1. Simple circuit consisting of a source, a resistor, and a diode.
For forward bias we have 2 models for the diode.
1) Ideal Diode: Short Circuit
2) Exponential Model
In this section, we will:
1) Address the suitability of these two models.
2) Develop new models.
This will help us to:
1) Analyze diode circuits efficiently.
2) Provide a foundation for modeling trasistor operation. R ID + VD -
VDD
30-Jul-19
Carl Leuschen

2. A) The Exponential Model 1) The most accurate model.
2) The hardest to analyze.
Let’s analyze a circuit using the exponential model.
IS, n, VT are know quantities.
We have two equaition with two unknowns.
Q. How do we solve for ID and VD? + VD - R
VDD ID
30-Jul-19
Carl Leuschen

3. a) Plot the two equations on the i-v plane.
Graphically
a) Plot the two equations
on the i-v plane.
Eq. 1: diode characteristic curve.
Eq. 2: load line.
b) The solution is the intersection, Q, of the two curves and is referred to as the operating point.
Graphical analysis becomes very difficult for complex circuits.
Too difficult to be justified for practical use.
Diode Characteristic
VDD/R
-1/R
Operating Point (Q) ID
Load Line v VD
VDD
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Carl Leuschen

4. Remember the two equations: Diode Characterisitc Load Line
Iterative analysis
Remember the two equations: Diode Characterisitc
Load Line
Q. What do we know about a diode in forward bias?
A. The voltage across the diode is between 0.6 and 0.8 Volts.
Approach: 1) Assume VD is 0.7 Volts.
2) Solve for ID using circuit equation (load line).
3) Caluculate new value of VD or ΔVD using the
Diode Characteristic curve.
4) Go back to step 2 and repeat until the answer converges.
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Carl Leuschen

5. 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current.
Example:
0.7 V, and ΔV=0.1V for every x10 change in current.
VDD = 5 V, and R=1 kΩ. R ID + VD -
VDD
30-Jul-19
Carl Leuschen

6. 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current.
Example:
0.7 V, and ΔV=0.1V for every x10 change in current.
VDD = 5 V, and R=1 kΩ. R ID + VD -
VDD
30-Jul-19
Carl Leuschen

7. However, this approach can be easily iterated using computer analysis.
For complex circuits this approach is very time consuming to do by hand, because you must re-analyze the circuit for each iteration.
However, this approach can be easily iterated using computer analysis.
Conclusions
the graphical and iterative solution methods using the exponential model are inefficient.
For effective circuit analysis a simple model for the junction diode is needed.
30-Jul-19
Carl Leuschen

8. The Piecewise Linear Model.
Consider the i-v relationship for the following configuration.
An ideal diode, a battery, and a resistor.
F.B. → VD(ideal)=0 → short → ID(ideal)≥0
D: Ideal
VD0
VD0 rD rD
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Carl Leuschen

9. R.B. → ID(ideal)=0 → open → VD(ideal)≤0rD rD
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Carl Leuschen

10. The Piecewise Linear Model. An ideal diode, a battery, and a resistor.
D: Ideal i
VD0 rD v
VD0
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Carl Leuschen

11. the non-linear on/off behavior. voltage source:
Approximate the diode exponential i-v relationship using a model containing:
ideal diode:
the non-linear on/off behavior.
voltage source:
to model the potential needed to turn the diode on.
resistor:
to model the exponential slope.
The model is called the:
Battery plus Resistance Model or
Piecewise Linear Model v
VD0
D: Ideal
VD0 rD
30-Jul-19
Carl Leuschen

12. 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current.
Example:
0.7 V, and ΔV=0.1V for every x10 change in current.
VDD = 5 V, and R=1 kΩ.
For these characteristics, VD0=0.65 V and rD=20 Ω. R ID R ID
D: Ideal
VD0 rD + VD -
VDD
VDD
30-Jul-19
Carl Leuschen

13. 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current.
Example:
0.7 V, and ΔV=0.1V for every x10 change in current.
VDD = 5 V, and R=1 kΩ.
For these characteristics, VD0=0.65 V and rD=20 Ω. R ID R ID + VD -
VDD
VDD
VD0 rD
30-Jul-19
Carl Leuschen

14. Q. How are values for VD0 and rD determined?
A. Minimize error within a certain range of operating conditions.
A. Taylor Expansion – 1 point and a slope.
A. Given 2 points we can find a line.
30-Jul-19
Carl Leuschen

15. Battery plus Resistance Model
rD and VD0 is chosen for a given range of operating values.
Q. What if we zoom out to a very wide range of values?
Q. What does the exponential i-v relationship look like?
For the Battery plus Resistance model, the resistance goes to zero.
Constant Voltage Drop Model
A forward bias diode exhibits a constant drop of 0.7 Volts.
One of the most frequently used diode model for initial designs.
Can be used without detailed information about the diode. i i
D: Ideal VD v v
VD0 VD
30-Jul-19
Carl Leuschen

16. 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current.
Example:
0.7 V, and ΔV=0.1V for every x10 change in current.
VDD = 5 V, and R=1 kΩ.
VD=0.7 R ID R ID + VD - D
VDD
VDD VD VD
30-Jul-19
Carl Leuschen

17. 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current.
Example:
The Ideal Diode Model
0.7 V, and ΔV=0.1V for every x10 change in current.
VDD = 5 V, and R=1 kΩ. R ID R ID + VD - D
VDD
VDD
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Carl Leuschen

18. Exponential Battery Plus Resistance Constant Voltage Drop Ideal
Complex Simple
Exponential Battery Plus Resistance Constant Voltage Drop Ideal
D: Ideal
VD0 rD
D: Ideal VD
VD=0.762V
ID= mA
VD=0.735V
ID= 4.26mA
VD=0.7V
ID= 4.3mA
VD=0.0V
ID= 5mA
30-Jul-19
Carl Leuschen