UNIT V RECTIFIERS AND POWER SUPPLIES Half-wave, full-wave and bridge rectifiers with resistive load. Analysis for Vdc and ripple voltage with C,CL, L-C and C-L-C filters. Voltage multipliers Zenerdiode regulator. Electronically regulated d.c power supplies. Line regulation, output resistance and temperature coefficient. Switched mode power supplies. Power control using SCR.

Objectives:

• Introduction to Power Supply • Rectifier with filter • Voltage regulators • Switching regulator

Electronic systems invariably require a source of d.c. power and, except in the

case of battery-powered (portable) equipment, the a.c. mains (240 V rms, 50 Hz) must be converted to the appropriate d.c. voltage. This is performed in a power supply unit (PSU) which usually comprises a transformer, rectifier, filter and regulator as shown in fig.5.1

Fig (5-1) Block diagram showing parts of a power supply. 1-The ac voltage, typically 120 volts rms, is connected to a transformer which steps that voltage up, or down to the level for the desired dc output. 2-A diode rectifier then provides a half wave or, more typically, full-wave-rectified voltage which is applied to a filter to smooth the varying signal. A rectifier circuit is necessary to convert a signal having zero average value to one that has a nonzero average. The resulting dc signal is not pure dc or even a good representation of it. 3-A simple capacitor filter is often sufficient to provide this smoothing action. 4-The resulting dc voltage with some ripple or ac voltage variation is then provided as input to an IC regulator.

5-IC regulator that provides as output a well-defined dc voltage level with extremely low ripple voltage over a range of load. 5.1Rectifier: 5.1.1 Half-Wave Rectifier Half-wave rectification: is the process of removing one half of the input signal to establish a dc level. The Circuit of the fig(5-2) called a half wave rectifiers will generates a waveform Vo that will have an average value of particular use in the ac– to–dc conversion process.

Fig (5-2) half wave Rectifier During the interval (t=0 to T/2) the polarity of the input voltage Vi is shown in fig (5-3).

Fig (5-3) conducting region (0 toT/2) The result that for period 0 to T/2, Vo=Vi. For period T/2 to T, the polarity of the input voltage Vi is shown in fig(2-18 ) and the ideal diode produces in off state, Vo=0V.

Fig (5-4) Non conducting region (T/2 to T) An average value determined by average value determined by average dc value = 0.318Vm

Fig (5-5) half wave rectified signal. The effect of using a silicon diode with VT=0.7V is shown by fig (5-6) for the forward bias. The input must now be at least 0.7V before the diode conducts. When conducting Vo =Vi – VT If Vm > VT … i.e. Vdc= 0.318Vm if Vm is close to VT … i.e. Vdc ≈ 0.318(Vm – VT)

Fig (5-6) Effect of VT on Half-wave rectified signal Eexample 1: for Half wave Rectifier: (a) Sketch the output vo and determine the dc level of the output for the network of Fig (5-7),(b) Repeat part (a) if the ideal diode is replaced by a silicon diode.

Fig (5-7) Solution: (a) In this situation the diode will conduct during the negative part of the input and vo will appear as shown in fig (5-8)

Fig (5-8) For the full period, the dc level is Vdc = -0.318Vm = -0.318(20) = -6.36 V (b) Using a silicon diode, the output has the appearance of Fig (5-9)

Fig (5-9) Vdc = -0.318(Vm - 0.7) = -0.318(19.3) = -6.14 V 5.1.2 Full-Wave Rectification The dc level obtained from a sinusoidal input by half wave rectifier can be improved using a process called Full-Wave Rectification. Four diodes in a bridge configuration can be used as Full Wave Rectifier as shown in fig (5-10).

Fig (5-10) Full-wave bridge rectifier Fig (5-11) Network for period (0 to T/2) For the positive region of the input the conducting diodes are D2 & D3 while D1& D4 are in the off state as shown in fig (5-12).

Fig (5-12) For the negative region of the input the conducting diodes are D1 & D4 while D2 & D3 are in the off state as shown in fig (5-13).

