# Fundamental Physics/Electricity/Electric circuits/Transistor circuit

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

## Current amplifier

$i_{b}>0$ $i_{e}=\alpha i_{b}$ $i_{c}=\beta i_{b}$ Provided that, transistor should be turned ON with biased voltage at the base must be greater than diode's break over voltage

$v_{b}>v_{be}$ . $v_{be}=v_{d}$ ## Voltage amplifier

### Non inverting voltage amplifier

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

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

With

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

### Inverting voltage amplifier ${\frac {v_{o}}{v_{i}}}=1-{{\frac {R_{2}}{R_{2}+R_{1}}}{\frac {R_{4}}{R_{3}}}}$ With

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

With

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

## An electronics switch

Transistor is switch On when $V_{b}>V_{be}$ Transistor is switch Off when $V_{b} ## A buffer

$v_{c}=v_{b}$ 