BJTs: DC Analysis#
How do we use the nonlinear VI characteristic to perform circuit analysis?
We can use nodal analysis as we did with the diode but how do we solve the system?
MATLAB or something similar
Load Lines
Some linear assumptions
Let’s work on the linear assumption method first. We assume that the BJT is in the active region. In this region we can use as many of these three assumptions as necessary:
\(V_{BE}=0.7V\)
\(I_C=ßI_B\)
\(I_E=I_C+I_B\)
These assumptions can be made in another form: a D.C. equivalent circuit
Let’s practice this on a simple circuit
Example
Assume the BJT is operating in the active region. Find the Q-point and check the active assumption.
Solution
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rb=86e3;
Rc=150;
B=200;
syms Va Vb Vc Ib
e(1)=Va==5;
e(2)=Vb==0.7;
e(3)=((Va-Vb)/Rb)-Ib==0;
e(4)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-0
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-0
Analysis with Load Lines#
Use figures from above
Region |
BE Junction |
BC Junction |
|---|---|---|
Cut-off |
Reverse |
Reverse |
Active |
Forward |
Reverse |
Saturation |
Forward |
Forward |
Checking Assumption in Saturation#
How can we change a value in the circuit to cause it to operate in the saturation region?
Lower \(R_B\)
Raise \(R_C\)
The analysis remains largely the same. Only the values change.
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rb=1e3;
Rc=150;
B=200;
syms Va Vb Vc Ib
e(1)=Va==5;
e(2)=Vb==0.7;
e(3)=((Va-Vb)/Rb)-Ib==0;
e(4)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-0
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-0
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rb=86e3;
Rc=1e3;
B=200;
syms Va Vb Vc Ib
e(1)=Va==5;
e(2)=Vb==0.7;
e(3)=((Va-Vb)/Rb)-Ib==0;
e(4)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-0
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-0
Checking Assumption in Cut-off#
How can we change a value in the circuit to cause it to operate in the cut-off region?
Raise \(R_B\)
Lower \(R_C\)
The analysis remains largely the same. Only the values change.
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rb=100e6;
Rc=150;
B=200;
syms Va Vb Vc Ib
e(1)=Va==5;
e(2)=Vb==0.7;
e(3)=((Va-Vb)/Rb)-Ib==0;
e(4)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-0
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-0
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rb=86e3;
Rc=5;
B=200;
syms Va Vb Vc Ib
e(1)=Va==5;
e(2)=Vb==0.7;
e(3)=((Va-Vb)/Rb)-Ib==0;
e(4)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-0
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-0
The problems above are designed to improve your understanding of the regions of operation but the analysis is simple and repetitive. Let’s practice analyzing other circuits.
Examples#
%% Find node voltages
clear all
close all
clc
format short eng
format compact
R1=20e3;
R2=80e3;
Rc=2e3;
Re=1e3;
B=200;
syms Va Vb Vc Ve Ib
e(1)=Va==15;
e(2)=Vb-Ve==0.7;
e(3)=((Va-Vb)/R1)-Ib-((Vb)/R2)==0;
e(4)=Ib-((Ve)/Re)+B*Ib==0;
e(5)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ve,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ve=eval(sol.Ve)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-Ve
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-Ve
%% Find node voltages
clear all
close all
clc
format short eng
format compact
R1=100e3;
R2=100e3;
Rc=1e3;
Re=5e3;
B=200;
syms Va Vb Vc Ve Ib
e(1)=Va==15;
e(2)=Vb-Ve==0.7;
e(3)=((Va-Vb)/R1)-Ib-((Vb)/R2)==0;
e(4)=Ib-((Ve)/Re)+B*Ib==0;
e(5)=((Va-Vc)/Rc)-B*Ib==0;
sol=solve(e,Va,Vb,Vc,Ve,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ve=eval(sol.Ve)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-Ve
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-Ve
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rf=250e3;
Rc=4.7e3;
Re=1.2e3;
B=90;
syms Va Vb Vc Ve Ib
e(1)=Va==10;
e(2)=Vb-Ve==0.7;
e(3)=((Vc-Vb)/Rf)-Ib==0;
e(4)=Ib-((Ve)/Re)+B*Ib==0;
e(5)=((Va-Vc)/Rc)-B*Ib-((Vc-Vb)/Rf)==0;
sol=solve(e,Va,Vb,Vc,Ve,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ve=eval(sol.Ve)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Vbe=Vb-Ve
Vbc=Vb-Vc
%% check active assumption - collector/emitter
Vce=Vc-Ve
A PNP Example#
%% Find node voltages
clear all
close all
clc
format short eng
format compact
Rb=580e3;
Rc=5e3;
B=100;
syms Va Vb Vc Ve Ib
e(1)=Va==1.5;
e(2)=Ve-Vb==0.6;
e(3)=Ve==5;
e(4)=((Va-Vb)/Rb)+Ib==0
e(5)=-((Vc)/Rc)+B*Ib==0
sol=solve(e,Va,Vb,Vc,Ve,Ib);
Va=eval(sol.Va)
Vb=eval(sol.Vb)
Vc=eval(sol.Vc)
Ve=eval(sol.Ve)
Ib=eval(sol.Ib)
%% check active assumption - Junctions
Veb=Ve-Vb
Vcb=Vc-Vb
%% check active assumption - collector/emitter
Vec=Ve-Vc