Fig (5-13) The dc level for Full wave rectifier is twice that obtained for a half wave system i.e. average(d.c) level= 0.636Vm Over one full cycle the input and output voltage is shown in fig (5-14)

Fig (5-14) The effect of Vo has also doubled, as shown in fig (5-15) for silicon diodes during the conduction state (for positive region). i.e. Vd.c=0.636Vm (Vm >> 2VT) And if Vm is close to 2VT i.e. Vd.c=0.636(Vm- 2VT) A second popular full wave rectifier used only two diodes but requiring a centre tapped (CT) transformer to establish the input signal across each section of the secondary of the transformer as shown in fig (5-15).

Fig (5-15) During the positive portion of Vi applied to the transformer, the diode D1 is short circuit and the diode D2 is open circuit.

Fig (5-16) During the negative portion of Vi applied to the transformer, the diode D1 is open circuit and the diode D2 is short circuit as shown in fig (5-17).

Fig (5-17)

Example 2: for Full-wave rectifier. Determine the output wave-form for the network of Fig (5-18) and calculate the output dc level

Fig (5-18) Solution: The network will appear as shown in Fig (5-19) for the positive region of the input voltage,

Fig (5-19) Where vo=1/2vi or Vo(max) = 1/2Vi(max) = 1/2(10) = 5 volt For the negative region of the input voltage the network will be appear as shown in Fig (5-20).

Fig (5-20) The effect of removing two diodes from the bridge configuration was therefore to reduce the available dc level to the following: Vdc = 0.636(5) = 3.18 volt 5.2 Simple-Capacitor Filter A popular filter circuit is the simple-capacitor filter circuit, the capacitor is connected across the rectifier output this filtered voltage has a dc level with some ripple voltage riding on it .

Fig (5-21) Simple capacitor filters

. Fig (5-22) capacitor filter operation (a) full-wave rectifier voltage (b) filtered output voltage.

Fig(5-23)(a) capacitor filter circuit (b) output voltage waveform If no load were connected to the filter, the output waveform a constant dc level equal in value to the peak voltage Vm from the rectifier circuit. For the full-wave-rectified signal indicated in Fig (5-23) T1 is the time during which a diode of the full-wave rectifier conducts and charges capacitor up to the peak rectifier output voltage Vm . T2 is the time during which the rectifier voltage drops below the peak voltage, and the capacitor discharges through the load. The capacitor filter circuit provides a large dc voltage with little ripple for light loads and a smaller dc voltage with larger ripple for heavy loads.

Fig (5-24a) Approximate output voltage of capacitor filter circuit. Figure (5-24a) shows the output waveform approximated by straight line charge and discharge. From an analysis of this voltage waveform the following relation can be:

Since the form of the ripple waveform for half-wave is the same as for full-wave. Ripple Voltage Vr(rms)

Assuming a triangular ripple waveform approximation as shown in fig (5-24b)

Fig (5-24b) Approximate triangular ripple voltage for capacitor filter

Fig (5-24c) Ripple voltage During capacitor-discharge the voltage change across C is:

f = frequency of the sinusoidal ac power supply voltage usually 60 Hz Idc = average current drawn from the filter by the load, C = filter capacitor value. For light loads Vdc ≈ Vm

f = 60 Hz

Idc in mA , C in µf, RL in kΩ. Example 3: Calculate the ripple voltage of a full-wave rectifier with a 100 µF filter capacitor connected to a load of 50 mA. Solution:

DC Voltage, Vdc:

f = 60 Hz

Vm = the peak rectified voltage, V Idc = the load current, mA C = the filter capacitor, µf Example 4: The peak rectified voltage for the filter circuit of Ex 3: is 30 volts, calculate the filter dc voltage. Solution:

Note: The larger value of average current drawn from the filter the less value of output dc voltage The larger the value of the filter capacitor the closer the output dc voltage approaches the peak value of Vm. Filter-Capacitor Ripple Using the definition of ripple, we obtain the expression for the ripple factor of a full wave capacitor filter

Since Vdc and Idc relate to the filter load RL , we can also express the ripple as

Idc in mA, C in µf,Vdc in volts,RL in kΩ. larger load current, larger ripple factor and inversely with the capacitor size. 5.2.1 RC Filter It is possible to further reduce the amount of ripple across a filter capacitor while reducing the dc voltage by using an additional RC filter section as shown in Fig (5-25)

Fig (5-25) RC filter stage The purpose of the added network is 1-to pass the dc component of the voltage developed across the first filter capacitor Cl . 2-and to attenuate the ac component of the ripple voltage developed across Cl .

