Diodes: Frequency Response#

The simple resistor model we used in the previous chapter is as simple as it can be. Unfortunately, it neglects the frequency response of the diode. Here, I will introduce a more complete model of a diode.

More Complete AC Model#

Other phenomena affect the diodes response to an AC waveform. The phenomena listed here can be incorporated into the diode model pictured below.

  1. Diode Resistance (\(R_D\)) - Equivalent resistance around the Q-point

  2. Serial Resistance (\(R_S\)) - A small resistance in series with the diode

  3. Diffusion Capacitance (\(C_D\)) - Capacitance caused by the diffusion of charge carriers across the depletion region

  4. Junction Capacitance (\(C_J\)) - Capacitance caused by the the depletion region acting as a dielectric

_images/diodes-ACmodel.svg

Simplified AC Model#

Two of the components in the model above tend to dominate the response. Because of this we can still get good estimate of the response using a simplified model by making two assumptions:

  1. The junction capacitance (\(C_J\)) is much smaller than the diffusion capacitance (\(C_D\)). Therefore the parallel combination of the capacitances can be approximated as simply \(C_D\).

  2. The series resistance (\(R_S\)) is much smaller than the diode resistance (\(R_D\)). Therefore the series resistance can be neglected.

The simplified model can be drawn as shown here:

_images/diodes-ACmodel-simp.svg

Frequency response example#

For this example let’s reuse the example from the previous chapter. However, this time let’s allow the frequency of the small signal AC input to vary in frequency. The capacitances will now cause the output to vary with frequency. We’ll develop a plot of the output magnitude and phase shift against the varying frequency.

Example

_images/diodes-freq-resp.svg

Solution

_images/diodes-freq-resp-Model.svg 
clear all
close all
clc
format short eng
format compact

R=5e3;
Vt=26e-3;

%% DC analysis
syms Va Vb Id
e(1)=Va==5;
e(2)=Va-Vb==0.6;
e(3)=Id-((Vb)/R)==0;
sol=solve(e,Va,Vb,Id);
Id=eval(sol.Id);
Rd=Vt/Id;

%% AC Analysis
Cj=0.4e-12;
Cd=3e-12;
Rs=100e-3;
Vi=0.1;

syms Va Vb Vc Id s
Zcj=1./(s*Cj);
Zcd=1./(s*Cd);
e(1)=Va==Vi;
e(2)=((Va-Vb)/Rd)+((Va-Vb)/Zcd)+((Va-Vb)/Zcj)-((Vb-Vc)/Rs)==0;
e(3)=((Vb-Vc)/Rs)-((Vc)/R)==0;
e(4)=Id==((Vb-Vc)/Rs);
sol=solve(e,Va,Vb,Vc,Id);
Id=sol.Id;

%% Plot the response
f=logspace(6,12,1000);
w=2*pi*f;
Id_f=eval(subs(Id,s,j*w))

figure
subplot(2,1,1)
semilogx(f,abs(Id_f))
grid on
ylabel('Current(A)')
    
subplot(2,1,2)
semilogx(f,angle(Id_f)*(180/pi))
grid on
ylabel('Phase Shift(°)')
xlabel('Frequency(Hz)')