Review Half Wave Full Wave Rectifier Center tapped
- Slides: 22
Review • Half Wave • Full Wave Rectifier • Center tapped • Bridge • Rectifier Parameters • PIV • Duty Cycle
Filters Ø A capacitor is added in parallel with the load resistor of a half-wave rectifier to form a simple filter circuit. At first there is no charge across the capacitor Ø During the 1 st quarter positive cycle, diode is forward biased, and C charges up. Ø VC = V O = V S - VD Ø As VS falls back towards zero, and into the negative cycle, the capacitor discharges through the resistor R. The diode is reversed biased ( turned off) Ø If the RC time constant is large, the voltage across the capacitor discharges exponentially.
Filters Ø During the next positive cycle of the input voltage, there is a point at which the input voltage is greater than the capacitor voltage, diode turns back on. Ø The diode remains on until the input reaches its peak value and the capacitor voltage is completely recharged.
Vpeak VM Quarter cycle; capacitor charges up Capacitor discharges through R since diode becomes off VC = VMe – t / RC Input voltage is greater than the capacitor voltage; recharge before discharging again NOTE: Vm is the peak value of the output voltage Since the capacitor filters out a large portion of the sinusoidal signal, it is called a filter capacitor.
Ripple Voltage, and Diode Current ØVr = ripple voltage Tp ØVr = VM – VMe -T’/RC T’ where T’ = time of the capacitor to discharge to its lowest value Vr = VM ( 1 – e -T’/RC ) Expand the exponential in series, Vr = ( VMT’) / RC Figure: Half-wave rectifier with smoothing capacitor.
• If the ripple is very small, we can approximate T’ = Tp which is the period of the input signal • Hence for half wave rectifier Vr = ( VMTp) / RC l For full wave rectifier Vr = ( VM 0. 5 Tp) / RC
Example Consider a full wave center-tapped rectifier. The capacitor is connected in parallel to a resistor, R = 2. 5 k. The input voltage has a peak value of 120 V with a frequency of 60 Hz. The output voltage cannot be lower than 100 V. Assume the diode turn-on voltage, V = 0. 7 V. Calculate the value of the capacitor. VM = Vo peak = 120 - VD 120 – 0. 7 = 119. 3 V Vr = 119. 3 – 100 = 19. 3 V 19. 3 = 119. 3 / (2*60*2500*C) C = 20. 6 F
Example Consider a full wave bridge rectifier. The capacitor C = 20. 3 F is connected in parallel to a resistor, R = 10 k. The input voltage, vs = 50 sin (2 (60)t). Assume the diode turn-on voltage, V = 0. 7 V. Calculate the value of the ripple voltage. Frequency = 60 Hz VM = Vo peak = 50 – 2 VD 50 – 1. 4 = 48. 6 V Vr = 48. 6 / (2*60*10 x 103*20. 3 x 10 -6) Vr = 2 V
The full-wave rectifier circuit is shown in the figure below. The output peak current of the circuit is 200 m. A when the peak output voltage is 12 V. Assume that input supply is 120 V(rms), 60 Hz and diode cut-in voltage Vγ = 0. 7 V. Find the required value of C for limiting the output ripple voltage, Vr = 0. 25 V. Answer: C = 6. 67 m. F
RECALL: Zener Effect and Zener Diode Ø The applied reverse biased voltage cannot increase without limit since at some point breakdown occurs causing current to increase rapidly. Ø The voltage at that point is known as the breakdown voltage, VZ Ø Diodes are fabricated with a specifically design breakdown voltage and are designed to operate in the breakdown region are called Zener diodes. Circuit symbol of the Zener diode: NOTE: 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 Ø Such a diode can be used as a constant-voltage reference in a circuit. Ø The large current that may exist at breakdown cause heating effects and catastrophic failure of the diode due to the large power dissipated in the device. Ø Diodes can be operated in the breakdown region by limiting the current to a value within the capacities of the device.
Basis For Comparison PN Junction Diode Zener Diode Definition It is a semiconductor diode which conducts only in one direction, i. e. , in forward direction. The diode which allows the current to flow in both the direction i. e. , forward and reverse, such type of diode is known as the Zener diode. Reverse Current Effect Damage the junction. Do not damage the junction. Doping Level Low High Breakdown Occurs in higher voltage. Occur in lower voltage. Symbol
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