• This action would reduce the amount of ripple in relation to the dc level, providing better filter operation than for the simple-capacitor filter.

• Since the rectifier feeds directly into a capacitor, the peak currents through the

diodes are many times the average current drawn from the supply.

• The voltage developed across capacitor Cl is then further filtered by the resistor-

capacitor section (R,C2) providing an output voltage having less percent of ripple than that across Cl.

• The load RL draws dc current through resistor R with an output dc voltage across

the load being less than that across Cl due to the voltage drop across R.

• Fig (5-26) Full wave rectifier and RC filter circuit. 5.2.1.1 DC Operation of RC Filter Section

The equivalent circuit used for considering the dc voltage and current in the filter and load shown in fig (5-27). The two filter capacitors are open circuit for dc.

Fig (5-27) (a) dc equivalent circuit (b) ac equivalent circuit Fig (5-27a) show that the voltage Vdc across capacitor Cl is attenuated by a resistor-divider network of R and RL , the resulting dc voltage across the load being V'dc

Example 5: The addition of an RC filter section with R=120Ω, reduces the dc voltage across the initial filter capacitor from 60 V (Vdc). If the load resistance is 1kΩ, calculate the value of the output dc voltage (Vdc) from the filter circuit. Solution:

The drop across the filter resistor and the load current drawn:

5.2.1.2 AC Operation of RC Filter Section

Fig (5-27b) is the equivalent circuit for analyzing the ac operation of the filter circuit. Vr (rms) is the input to the filter stage, it is a ripple or ac signal part of the voltage across C1.Both the RC filter stage R, C2 and the load resistance RL affect the ac signal at the output.

• For C2 = 10µF at a ripple voltage frequency f = 60Hz, the ac impedance of the capacitor is:

This capacitive impedance is in parallel with the load resistance RL

• If RL= 2 kΩ, the parallel combination of the two components would an impedance Z

This is close to the value of the capacitive impedance (Xc) alone. Note:

• In parallel combination we can neglecting the loading RL if RL > 5Xc • The ripple frequency = 60 Hz for the ripple voltage from a half-wave rectifier. • The ripple frequency = 120 Hz for the ripple voltage from a full-wave rectifier

• Xc = 1/ wC • We have value of w = 377 for 60 Hz and of w = 754 for 120 Hz.

Example6: Calculate the impedance of a 15µF capacitor used in the filter section of a circuit using full wave rectification. Solution:

Using the simplified relation that the parallel combination of the load resistor and the capacitive impedance equals, approximately, the capacitive impedance, the ac attenuation in the filter is

If RL > 5Xc then :

5.3 Voltage-Multiplier Circuits 5.3.1 Voltage Doubler

• A modification of the capacitor filter circuit allows building up a larger voltage than Vm

• The use of this type of circuit allows keeping the transformer peak voltage rating low while stepping up the peak output voltage to two, three, four, or more times Vm

Fig (5-28) half wave voltage doubler

1. During the positive-voltage half-cycle across the transformer, 2. Dl conducts D2 is cut off, charging Cl up to the Vm 3. During the negative half-cycle of the secondary voltage, 4. Dl is cut off and D2 conducts charging C2 5.

Fig (5-29) Double operation, (a) positive half-cycle (b) negative half-cycle We can sum the voltages around the outside loop (Fig. 5-29b):

From which

On the next positive half-cycle:

• D2 is none conducting and C2 will discharge through the load. • If no load is connected across C2 both capacitors stay charged Cl to Vm and C2 to

2Vm. • If there is a load connected to the output , the voltage across C2 drops during the

positive half-cycle and the capacitor is recharged up to 2Vm during the negative half-cycle.

