COMMON DIODE APPLICATIONSBasic Power Supply Circuits

Outline Transformers Half-wave rectifiers Full-wave rectifiers Full-wave bridge rectifiers Working with rectifiers Filters Voltage regulators

Introduction

Introduction Power supply A group of circuits that convert ac energy povided by the wall outlet to dc energy

Two basic types Linear power supply Switching power supply

Linear power supply One that provides a constant current path between its input and its load

Switching power supply Provides an intermittent current path between its input and load

Linear power supply Three groups: Rectifier Filter Voltage regulator Rectifier Diode circuit that converts ac to pulsating dc

Filter Circuit that reduces the variations in the output of a rectifier

Voltage regulator Circuit designed to maintain a constant power supply output voltage

Transformers

Transformers Transformer is used to connect the power supply to the ac line input Transformers are made up of inductors, not electrically connected, whereby alternating voltage applied in the primary induces alternating voltage in the secondary Transformer provides ac coupling from primary to secondary while providing physical isolation between the two circuits Three types: Step-up Step-down isolation

Primary (input)

secondary (output)

Transformer symbol

Three types of transformers Step-up transformer Provides secondary voltage that is greater than the primary voltage

Step-down transformer Provides secondary voltage that is less than the primary voltage

Isolation transformer Provides an output voltage that is equal to the input voltage Used for protecting the power supply from problems originating from the ac line

Turns Ratio Turns ratio Ratio of number of turns in the primary to the number of turns in the secondary Equal to the voltage ratio of the componentWhere: NS NP = VS VP NS = number of turns in the secondary NP = number of turns in the primary VS = secondary voltage VP = primary voltage

Current Ratio Ideally, transformers are 100% efficient Ideal transformers transfer 100% of its input power to the secondary: PS = PP ISVS = IPVP

IP IS

=

VS VP

The current ratio is the inverse of the voltage ratio.

Example1. Given: Step-down transformer with 120vac input, turns ratio= 1:4 Required: secondary voltage Solution: VS = (NS/NP)VP = (120vac) = 30 vac

2. Given: Turns ratio= 1:4 fuse with IP limit of 1 A Required: secondary current limit

Solution: IS = (NP/NS)IP = (1 A) = 250 mA

Half-Wave Rectifiers* Diode placed in series between a transformer (or ac line input) and its load * eliminates either the positive or negative alternation of the input

Half-wave rectifier The output from a positive HW rectifier is a series of positive pulses The output from a negative HW rectifier is a series of negative pulses.

Basic circuit operationDiode VD1 VL condition (ideal) (ideal) Forward 0 v biased Reverse VS biasedCircuit recognition: 1. When the diode points toward the load (RL), the output from the rectifier will be positive. 2. When the diode points toward the source, the output from the rectifier will be negative. Positive HW rectifier

Diode VD1 VL condition (ideal) (ideal) Reverse VS biased Forward 0 v biased 0v VS

Negative HW rectifier

VS 0v

CalculationsVpk = 2 Vrm srms value corresponds to the equivalent dc value that will produce the same heating effect Peak value maximum value in a sine wave

Half Wave Rectifier Vs NsVp/Np Vs(pk) VL(pk) Vdc IL Idiode PIV fout 2 Vs Vs(pk) - VB VL(pk) / Vdc / RL IL Vs(pk) fin

Vave = Vpk / Average voltage value that will be measured with a dc voltmeter - dc equivalent of an ac (or other) waveform

Why use peak values? Knowing peak values of the circuit voltages allows us to effectively analyze the circuit operation with an oscilloscope Average (dc) voltage and current values are commonly defined in terms of (and are calculated using) peak values.s

ExampleGiven: Turns ratio= 3:1, HW rectifier, Si diode, 120v 60 Hz power line, load of 100 ohms

Half Wave Rectifier Vs NsVp/Np Vs(pk) VL(pk) Vdc IL Idiode PIV fout 2 Vs Vs(pk) - VB VL(pk) / Vdc / RL IL Vs(pk) fin

Required: Vs, Vdc , IL, Idiode , PIV, and foutAns: 40 vac , 17.76 v, 177.6 mA, 177.6 mA, 56.56 v, 60 Hz

Full Wave RectifierCenter-tapped

Full Wave Rectifier: CTCenter-tapped transformer a transformer with an output lead connected to the center of the secondary winding - the voltage is always half the total secondary voltage

FW rectifier : CT

Full-wave center-tap rectifier: Top half of secondary winding conducts during positive halfcycle of input, delivering positive half-cycle to load.. Full-wave center-tap rectifier: During negative input halfcycle, bottom half of secondary winding conducts, delivering a positive half-cycle to the load.

