Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 4 M

Year: Nov 2013

• Now given diode is practical diode hence it starts to conduct at voltage greater than cut-in voltage let it be $0.7V$

• During first half or during positive cycle the diode $D_1$ is forward bias and $D_2$ is reverse biased Using transition model, replace diode by a voltage source of $0.7V$

  • Input voltage $(V_i)$

    -$V_i+0.7+4=0 hence V_i=4.7V$

  • The diode will start conducting after i/p voltage is greater than 4.7V and for less than 4.7V the diode will act as open circuit.

  • Output voltage is

    -$V_o +4+0.7=0 hence V_o=4.7V$

  • For voltage less than $4.7V$ diode is reverse biased, $o/p$ is same as $i/p$ i.e. sinewave. When i/p voltage is greater than 4.7V diode is forward biased and output is straight line at $V_o=4.7V$ as shown in Fig3

Fig1 clipper circuit working during positive half of wave

• During $2^{nd}$ half of wave or during negative cycle the diode D2 is forward bias and D1 is reverse biased Using transition model, replace diode by a voltage source of 0.7V

  • Input voltage $V_i$

    +$V_i$+0.7+3=0 hence $V_i$= -3.7V

  • The diode will start conducting after i/p voltage is greater than -3.7V and for less than -3.7V the diode will act as open circuit.

  • Output voltage is.

    -3-0.7=$V_O$hence $V_O$=-3.7V

  • For voltage less than -3.7V diode is reverse biased, o/p is same as i/p i.e. sinewave. When i/p voltage is greater than -3.7V diode is forward biased and output is is straight line at $V_O$=-3.7V as shown in Fig3

Fig2 clipper circuit working during negative half of wave

Fig3 clipper outputwaveform