• Another doubler circuit is the full-wave doubler of Fig (5-30).

Fig (5-30) full wave voltage doubler

Fig (5-31) Alternate half-cycle of operation for full wave voltage doubler

• During the positive half-cycle of transformer secondary voltage • D1 conducts charging Cl to a peak voltage Vm , D2 is non conducting at this time. • During the negative half-cycle • D2 conducts charging C2 while Dl is non conducting.

One difference is that the effective capacitance is that of Cl and C2 in series, which is less than the capacitance of either Cl or C2 alone. The lower capacitor value will provide poorer filtering action than the single-capacitor filter In summary: The half-wave or full-wave voltage doubler provide twice the peak voltage 2Vm 2-Voltage Tripler and Quadrupler Fig (5-32) shows an extension of the half-wave voltage doubler, which develops three and four times Vm It should be obvious from the pattern of the circuit connection how additional diodes and capacitors may be connected so that the output voltage may also be five, six, seven, etc.,times Vm

Fig (5-32) voltage tripler and quadrupler.

• During the positive half-cycle • Cl charges through D1 to a peak voltage Vm • During the negative half-cycle • C2 charges to twice the peak voltage 2Vm • During the positive half-cycle: • D3 conducts and the voltage across C2 charges C3 to the same 2Vm • On the negative half-cycle • D2 and D4 conduct with C3 charging C4 to 2Vm. •

The voltage across C2 is 2Vm The voltage across Cl and C3 is 3Vm The voltage across C2 and C4 is 4Vm. If additional sections of diode and capacitor are used, each capacitor will be charged to 2Vm Measuring from the top of the transformer winding, will provide odd multiples of Vm Measuring from the bottom of the transformer will provide even multiples Vm. SUMMARY 1- The depletion region is a region adjacent to the p-n junction containing no majority carriers. 2- Forward bias permits majority carrier current through the p-n junction 3- Reverse bias prevents majority carrier current. 4- A p-n structure is called a diode. 5- Reverse breakdown occurs when the reverse-biased voltage exceeds a specified value 6- The single diode in a half-wave rectifier conducts for half of the input cycle.

7- The output frequency of a half-wave rectifier equals the input frequency. 8- The average (dc) value of a half-wave rectified signal is 0.318 or 1/π times its peak value. 9- Each diode in a full-wave rectifier conducts for half of the input cycle. 10-The output frequency of a full-wave rectifier is twice the input frequency. 11- The basic types of full-wave rectifier are center-tapped and bridge. 12-The output voltage of a center-tapped full-wave rectifier is1/2 of the total secondary voltage. 13-The output voltage of a bridge rectifier equals the total secondary voltage. 14-A capacitor filter provides a dc output approximately equal to the peak of the input. 15- Ripple voltage is caused by the charging and discharging of the filter capacitor. 16- The smaller the ripple, the better the filter. 17-Diode limiters cut off voltage above and below specified levels. Limiters are also called clippers. 18-Diode clampers add a dc level to an ac signal. 19-The zener diode operates in reverse breakdown. 20-A zener diode maintains an essentially constant voltage across its terminals over a specified range of zener currents. 21- Zener diodes are used as shunt voltage regulators. 22-Regulation of output voltage over a range of load currents is called load regulation. 23-The smaller the percent regulation, the better. 24-The characteristics of an ideal diode are a close match with those of a simple switch except for the important fact that an ideal diode can conduct in only one direction 25-The ideal diode is a short in the region of conduction and an open circuit in the region of non conduction 26-The region near the junction of a diode that has very few carriers is called the depletion region. 27-In the absence of any externally applied bias. The diode current is zero. 28-In the forward-bias region the diode current will increase exponentially with in crease in voltage across the diode. 29-In the reverse-bias region the diode current is the very small reverse saturation current until Zener breakdown is reached and current will flow in the opposite direction through the diode. 30-The threshold voltage is about 0.7 V for silicon diodes and 0.3 V for germanium diode 31-Rectification is a process whereby an applied waveform of zero average value changed to one that has a dc level. 32-Clippers are networks that "clip" away part of the applied signal either to create a specific type of signal or to limit the voltage that can be applied to a network 33-Clampers are networks that "clamp" the input signal to a different dc level, in any event, the peak-to-peak swing of the applied signal will remain the same 34- Zener diodes are diodes that make effective use of the Zener breakdown potential of an ordinary p-n junction characteristic to provide a device of wide importance and application.

