Examples to Power Supply

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Lucy Terry
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1 Examples to Power Supply Example-1: A center-tapped full-wave rectifier connected to a transformer whose each secondary coil has a r.m.s. voltage of 1 V. Assume the internal resistances of the diode and load resistance are 50 Ω and 1k Ω, respectively. Find: 1- The load current measured by a voltmeter. - The D.C. load current. Solution V PS = V rms = 1 V = V I P = V PS R S + R L = V 50 Ω Ω = ma The load current measured by a voltmeter is I rms = I P = ma = 11.4 ma I dc = I P π = ma π = 10.8 ma Example-: In the center-tapped full-wave rectifier shown below find: 1- The dc output voltage. - PIV. 3- Rectification efficiency. 4- Output frequency. Assume the diode resistance is 0 Ω and the frequency of the input is 50 Hz. 1

2 Solution 1 - V PS = V n rms = n 1 0 V 10 = V I dc = V PS π (R S + R L ) = V π 0 Ω Ω = ma V dc = I dc R L = A 1000 Ω = V PIV = V PS = 6. V 4 η FW η FW η FW R S Ω 1000 Ω Ω 1000 Ω R L η FW 79.4 % 100% 100% f out = f in = 50 Hz = 100 Hz

3 Example-3: In the full-wave bridge rectifier shown below find: 1- The dc output voltage. - PIV. 3- Rectification efficiency. Assume the diode resistance is 0 Ω. Solution n V PS = V rms = 0 V n 1 10 = 6. V V PS I dc = π (R S + R L ) = 6. V π 0 Ω Ω = ma V dc = I dc R L = A 1000 Ω = V PIV = V PS = 6. V Ripple Factor The output of a rectifier consists of a d.c. component and an a.c. component. The a.c. component is undesirable and causes pulsations in the rectifier output. Thus, the smaller this component, the more effective is the rectifier. The ratio of r.m.s. value of a.c. component to the d.c. component in the rectifier output is called ripple factor and calculated as Ripple factor = R = V rms V ac 3

4 V rms V dc 1 Half-wave rectifier R = V Ps / V Ps /π 1 = 1.1 Thus, AC component is more than DC component. Full-wave rectifier R = V Ps / V Ps /π 1 = 0.48 Thus, AC component is less than DC component. Voltage Filtration Since the output of a rectifier has pulsating, the A.C. component is undesirable and must be omitted. To do so, a filter circuit is used which finally removes or reduces value of the A.C. component. (A) RC Filter A capacitor C is placed at the output of a rectifier and in parallel with load R L. During the positive-quarter cycle of the rectifier voltage output, the capacitor charges and also a current flows to the load. At the end of the quarter cycle the voltage becomes lower than the voltage across the capacitor, it will then discharge through the load. X C = 1 πfc 4

5 RC Filter (B) LC Filter It is a compensation of a coil L and a filter capacitor C connected in series with the rectifier output and across the load. A single filter section or a several identical sections can be used to reduce the pulsations. X L = πfl The most of the A.C. component appears across the coil while whole of D.C. component passes through the coil to load. This results in the reduced pulsations at the load. LC Filter 5

6 Filtered half-wave rectifier waveform (B) LC Filter It is a compensation of a coil L and a filter capacitor C connected in series with the rectifier output and across the load. A single filter section or a several identical sections can be used to reduce the pulsations. X L = πfl The most of the A.C. component appears across the coil while whole of D.C. component passes through the coil to load. This results in the reduced pulsations at the load. (C) π-section Filter It consists of two filter capacitors C1 and C connected with a coil L resulting in a π section. The section is connected across the output of the power supply. 6

7 (a) The filter capacitor C1 offers low reactance to A.C. component of rectifier output of A.C. component while the D.C. component goes to the coil L. (b) The coil L blocks the A.C. component but it offers almost zero reactance to the D.C. component. (c) The filter capacitor C bypasses the A.C. component which the coil has failed to block. Therefore, only D.C. component appears across the load. Example-4: For the circuit shown below, the D.C. resistance of the choke is 100 Ω. Find the output D.C. voltage, if the peak voltage applied to the choke is 5 V. Solution R L V dc = V dc(in) R L + X L V dc = V p π R L R L + X L V dc = 5 π 1 kω 1 kω kω V dc = V 7

8 Voltage Stabilization Till now the power supply that are studied serves as an appropriate D.C. source output. But, there is a problem that is the output may changes with the variations occurring in the input. If the input voltage of the power supply is increased due to an error in the main source, the output would also change causing various problems in the loads. Therefore, regulators are usually used to protect the loads against the change in the output voltage. Zener diode is used as a voltage regulator. Zener Diode The zener diode operates in the breakdown region of a pn-junction diode. Therefore, the breakdown voltage is called zener voltage and the current passes through it is known as zener current. This voltage is doping amount dependent. If the diode is heavily doped, depletion region would be small so that the breakdown may occur at a low reverse voltage. I-V Characteristics of Zener Diode I-V Characteristics of Zener Diode 1- A zener diode shows sharp breakdown voltage V Z because it is different from the pn junction diode with doping. - It is connected in reverse-bias configuration, the cathode is connected to the positive terminal. 8

9 Electronic circuit of a zener diode 3- It behaves like pn-junction diode, if being forward biased. 4- The zener diode is not easily damaged while it operates in the breakdown region because a limited current called I Z will pass through it, which is always lower than the current that causes. However, the zener diode current is determined between two the maximum and minimum values I Zmax I Z I Zmin. 5- If V out > V Z, then I Z will flow in the zener diode decreasing the output voltage so that V L = V Z and so I L = I out I Z. Equivalent circuit for a zener diode in the case of closed switch if V out > V Z 6- If V out < V Z, then I Z will not flow in the zener diode and therefore the output voltage will present on the load, V L = V out and so I L = I out. 9

10 Equivalent circuit for a zener diode in the case of opened switch if V out < V Z Example 5 For the circuit shown below find: 1- The output voltage - The voltage drop across series resistance 3- The current through zener diode. Solution 10

Positive Voltage Regulator There are also integrated circuit can be used as voltage regulators. The output of this regulator is fixed to a positive output voltage.

11 Positive Voltage Regulator There are also integrated circuit can be used as voltage regulators. The output of this regulator is fixed to a positive output voltage. The most used series is 7800 series and the ICs will be like IC 7805 and IC 781, which provide +5V and +1V, respectively. The figure below shows an IC 7805 connected to a 1V to provide a fixed 5V positive regulated output voltage. Electronic circuit of IC 7805 Regulator 11

Sheet 2 Diodes. ECE335: Electronic Engineering Fall Ain Shams University Faculty of Engineering. Problem (1) Draw the

Sheet 2 Diodes. ECE335: Electronic Engineering Fall Ain Shams University Faculty of Engineering. Problem (1) Draw the Ain Shams University Faculty of Engineering ECE335: Electronic Engineering Fall 2014 Sheet 2 Diodes Problem (1) Draw the i) Charge density distribution, ii) Electric field distribution iii) Potential distribution,