Files 3-Handouts Handout 2d
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Transcript of Files 3-Handouts Handout 2d
8/3/2019 Files 3-Handouts Handout 2d
http://slidepdf.com/reader/full/files-3-handouts-handout-2d 1/13
Diode Circuits Analysis
Graphical Method The intersection of the diode and theresistor current equations is calculated
Gives good understanding to the circuit operation Time consuming Not suitable for large circuits
Analyt ical Method Both diode and resistor current equations
are solved simultaneously or iteratively Faster than graphical method More accurate Time consuming
Not suitable for large circuits
Diode Models Piecewise-linear diode model is used to
replace the diode Least time consuming method Suitable for larger circuits
Accuracy depends on the used model
A simple diode circuit.
R
V V I
e I I
DS
R
nV
V
S D
T
D
−=
⎟⎟ ⎠ ⎞
⎜⎜⎝ ⎛ −=
EquationCurrentResistor
EquationCurrentDiode
1
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Graphical MethodStep 1 : Locating the Operating Points Step 2 : KVL & KCL Constraints
Step 3 : Enforcing KVL & KCL Solution
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Graphical Method (cont.)
Gives good understanding tothe circuit operation
Not suitable for large circuits
General Problem Basic Load Line Construction
Alternative Construction
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Graphical Method (Example)Problem: Find Q-point
Given data: V s =10 V, R =10k Ω. Analysis:
To define the load line we use,V D = 0
V D = 5 V, I
D =0.5 mA
These points and the resultingload line are plotted.Q-point isgiven by the intersection of theload line and the diodecharacteristic:
Q-point = (0.95 mA, 0.6 V)
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Analytical Method
R
V V I
e I I
DS
R
nV
V
S D
T
D
−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −=
EquationCurrentResistor
EquationCurrentDiode
1
Example:Is=10-15 A, n=1, Vs=5 V, R=1K
Step 1:
Assume VD=VDo=0.7VID1=(Vs-VDo)/R=4.3 m.A
VD1=0.025ln(ID1 /Is)=0.7272 V
Step 2:ID2=(Vs-VD1)/R=4.2728 m.A VD2=0.025ln(ID2 /Is)=0.7271 V
Exact Value:ID=4.2729 m.A, VD=0.7271 V
8/3/2019 Files 3-Handouts Handout 2d
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Diode Models
Approximating the diode forwardcharacteristic with two straight lines.
Piecewise-linear model of the diodeforward characteristic and its equivalentcircuit representation.
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Diode Models (cont.)
Development of the constant-voltage-dropmodel of the diode forward characteristics. A vertical straight line (b) is used toapproximate the fast-rising exponential.
The constant-voltage-drop model of the diode forwardcharacteristic and its equivalent circuit representation.
Also Ideal diode model may be used
which means short circuit in theforward direction and open circuitin the reverse direction.
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Which Model to Use?
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Analysis using Diode Models
Diode is assumed to be either on or off.
Analysis is conducted as follows:
1. Select diode model.2. Identify anode and cathode of diode and label v D and i D .3. Guess diode’s region of operation from circuit.4. Analyze circuit using diode model appropriate for
assumed operating region.5. Check results to check consistency with assumptions.
(For forward assumption check that iD > 0, for reverse
biased check that vD < VDO or zero for ideal diode model)
8/3/2019 Files 3-Handouts Handout 2d
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Since source is forcing positive currentthrough diode assume diode is on.
Q-point is(1 mA, 0V)
Since source is forcing currentbackward through diode assume diode
is off. Hence I D =0 . Loop equation is:
Q-point is (0, -10 V)
Analysis using Ideal Model for Diode: Example
0
m1k 10
V)010(
≥
=Ω
−=
D
D
I
A I
∵our assumption is right.
V10
010104
−=∴
=++
D
D D
V
I V
our assumption is right.
Find the Q-Point (ID, VD) for the following diodes assuming ideal diode model.
8/3/2019 Files 3-Handouts Handout 2d
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Analysis using Constant Voltage Drop Model for Diode
Analysis:Since 10V source is forcing positive currentthrough diode assume diode is on.
v D = V on for i D >0 andi D
= 0 for v D
< V on
.
mA94.0k 10
V)6.010(
k 10
V)10(
=Ω
−=
Ω
−= on
D
V I
Find the Q-Point (ID, VD) for the following diode assuming CVD, with VDO=0.6V.
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Two-Diode Circuit Analysis
Analysis: Since 15V source is forcing positivecurrent through D 1 and D 2 and -10V source isforcing positive current through D 2, assume bothdiodes are on.
Since voltage at node D is zero due to shortcircuit of ideal diode D 1,
Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)
But, I D1 <0 is not allowed by diode, so try again.
mA50.1k 10
V)015(1 =Ω−= I mA00.2k 5
V)10(02 =Ω−−= D
I
211 D I
D I I += mA50.025.1
1−=−=∴
D I
Find the Q-Point (ID, VD) for the following diodes assuming ideal diode model.
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Two-Diode Circuit Analysis (cont.)
Since current in D 1 is zero, I D2 = I 1,
Q-points are D 1 : (0 mA, -1.67 V):off
D 2 : (1.67 mA, 0 V) :on
Since current in D 2 but thatin D 1 is invalid, the second
guess is D 1 off and D 2 on.
V67.17.16151
000,10151
mA67.115,000
V251
0)10(2
000,51
000,1015
−=−=−=
=Ω=∴
=−−−−
I DV
I
D I I