Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 5 M

Year: Dec 2013

Fig1 Circuit diagram

• Consider the above circuit,

  Input voltage is $v_D(t)= V_D+v_d(t)$

• DC analysis:

  The purpose of DC analysis is to find the operating point of diode.

  To do DC analysis, turn off the small AC signal

  The voltage $v_D(t)=V_D$

  And the current $i_D(t)=I_D=I_Sexp(\frac{V_D}{_ηv_T })$

Fig2 characteristics of diode

• When small AC signal is applied

  Total Voltage across diode is $v_D(t)=V_D+v_d(t)$

  Total instantaneous current is $i_D(t)= I_Sexp(\frac{V_D+v_d (t)}{_ηv_T })$

  $i_D(t) =I_Sexp(\frac{V_D}{_ηv_T})× exp(\frac{v_d (t)}{_ηv_T })$

  $i_D(t)=I_D× exp(\frac{v_d (t)}{_ηv_T})$

  • When $v_d(t)\lt\lt_ηv_T$ hence $( \frac{v_d (t)}{_ηv_T}\lt\lt1)$ therefore we can ignore powers of $\frac{v_d (t)}{_ηv_T }$

    $exp\frac{v_d (t)}{_ηv_T})≈1+\frac{v_d (t)}{_ηv_T }$

    $i_D(t)≈I_D(1+\frac{v_d (t)}{_ηv_T })$

    $i_D(t)≈I_D+\frac{I_D}{_ηv_T}v_d(t)$

    $i_D(t)=I_D+i_d(t)$

    $i_d(t)= \frac{I_D}{_ηv_T }v_d(t)$

  • Since \frac{I_D}{_ηv_T } is a ratio of current and voltage hence we can replace it by resistance

    Inverse of $\frac{I_D}{_ηv_T }=\frac{_ηv_T}{I_D} =r_d$

• Small signal AC equivalent circuit is

Fig3 small AC signal equivalent circuit