For Zener conduction, the direction of conventional flow is opposite to the arrow in the symbol. The polarity under conduction is also opposite to that of the conventional diode. 35-To determine the state of a Zener diode in a dc network, simply remove the Zener from the network, and determine the open-circuit voltage between the two points where the Zener diode was originally connected. If it is more than the Zener potential and has the correct polarity, the Zener diode is in the "on" state. 36-A half-wave or full-wave voltage doubler employs 2 capacitors a tripler, 3 capacitors a quadrupler, 4 capacitors. for each, the number of diodes equals the number of capacitors Power Supply

• Describe the basic concept of voltage regulation • Discuss the principles of series voltage regulators • Discuss the principles of shunt voltage regulators • Discuss the principles of switching regulators • Discuss integrated circuit voltage regulators

• Discuss applications of IC voltage regulators

TYPES OF POWER SUPPLY

• Linear Power Supply. • Non-Linear Power Supply

LINEAR POWER SUPPLY • Used power devices that operated at linear/active region. • Dissipates more power.

NON-LINEAR POWER SUPPLY • Used power devices that operated at saturation and cutoff alternately. • Dissipates less power. • Also named as switching power supply or switching regulator.

• These power supplies were constructed using discrete components, integrated

circuits or combination of both. • Discrete power transistor, op-amp and comparator were used to complete the

circuit. Voltage Regulation Another factor of importance in a voltage supply is the amount of change in the output dc voltage over the range of the circuit operation. This voltage change with respect to either the loaded or unloaded is described by a factor called voltage regulation.

Example 2: A dc voltage supply provides 60V when the output is unloaded. When full-load current is drawn from the supply, the output voltage drops to 56V. Calculate the value of voltage regulation Solution:

The output voltage from most supplies decreases as the amount of current drawn from the voltage supply is increased.

The smaller the voltage decreases, the smaller the percent of V.R. and the better the operation of the voltage supply circuit. Basic categories: (i) Load Regulation. • Output voltage nearly constant when load change. (ii) Line Regulation. • Output voltage nearly constant line voltage change when change. LOAD REGULATION • Load regulation is a measure of how well a power supply is able tomaintain the dc output voltage between no load and full load with the input voltage constant. • For real power supply, output voltage will drop when load current increases.

Load regulation can be expressed as a percentage change in load voltage. Load Regulation =

• Load regulation can also be expressed in terms of percent change in the output per mA change in load current (%/mA).

EXAMPLE: A regulated power supply with an output resistance of 1 Ω deliver a full load current of 1A to a 25 Ω load. What is the load regulation?

Solution:

LINE REGULATION • Line regulation is a measure of how well a power supply is able to maintain the dc output voltage for a change in the ac input line voltage. • When the dc input (line) voltage changes, the voltage regulator must maintain nearly constant output voltages. Line regulation can be expressed as:

Line Regulation = • Line regulation can also be expressed in terms of percent change in VO per volt change on the Vi (%/V).

Line Regulation = EXAMPLE: When the input to a particular voltage regulator decreases by 5V, the output decreases by 0.25V. The nominal output is 15V. Determine the line regulation in %/V.

Line Regulation = Note: For ideal voltage regulation, both categories will give zero percent regulation (0%) ZENER REGULATOR Zener Region When the applied reverse potential becomes more and more negative, a few free minority carriers have developed sufficient velocity to liberate additional carriers through ionization. When VZ decreases to very low levels,this mechanism, called Zener breakdown, will contribute to the sharp change in the characteristic. It occurs because there is a strong electric field in the region of the junction that can destroy the bonding forces within the atom and "generate" carriers. Diodes employing this unique portion of the characteristic of a p-n junction are called Zener diodes. Zener Diode The Zener diode is a device that is designed to make full use of this Zener region. zener region occurs at a reverse bias potential of VZ.