Dual Polarity FW CT Rectifier

The full-wave center-tapped rectifier polarity at the load may be reversed by changing the direction of the diodes. Furthermore, the reversed diodes can be paralleled with an existing positive-output rectifier. The result is dual-polarity fullwave center-tapped rectifier

CalculationsHalf Wave Rectifier Vs NsVp/Np Vs(pk) VL(pk) Vdc IL Idiode PIV fout 2 Vs Vs(pk) - VB VL(pk) / Vdc / RL IL Vs(pk) FW CT Vs NsVp/Np Vs(pk) VL(pk) Vdc IL Idiode PIV 2 Vs Vs(pk) /2 - VB 2 VL(pk) / Vdc / RL IL /2 Vs(pk) - VBBecause each diode supplies half rectified waveform to the load

f FW rectifier has twice the output frequency of the HW fout 2 fin Thein rectifier.

ExampleFW CT Vs NsVp/Np Vs(pk) VL(pk)Given: Turns ratio= 3:1, 120vac , 60Hz, RL = 100 ohms Required: Vdc , IL, Idiode , PIV for D1, fout Ans: 17.54v, 175.4 mA, 87.7 mA, 55.86 v, 120 Hz

2 Vs Vs(pk) /2 - VB 2 VL(pk) / Vdc / RL IL /2 Vs(pk) - VB 2 fin

Vdc IL Idiode PIV fout

Limitation One disadvantage of this full-wave rectifier design is the necessity of a transformer with a center-tapped secondary winding. If the circuit in question is one of high power, the size and expense of a suitable transformer is significant. Consequently, the center-tap rectifier design is only seen in low-power applications.

Full Wave RectifiersBridge Type

FW Bridge Rectifier Why are bridge rectifiers preferred over other FW rectifiers? It does not require the use of center-tapped transformer and therefore can be coupled directly to the ac power line (if desired). When connected to a transformer with the same secondary voltage, it produces nearly twice the peak output voltage of the conventional FW rectifier. This results in a higher dc output voltage from the supply.

FW Bridge Rectifier

When a bridge rectifier, as opposed to the 2 diode FW rectifier, is used, the same dc output voltage can be obtained with a transformer having a urns ratio Np/Ns. Meaning, fewer turns are needed in the secondary transformer. Hence, bridge rectifiers may be smaller, lighter, and probably costs less.

FW bridge

Circuit operationFull-wave bridge rectifier: Electron flow for positive half-cycles.

Full-wave bridge rectifier: Electron flow for negative half=cycles.

Alternate layout style

CalculationsVs Vs(pk) VL(pk) Vdc IL Idiode PIV fout HWFW CT NsVp/Np NsVp/Np 2 Vs Vs(pk) - VB VL(pk) / Vdc / RL IL Vs(pk) fin 2 Vs Vs(pk) /2 - VB 2 VL(pk) / Vdc / RL IL /2 Vs(pk) - VB 2 fin FW Bridge NsVp/Np 2 Vs Vs(pk) 2 VB 2 VL(pk) / Vdc / RL IL /2 Vs(pk) - VB 2 fin

ExampleFW Bridge Vs NsVp/Np Vs(pk) VL(pk) VdcGiven: Turns ratio= 3:1, 120vac , 60Hz, RL = 100 ohms Required: Vdc , IL, Idiode , PIV for each diode, fout Ans: 35.08 v, 350.8 mA, 175.4 mA, 55.86 v, 120 Hz

2 Vs Vs(pk) 2 VB 2 VL(pk) / Vdc / RL IL /2 Vs(pk) - VB 2 fin

IL Idiode PIV fout

Review Rectification is the conversion of alternating current (AC) to direct current (DC). A half-wave rectifier is a circuit that allows only one half-cycle of the AC voltage waveform to be applied to the load, resulting in one nonalternating polarity across it. The resulting DC delivered to the load pulsates significantly. A full-wave rectifier is a circuit that converts both half-cycles of the AC voltage waveform to an unbroken series of voltage pulses of the same polarity. The resulting DC delivered to the load doesn't pulsate as much.

FiltersCapacitive Filter

Filters Filters reduce the variations in the rectifier output signal. Since our goal is to produce a constant dc output voltage, it is necessary to remove as much of the rectifier output variation as possible. Also known as peak detectors

Concepts Ripple voltage (Vr or Vripple) The variation in the output voltage from a filter

Power supplies are designed to produce as little ripple voltage as possible. Too much ripple in the output can have adverse effects.

Basic Capacitive Filter Capacitive filter Most basic filter type and the most commonly used A capacitor connected in parallel with the load resistance The filtering action is based on the charge/discharge action of the capacitor.

Operation A peak detector is a series connection of a diode and a capacitor outputting a DC voltage equal to the peak value of the applied AC signal During the + half-cycle of the input, D1 conducts and the capacitor charges rapidly. As the input starts to go negative, D1 turns off, and the capacitor slowly discharges through the load resistance. As the output from the rectifier drops below the charged voltage of the capacitor, the capacitor acts as the voltage source for the load. It is the difference between the charge and discharge times of the capacitor that reduces the variations in the rectifier output voltage.

Peak detector: Diode conducts on positive half cycles charging capacitor to the peak voltage (less diode forward drop).

