**1 Answer**

written 6.0 years ago by | modified 5.9 years ago by |

• In a small signal model, we can take a nonlinear circuit or device and make approximations which linearize it about an DC operating point.

• To do this, we write all voltages and currents as the sum of a DC value for each, plus a small extra amount which is the small signal.

• Consider the above circuit,

Input voltage is $vD(t)=VD+vd(t)$

• 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 $vD(t)=VDvD(t)=VD$

And the current $iD(t)=ID=ISexp(VDηvT)$

• When small AC signal is applied

Total Voltage across diode is $vD(t)=VD+vd(t)vD(t)=VD+vd(t)$

Total instantaneous current is $iD(t)=ISexp(VD+vd(t)ηvT)iD(t)=ISexp(VD+vd(t)ηvT)$

$iD(t)=ISexp(VDηvT)×exp(vd(t)ηvT)iD(t)=ISexp(VDηvT)×exp(vd(t)ηvT)$

$iD(t)=ID×exp(vd(t)ηvT)iD(t)=ID×exp(vd(t)ηvT)$

When $vd(t)\lt\ltηvTvd(t)\lt\ltηvT$ hence $(vd(t)ηvT\lt\lt1)(vd(t)ηvT\lt\lt1)$ therefore we can ignore powers of $vd(t)ηvTvd(t)ηvT$

$expvd(t)ηvT)≈1+vd(t)ηvTexpvd(t)ηvT)≈1+vd(t)ηvT$

$iD(t)≈ID(1+vd(t)ηvT)iD(t)≈ID(1+vd(t)ηvT)$

$iD(t)≈ID+IDηvTvd(t)iD(t)≈ID+IDηvTvd(t)$

$iD(t)=ID+id(t)iD(t)=ID+id(t)$

$id(t)=IDηvTvd(t)id(t)=IDηvTvd(t)$

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

Inverse of $IDηvT=ηvTID=rdIDηvT=ηvTID=rd$

• Small signal AC equivalent circuit is