Fig (2-70) Zener diodes (a) Zener potential (b) characteristic and notation

Fig (2-76)

Example 1: (a) For the network of Fig (2-77), determine the range of RL and IL that will result in VRL being maintained at 10 V.

(b) Determine the maximum wattage & rating of the diode as a regulator.

Fig (2-77) Solution: (a)To determine the value of RL that will turn the Zener diode

b- Fixed RL, Variable Vi For fixed value of RL in Fig (2-79) the voltage Vi must be sufficiently to turn the Zener diode on. The turn-on voltage is determined by:

[2-4] [2-5]

[2-6] Power supply using zener regulator:

• Output voltage constant as long as VIN > VZ • Changes in IL will caused IZ to change in equal & opposite direction • When IZ changes, VL will also changes • The larger IZ change, the larger VL will change • Higher power dissipation in zener • Unable to control the changes in current

Add a series-pass transistor to greatly improve the efficiency and power-handling capability as well as to control the changes in output current.

EXAMPLE: If = 50, determine: (a) output voltage, VO (b) voltage, VCE1, (c) current, IS (d) current, IZ

A Darlington pair transistor (a very high βDC) can be used to increase the current gain. This will reduce the base current and the zener power rating will be low.

REGULATOR CIRCUIT WITH FEEDBACK

Any change in VO must cause a change in VBE1 to maintain the equality. If VO decreases, VBE1 must q y O BE1 increase since VZ is constant. Similarly if VO increases, VBE1 must decrease. SIMPLE SERIES VOLTAGE REGULATOR

BASIC OP-AMP SERIES REGULATOR

EXAMPLE:

For the series regulator circuit shown below: (a) What is the output voltage? (b) If the load current is 200mA what is the power dissipated by Q1?

PROTECTION CIRCUIT 2 types of current limiting circuit:

• Linear/Constant Current Limiting • Fold-back Current Limiting

LINEAR CURRENT LIMITING CIRCUIT

VO is constant until IL(max) is reached. When IL > IL(max), VO decreases and IL will slightly greater than IL(max). This value I L(max) of IL will remain constant even when RL is short circuit. SERIES REGULATOR WITH CONSTANT CURRENT LIMITING Current limiting prevents excessive load current. Q2 will conduct when the current through RSC develops 0.7V across Q2’s VBE. This reduces base current to Q1, limiting the load current.

The current limit is: EXAMPLE:

A series regulator circuit shown above maintains a constant output voltage of 25 V. What is the value of resistor, RSC in order to limit the maximum current, IL(max) to 0.5 A?) With the calculated value of RSC, what is the value of VO when RL = 100 Ω and RL = 10 Ω?

FOLDBACK CURRENT LIMITING CIRCUIT

During short circuit where VO = 0 V and IL = ISC, PD =(Vi–VO)ISC = (20 – 0)1A= 20 W (for constant current limiting) PD =(Vi–VO)ISC = (20 – 0)0.5A=10 W (for foldback current limiting) During maximum operation where VO = 15 V and IL = 1 A, PD= (Vi–VO)IL= (20 – 15)1 A = 5 W (for both current limiting) During short circuit condition, a regulator with constant current limiting has to dissipate 20 W of power in transistor Q1 compared to regulator with foldback current limiting i.e only 10 W. SERIES REGULATOR WITH FOLDBACK CURRENT LIMITING Fold-back current limiting drops the load current well below the peak during overload conditions. Q2 conducts when VR4 + VBE2 = VRSC and begins current limiting.

