# Fundamental Physics/Electricity/Electric circuits/Transistor circuit

With the right connection, by connecting transistor with resistor(s) Transistor can acts as

## Current amplifier

${\displaystyle i_{b}>0}$
${\displaystyle i_{e}=\alpha i_{b}}$
${\displaystyle i_{c}=\beta i_{b}}$

Provided that, transistor should be turned ON with biased voltage at the base must be greater than diode's break over voltage

${\displaystyle v_{b}>v_{be}}$ . ${\displaystyle v_{be}=v_{d}}$

## Voltage amplifier

### Non inverting voltage amplifier

${\displaystyle {\frac {v_{o}}{v_{i}}}={\frac {R_{2}}{R_{2}+R_{1}}}{\frac {R_{4}}{R_{3}}}}$

With

${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{2}=0}$ . ${\displaystyle v_{o}=0}$ . Transistor acts as Switch OFF (open circuit)
${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{1}=0}$ . ${\displaystyle v_{o}=1}$ . Transistor acts as Switch ON (close circuit)

With

${\displaystyle R_{1}=0}$ and ${\displaystyle R_{4}=nR_{3}}$ . ${\displaystyle v_{o}=nv_{i}}$ . Transistor acts as Non inverting amplifier

### Inverting voltage amplifier

${\displaystyle {\frac {v_{o}}{v_{i}}}=1-{{\frac {R_{2}}{R_{2}+R_{1}}}{\frac {R_{4}}{R_{3}}}}}$

With

${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{2}=0}$ . ${\displaystyle v_{o}=1}$ . Transistor acts as Switch ON (close circuit)
${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{1}=0}$ . ${\displaystyle v_{o}=0}$ . Transistor acts as Switch OFF (open circuit)

With

${\displaystyle R_{1}=0}$ and ${\displaystyle R_{4}=(n+1)R_{3}}$ . ${\displaystyle v_{o}=-nv_{i}}$ . Transistor acts as Inverting amplifier

## An electronics switch

Transistor is switch On when ${\displaystyle V_{b}>V_{be}}$
Transistor is switch Off when ${\displaystyle V_{b}

## A buffer

${\displaystyle v_{c}=v_{b}}$