**1 Answer**

written 2.8 years ago by |

Since the input voltage contains a dc component with an ac signal superimposed,the diode current will also contain a dc component with an ac signal superimposed.Let us consider $I_{DQ}$ is the dc component of the voltage.For this situation we consider the ac component to be a comparatively small value.

The relationship between the diode current and diode voltage can be given by:

$ i_D \approx I_S\:e^{v_{D}\over V_T}= I_S \:e^{\dfrac{V_{DQ}+v_d}{V_T}}$,where $V_{DQ} $ is the dc component of voltage and $\LARGE v_d$ is the ac component.

We can write the above equation as:

$ i_D =I_S \bigg[e^{\dfrac{V_{DQ}}{{}V_T}}\bigg]\cdot \bigg[e^{{v_d}\over{V_T}}\bigg]$ ....(***)

If the ac signal is small,then (v_d<<V_T),we can then expand the exponential function into a linear series, $e^{\dfrac{v_d}{V_T}}=1+\dfrac{v_d}{ V_T}$ ...(!!!)</p>

Also lets us say,$I_S\:e^{{I_{DQ}}\over V_T}=I_{DQ}$

From (***) and (!!!), we get:

$ i_D=I_{DQ}\:\bigg(1+{\dfrac{v_d}{V_T}}\bigg)=I_{DQ}+{\dfrac{I_{DQ}}{{V_T}}\cdot v_d }=I_{DQ}+i_d$

where $\LARGE i_D$ is the ac component of the diode current.The relationship between components of diode voltage and current will be:

$i_d =\bigg(\dfrac{I_{DQ}}{V_T}\bigg)\cdot v_d = {g_d} \cdot {v_d}$

where $\LARGE {g_d}$ is the diffusion conductance

Also $r_d = \dfrac{1}{g_{d}} = \dfrac{V_T}{I_{DQ}}$

These are the diode equations and parameters.