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. Forward-Biased pn Junction
+ve terminal is applied to the p-region of the pn junction and vice versa.
Direction of the applied electric field EA is the opposite as that of the E-field in the space-charge region.
The net result is that the electric field in the space-charge region lower than the thermal equilibrium value causing diffusion of charges to begin again.
The diffusion process continues as long as VD is applied.
Creating current in the pn junction, iD.

4. Forward-Biased pn Junction++
- - W E p n
Equilibrium
- - ++ p n
Forward Biased WF
Width reduces, causing diffusion of carriers current flows

5. Ideal Current-Voltage Relationship
Current ID equation of a pn junction diode:
IS = the reverse-bias saturation current (for silicon to A)
VT = the thermal voltage (0.026 V at room temperature)
n = the emission coefficient (1 ≤ n ≤ 2)

6. Ideal Current-Voltage Relationship
Example
Determine the current in a pn junction diode.
Consider a pn junction at T = 300 K in which IS = 1.4 x A and n = 1.
Find the diode current for vD = V and vD = V.
Very small current

7. PN Junction Diode
The basic PN junction diode circuit symbol, and conventional current direction and voltage polarity.
The graphs shows the ideal I-V characteristics of a PN junction diode.
The diode current is an exponential function of diode voltage in the forward-bias region.
The current is very nearly zero in the reverse-bias region.

8. PN Junction Diode Temperature Effects
Both IS and VT are functions of temperature.
The diode characteristics vary with temperature.
For silicon diodes, the change is approximately 2 mV/oC.
Forward-biased PN junction characteristics versus temperature.
The required diode voltage, V to produce a given current decreases with an increase in temperature.
V2
V1

9. Analysis of PN Junction Diode in a Circuit

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

11. 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.

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

13. 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.

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

15. 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.

16. Why do you need to use these models?

17. 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

18. 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

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

20. VD 2 Use KVL: 2ID + VD – 5 = 0 ID = -VD + 5 = - VD + 2.5
The value of ID VD 2
Use KVL:
2ID + VD – 5 = 0
ID = -VD = - VD + 2.5

21. EXAMPLE
A diode circuit and its load line are as shown in the figure below
Design the circuit when the diode is operating in forward bias condition.
Determine the diode current ID and diode forward resistance rf in the circuit using a piecewise linear model. Consider the cut-in voltage of the diode, Vγ = 0.65V.

22. IDR + VD – 5 = 0
ID = -VD/R + 5/ R
5/ R = 2.5 mA
R = 2 kΩ
at VD = 0.7V
ID = (5 – 0.7) / 2 = 2.15 mA
Now, VD = Vγ + ID rf
0.7 = rf
rf = 0.05 / 2.15 mA = Ω