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

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

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

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 = 0.026 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

Chapter 3 Diode Circuits

Voltage Regulator

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

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

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 = 26.3 mA IZ = 18.7 mA II = 45 mA ANSWER: Part (b) PZ = IZ VZ IZ = 40 mA IL = 45 -40 = 5 mA  RL = 2 k

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

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

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 + 100 mA) 0.5 IZmax = (1.1) (0.1 IZmax + 100 mA) 0.5 IZmax = 0.11 Izmax + 110 0.39 IZmax = 110 IZmax = 282.05 mA

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

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 + 280 mA) 4.4 IZmax + 246.4 = (8.4) (0.1 IZmax + 280 mA) 4.4 IZmax + 246.4 = 0.84 Izmax + 2352 3.56 IZmax = 2105.6 IZmax = 591.46 mA PZmax = 5.6 x 591.46 = 3.312 W Ri = 8.4 / 647.46 = 13 