1. Recall-Lecture 6 Diode AC equivalent circuit – small signal analysis
During AC analysis the diode is equivalent to a resistor, rd
IDQ
VDQ
+ - rd id
DC equivalent
AC equivalent

2. 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

3. Zener effect and Zener diode
When a Zener diode is reverse-biased, it acts at the breakdown region, when it is forward biased, it acts like a normal PN junction diode
Avalanche Effect
Gain kinetic energy – hit another atom –produce electron and hole pair

4. CHAPTER 2 Forward Biased, DC Analysis AC Analysis Reverse Biased
Model 1
V = 0
Model 2 V
Model 3
V and rf
Load Line  ID vs VD
At 300K VT = V
Forward Biased, DC Analysis
AC Analysis
Thermal equilibrium, depletion region
Reverse Biased
Must perform DC Analysis first to get DC diode current, ID
PN junction
Group 5
N-type
P-type
Group 3
Calculate
rd = VT / ID
Insulator
Conductor
Semiconductor
Extrinsic
Semiconductor: Group 4 eg. Silicon and Germanium
photodiode
Intrinsic
Other types of diode
Solar cells
Materials
Bandgap Energy
LED
CHAPTER 2
Zener Diode

5. Chapter 3 Diode Circuits

6. Voltage Regulator

7. A voltage regulator supplies constant voltage to a load.
Voltage Regulator - Zener Diode
A voltage regulator supplies constant voltage to a load.

8. The breakdown voltage of a Zener diode is nearly constant over a wide range of reverse-bias currents.
This make the Zener diode useful in a voltage regulator, or a constant-voltage reference circuit.
3. The remainder of VPS drops across Ri
2. The load resistor sees a constant voltage regardless of the current
1. The Zener diode holds the voltage constant regardless of the current

9. Example
A Zener diode is connected in a voltage regulator circuit. It is given that VPS = 20V, the Zener voltage, VZ = 10V, Ri = 222  and PZ(max) = 400 mW.
Determine the values of IL, IZ and II if RL = 380 .
Determine the value of RL that will establish PZ(max) = 400 mW in the diode.
ANSWER:
Part (a)
IL = mA
IZ = 18.7 mA
II = 45 mA
ANSWER:
Part (b)
PZ = IZ VZ
IZ = 40 mA
IL = = 5 mA
 RL = 2 k

10. For proper function the circuit must satisfied the following conditions.
The power dissipation in the Zener diode is less than the rated value
When the power supply is a minimum, VPS(min), there must be minimum current in the Zener diode IZ(min), hence the load current is a maximum, IL(max),
When the power supply is a maximum, VPS(max), the current in the diode is a maximum, IZ(max), hence the load current is a minimum, IL(min)
AND
Or, we can write

11. Considering designing this circuit by substituting IZ(min) = 0
Considering designing this circuit by substituting IZ(min) = 0.1 IZ(max), now the last Equation becomes:
Maximum power dispassion in the Zener diode is
EXAMPLE 1
Consider voltage regulator is used to power the cell phone at 2.5 V from the lithium ion battery, which voltage may vary between 3 and 3.6 V. The current in the phone will vary 0 (off) to 100 mA(when talking). Calculate the value of Ri and the Zener diode power dissipation

12. Solution:
The stabilized voltage VL = 2.5 V, so the Zener diode voltage must be VZ = 2.5 V. The maximum Zener diode current is
The maximum power dispassion in the Zener diode is
The value of the current limiting resistance is
IZ(min) = 0.1 IZ(max),
(3 – 2.5) (IZmax + 0) = (3.6 – 2.5) (0.1 IZmax mA)
0.5 IZmax = (1.1) (0.1 IZmax mA)
0.5 IZmax = Izmax
0.39 IZmax = 110
IZmax = mA

13. Example 2 Range of VPS : 10V– 14V RL = 20 – 100  VZ = 5.6V
Find value of Ri and calculate the maximum power rating of the diode

14. Solution:
The stabilized voltage VL = 2.5 V, so the Zener diode voltage must be VZ = 2.5 V. The maximum Zener diode current is
The maximum power dispassion in the Zener diode is
The value of the current limiting resistance is
IZ(min) = 0.1 IZ(max),
(10 – 5.6) (IZmax + 56 mA) = (14 – 5.6) (0.1 IZmax mA)
4.4 IZmax = (8.4) (0.1 IZmax mA)
4.4 IZmax = Izmax
3.56 IZmax =
IZmax = mA
PZmax = x = W
Ri = / = 13 