ENGINEERING-25
MidTerm Exam Review
+ VD −
Problem
A source of electrical power, the
Voltage Source VS, connects to
a Diode and a load-Resistor as
shown in the diagram at right.
The Diode absorbs some of the
voltage supplied by the source,
Load Resistor absorbs the rest.
This balance between powerSupplied and power-Absorbed
can
be
described
using
“Kirchoff’s Voltage Law” as
VS
ID
R
+
VR
−
VS  VD  VR
Next, the Resistor Voltage Drop, VR, can be related to the Diode electrical Current, ID, by
“Ohm’s Law”:
VR  RI D
And finally the Diode Current, ID, as function of the Voltage across the Diode, VD:
VD
 nV

th

I D  I 0 e  1




WHERE
 I0 ≡ The “Reverse Bias Saturation
Current” for the Diode
 n ≡ The Diode Ideality Factor
 Vth ≡ The “Thermodynamic Thermal
Voltage”
© Bruce Mayer, PE • Chabot College • Document1 • Page 1 of 7
Note that the Diode Eqn Models the REAL behavior of a semiconductor PN-Junction Diode as
discussed in Lecture-12 on Looping Structures
Please consider a Vsrc-Diode-Resistor series-circuit with these ParaMeter values:
I0 = 35µA (35x10−6 A)
R = 4.7 Ω
VS = 9.000 V
n = 2.1
Vth = 25.9 mV (0.0259 V)
Use any Combination of HandWork and MATLAB to compute to at least 4
significant figures the values for
 VD,
 VR
 ID
© Bruce Mayer, PE • Chabot College • Document1 • Page 2 of 7
The hand PreAnalysis to find a Dckt(VD) Function
© Bruce Mayer, PE • Chabot College • Document1 • Page 3 of 7
Once VDop has been Determined BACK SUBSTITUE into the
DIODE EQN and OHM’S LAW to Find VR & ID
© Bruce Mayer, PE • Chabot College • Document1 • Page 4 of 7
 VDop 
I Dop  I 0  e nVth  1




VRop  RI Dop
The MATLAB PLOT for visualization of the solution
 Note: SOLUTION at DVR = 0
20
DVR (soln @ DVR=0)
15
10
5
0
-5
0.55
0.56
0.57
0.58
0.59
0.6
0.61
0.62
vD
© Bruce Mayer, PE • Chabot College • Document1 • Page 5 of 7
0.63
0.64
0.65
The MATLAB OUTPUT
Showing The Solution in SI units of Volts & Amps
Vd_op =
0.5897
Id =
1.7894
Vr =
8.4103
Chk that KVL yields vD + vR = Vs = 9.00 volts
Vs_chk =
9.0000
© Bruce Mayer, PE • Chabot College • Document1 • Page 6 of 7
The MATLAB CODE
% Bruce Mayer, PE
% ENGR25 * 18Mar10
% file Diode_Prob_Transcendental_Voltage_100322.m
%
%
I0 = 35E-6 % Diode Reverse-Saturation Current in Amps
R = 4.7 % Load Resistance in ohms
n = 2.1 % Diode Quality Factor
Vth = 25.9/1000 % Thermal Voltage at 300K in volts
Vs = 9 % Supply potential in volts
%
% construct Function to ZERO using KVL
DVR = @(Vd) Vd + I0*(exp(Vd/(n*Vth))-1)*R - Vs
%
% quickly check by fplot - soln is where DVR= 0
fplot(DVR, [.55 0.65]), xlabel('vD'), ylabel('DVR (soln @
DVR=0)'), grid
%
%
% find the operating-point values
disp('Showing The Solution in SI units of Volts & Amps')
Vd_op = fzero(DVR, 0)
Id = I0*(exp(Vd_op/(n*Vth))-1)
Vr = Id*R
%
% Check KVL eqn => Vs = Vd + Vr
disp('Chk that KVL yields vD + vR = Vs = 9.00 volts')
Vs_chk = Vd_op + Vr
Print Date/Time = 10-Mar-16/19:37
© Bruce Mayer, PE • Chabot College • Document1 • Page 7 of 7