RC time constant Recall: a capacitor will charge (or discharge) in five time constants The amplitude of the ripple voltage at the output of a filter varies inversely with the values of filter capacitance and load resistance.one time constant: example: Given: C= 100uF, rB = 5 ohms, RL = 1 kilo-ohm charge time: T = 5rBC = 5(5)(100u) = 2.5 ms discharge time: T = 5RLC = 5(1k)(100u) = 500 ms Note: capacitor charges almost instantly yet barely starts to discharge before another charging voltage is provided by the rectifier.

= RC(dis)charging period: = 5RC where: R- resistance C- capacitance

consideration What limits the value of filter capacitance? The maximum allowable charge time for the component The amount of surge current (Isurge ) that the rectifier diodes can withstand The cost of larger-than-needed filter capacitors

The value of the filter capacitor affects both its charge time and its discharge time. If you increase the filter capacitance to increase its discharge time, you increase its charge time as well. This can cause a surge current problem within the power supply.

Surge Current surge current high initial current in the power supply

current-limiting resistor low-resistance, high wattage resistor (connected in series) that limits surge current, but also reduces the output voltage

Isurge

=

Vs(pk) Rw + rB

C = I (t) Vcwhere: C = capacitance, in farads I = dc charge/discharge current t= charge/discharge time Vc = change in capacitor voltage during charge/discharge

where: Vs(pk) peak secondary voltage Rw resistance of secondary windings rB diode bulk resistance

Filter Output VoltageVdc = Vpk Vripple 2 Vripple = IL t Cwhere: Vpk = peak rectifier output voltage Vripple = peak-to-peak value of ripple voltage IL = dc load current t = time between charging peaks C = capacitance, in farads

t = 1/f

example1. Given: HW rectifier C = 500uF; IL = 20 mA Required: ripple voltage

Solution: t = 1/f = 1/60 Hz = 16.7 ms Vripple = IL t /C = (20mA)(16.7ms)/500uF = 668 mVpp

2. Given: FW rectifier C = 500 uF; IL = 20 mA Required: ripple voltage

Solution: t = 1/f = 1/120 Hz = 8.33 ms Vripple = IL t /C = (20mA)(8.33ms)/500uF = 333 mVpp

Note: Since our goal is to have a steady dc voltage that has little ripple voltage as possible, the FW rectifier gets us much closer to our goal than does the HW rectifier.

example3. Given: FW rectifier (CT), 24 vac C = 470 uF RL = 1.2 kilo ohms Required: Vdc

Solution: Vs(pk) = 2 vac = 2 (24) = 33.9 v VL(pk) = Vs(pk) /2 VB = 33.9 /2 0.7 = 16.3v **assuming Vdc = VL(pk) = 16.3 v IL = Vdc / RL= 16.3/1.2k = 13.6 mA Vripple = IL t/ C = (13.6mA)(8.33ms) 470uF = 241 mVpp Vdc = VL(pk) Vripple /2 = 16.3 v 120.5 mV = 16.2 v

Voltage Regulatorusing zener diode

Zener Voltage RegulatorAc input Rectifier Filter IT Regulator Rs

IT = Vin Vz Rs IL = VZ IZ = IT - IL RL

D1

Load

Where: IT = the total current drawn through Rs ; IL = load current; IZ = zener current Vin = the input voltage Vz = the nominal (rated) zener voltage

Load regulation Load regulation = the ability of a regulator to maintain a constant load voltage despite anticipated variations in load current demand

Zener reduction of ripple voltage The zener regulator provides an added bonus: It reduces the amount of ripple voltage at the filter output. The zener impedance (Zz ) is a dynamic value; that is, the opposition that a zener diode presents to a change in voltage or current

Vr(out)

= (Zz || RL ) Vr (Zz || RL ) + Rs

Where: Vr(out) = the ripple present at the regulator output (Zz || RL ) = the parallel combination of zener impedance and load resistance Rs = the regulator series resistance Vr = the peak-to-peak ripple voltage present at the regulator input

Putting it all together The combination of the 4 circuits has converted an ac line voltage to a steady dc supply voltage that remains relatively constant when load current demands change. Procedure1. 2. 3. 4. 5. 6. 7. determine the value of Vs(pk) Determine the peak rectifier output voltage Determine the total current through the series resistor. This current value will be used when calculating the value of ripple voltage Determine the value of ripple voltage from the filter Find the Vdc at the output. Equate it to Vz rating of the zener diode. Approximate the final ripple output voltage Using Vz and RL, determine the load current

example

Determine the values of Vdc , Vr(out) , and IL for the power supply shown.

solutionVs(pk) Vpk Vr Vdc Vr(out) = 36 Vac / 0.707 = 51 V = Vs(pk) 1.4 V = 49.6 V // 2 diodes = IR t/ C = (261 mA)(8.33 ms)/2200 F = 988 mVpp = VZ = 30 V = (Zz || RL )Vr /[(Zz || RL ) + Rs ] IL = VZ / RL = 30 V/ 300 = 100 mA = 50 (988 mVpp )/ 125 = 395 mVpp

IR = (Vin VZ )/Rs = (49.6 V 30 V)/ 75 = 261 mA