VBE2 = VRSC – VR4 VR4 will increase or decrease if VO increases or decreases. At this instant, Q2 is still not conducting. VR4is found by applying the voltage-divider rule:

When IL increase to IL(max) or during overload, VR4 will drop because VO drops. A smaller value of VRSC is required to maintain VBE2 ≈ 0.7V. This means that less current is needed to maintain conduction in Q2 and the load current drops. At this point, current limiting occurs. IL will be limited and Q2 conducting (ON)

SHUNT-TYPE VOLTAGE REGULATOR BLOCK DIAGRAM

SHUNT VOLTAGE REGULATOR WITH OP-AMP

• Shunt regulators use a parallel transistor for the control element. If the output voltage changes, the op-amp senses the change and corrects the bias on on Q1. A decrease in output voltage causes a decrease in VB and an increase in Vc.

• Although it is less efficient than the series regulator, the shunt regulator has

inherent short-circuit protection. The maximum current when the output is shorted is VIN/R4.

SWITCHING REGULATORS To reduce power dissipation in pass transistor.

• Gives higher efficiency. • Able to supply very large load current with low voltage as required in the PC. • 3 basic configurations • step-down • step-up • inverting • Step-down switching regulator is widely used as the power supply in PC. • All switching regulators control the output voltage by rapidly switching the input

voltage on and off with a duty cycle that depends on the load. Because they use high frequency switching, they tend to be electrically noisy.

• An increase in the duty cycle increases the output voltage. Case 1

Case2

BASIC SWITCHING REGULATOR

A STEP-DOWN SWITCHING REGULATOR

• A step-down switching regulator controls the output voltage by controlling the duty cycle to a series transistor. The duty cycle changes depending on the load requirement.

• Because the transistor is either ON or OFF on all switching regulators, the power dissipated in the transistor is very small and the regulator is very efficient. The pulses are smoothed by an LC filter.

A STEP-UP SWITCHING REGULATOR In a step-up switching regulator, the control element operates as a rapidly pulsing switch to ground. The switch ON and OFF times are controlled by the output

voltage. Step-up action is due to the fact that the inductor changes polarity during switching and adds to VIN. Thus, the output voltage is larger than the input voltage.

AN INVERTING SWITCHING REGULATORS In a voltage-inverter switching regulator, the output is the opposite polarity of the input. It can be used in conjunction with a positive regulator from the same input source. Inversion occur because the inductor reverses polarity when the diode conducts conducts, charging the capacitor with the opposite polarity of the input.

THE OPERATION OF PULSE WIDTH MODULATOR (PWM)

EXAMPLE By assuming an ideal LC, (a) Explain the function of PWM, D1, L and C. (b) Explain the operation of the circuit if VOUT decreases. (c) Calculate VOUT. (d) If Vi increase to15 V, sketch the waveform at point B in order to maintain the value found in (c).

• PWM is used to produce pulse trains with pulse width depend on the changes

in output, VOUT. These pulse trains (at point B) will control the ON and OFF interval of Q1 thus will finally increase or decrease the value of VOUT.

• The diode D1 is used to eliminate the negative voltage.

• Inductor, L and capacitor, C is used as filter to average the switched voltage

thus produce VDC.

• When VOUT reduced, VR2 will also reduced thus Verror will increase because VZ is constant.

• PWM will produce pulse trains with large pulse width.

• Q1 will ON and OFF with large duty cycle thus increase the dc current

flowing through it.

• The increase of dc current in Q1 will then increase the VOUT that try to reduce previously

This regulating action maintains VOUT at an essentially constant level.

SWITCHING REGULATOR WITH PWM CONSTRUCTION

VOSC , VERROR AND PWM OUTPUT VOLTAGES

PWM GENERATION USING

SAWTOOTH GENERATOR AND VOLTAGE COMPARATOR

SWITCHING REGULATORS: ADVANTAGES

• Higher efficiency. • Light and compact. • Filtering is easy to achieve at high frequencies. • VO ≥ Vi SWITCHING REGULATORS: DISADVANTAGES

• Generate EMI (electromagnetic interference) where switching at high frequency for Q1 current will produce large magnetic fields which induced noise voltage around conductor. • Limited performance of power transistor (pass transistor) to switch ON and OFF at high speed. • Contain large noise and ripple in VO.