RecallLecture 6 Diode AC equivalent circuit small signal

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

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

Model 1 V = 0 Model 2 V Model 3 V and rf Load Line ID vs VD Forward Biased, DC Analysis At 300 K VT = 0. 026 V Reverse Biased Group 5 Materials CHAPTER 2 Semiconductor: Group 4 eg. Silicon and Germanium Bandgap Energy Must perform DC Analysis first to get DC diode current, ID PN junction N-type Insulator Conductor Semiconductor AC Analysis Thermal equilibrium, depletion region P-type Group 3 Calculate rd = VT / I D Extrinsic photodiode Intrinsic Other types of diode Solar cells LED Zener Diode

Chapter 3 Diode Circuits

Voltage Regulator

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

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

Example A Zener diode is connected in a voltage regulator circuit. It is given that VPS = 20 V, the Zener voltage, VZ = 10 V, Ri = 222 and PZ(max) = 400 m. W. a. Determine the values of IL, IZ and II if RL = 380 . b. Determine the value of RL that will establish PZ(max) = 400 m. W in the diode. ANSWER: Part (a) IL = 26. 3 m. A IZ = 18. 7 m. A II = 45 m. A ANSWER: Part (b) PZ = I Z VZ IZ = 40 m. A IL = 45 -40 = 5 m. A RL = 2 k

For proper function the circuit must satisfied the following conditions. 1. The power dissipation in the Zener diode is less than the rated value 2. 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), 3. 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. 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 m. A(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 IZ(min) = 0. 1 IZ(max), (3 – 2. 5) (IZmax + 0) = (3. 6 – 2. 5) (0. 1 IZmax + 100 m. A) 0. 5 IZmax = (1. 1) (0. 1 IZmax + 100 m. A) 0. 5 IZmax = 0. 11 Izmax + 110 0. 39 IZmax = 110 IZmax = 282. 05 m. A The maximum power dispassion in the Zener diode is The value of the current limiting resistance is

• Example 2 Range of VPS : 10 V– 14 V RL = 20 – 100 VZ = 5. 6 V 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 IZ(min) = 0. 1 IZ(max), (10 – 5. 6) (IZmax + 56 m. A) = (14 – 5. 6) (0. 1 IZmax + 280 m. A) 4. 4 IZmax + 246. 4 = (8. 4) (0. 1 IZmax + 280 m. A) 4. 4 IZmax + 246. 4 = 0. 84 Izmax + 2352 3. 56 IZmax = 2105. 6 IZmax = 591. 46 m. A The maximum power dispassion in the Zener diode is PZmax = 5. 6 x 591. 46 = 3. 312 W The value of the current limiting resistance is Ri = 8. 4 / 647. 46 = 13