3 years ago

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ra1l [238]

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3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$