1. Recall-Lecture 4 Current generated due to two main factors
Drift – movement of carriers due to the existence of electric field
Diffusion – movement of carriers due to gradient in concentrations

2. Recall-Lecture 4 Introduction of PN junction
Space charge region/depletion region
Built-in potential voltage Vbi
Reversed biased pn junction
no current flow
Forward biased pn junction
current flow due to diffusion of carriers.

3. Analysis of PN Junction Diode in a Circuit

4. CIRCUIT REPRESENTATION OF DIODE – Ideal Model
Reverse-bias
VD = - VS
Forward-bias
I-V characteristics of ideal model

5. EXAMPLE: Determine the diode voltage and current in the circuit using ideal model for a silicon diode. Also determine the power dissipated in the diode.

6. CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
Reverse-bias
VD = - VS
Forward-bias
I-V characteristics of constant voltage model

7. EXAMPLE: Determine the diode voltage and current in the circuit (using constant voltage model) for a silicon diode. Also determine the power dissipated in the diode. Consider the cut-in voltage V = 0.65 V.

8. CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
Reverse-bias
VD = - VS
Forward-bias
I-V characteristics of piecewise model

9. EXAMPLE: Determine the diode voltage and current in the circuit using piecewise linear model for a silicon diode. Also determine the power dissipated in the diode. Consider the cut-in voltage V = 0.65 V and the diode DC forward resistance, rf = 15 Ω.

10. Why do you need to use these models?

11. Diode Circuits: Direct Approach
Question
Determine the diode voltage and current for the circuit.
Consider IS = A.
VPS = IDR + VD
5 = (2 x 103) (10-3) [ e ( VD / 0.026) – 1 ] + VD
VD = V
And ID = 2.19 mA
ITERATION METHOD

12. PIECEWISE LINEAR MODEL 1 PIECEWISE LINEAR MODEL 2
IDEAL MODEL
PIECEWISE LINEAR MODEL 1
PIECEWISE LINEAR MODEL 2
DIRECT APPROACH
Diode voltage VD 0V
0.65 V
0.652 V
0.619 V
Diode current ID
2.5 mA
2.175 mA
2.159 mA
2.19 mA

13. DC Load Line A linear line equation ID versus VD
Obtain the equation using KVL

14. V 2 Use KVL: 2ID + VD – 5 = 0 ID = -VD + 5 = - VD + 2.5
The value of ID at VD = V V 2
Use KVL:
2ID + VD – 5 = 0
ID = -VD = - VD + 2.5

15. DIODE AC EQUIVALENT

16. Sinusoidal Analysis The total input voltage vI = dc VPS + ac vi
iD = IDQ + id
vD = VDQ + vd
IDQ and VDQ are the DC diode current
and voltage respectively.

17. The DC diode current IDQ in term of diode voltage VDQ
Total voltage
Total current
VDQ = DC voltage
vd = ac component
The DC diode current IDQ in term of diode voltage VDQ
If vd << VT , the equation can be expanded into linear series as:

18. During AC analysis the diode is equivalent to a resistor, rd
Therefore, the diode current-voltage relationship can be represented as
The relationship between the AC components of the diode voltage and diode current is
Or,
Where,
During AC analysis the diode is equivalent to a resistor, rd

19. IDQ
VDQ = V
+ - rd id
DC equivalent
AC equivalent

20. Example 1 DC Current DC Output voltage
Analyze the circuit (by determining VO & vo ).
Assume circuit and diode parameters of
VPS = 10 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.2 sin ωt
DC Current
DC Output voltage

21. vi vi

22. CALCULATE DC CURRENT, ID CALCULATE AC CURRENT, id
DC ANALYSIS
AC ANALYSIS
DIODE = MODEL 1 ,2 OR 3
CALCULATE rd
DIODE = RESISTOR, rd
CALCULATE DC CURRENT, ID
CALCULATE AC CURRENT, id

23. EXAMPLE 2 VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t (V).
Assume the circuit and diode parameters for the circuit below are
VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t (V).
Determine the current, IDQ and the time varying current, id
ANSWERS
IDQ = mA
Id = 9.97 sin t (µA)