The D.C Power Supply

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1 The D.C Power Supply Voltage Step Down Electrical Isolation Converts Bipolar signal to Unipolar Half or Full wave Smoothes the voltage variation Still has some ripples Reduce ripples Stabilize the output against load variations Block diagram of a dc power supply.

2 Rectifier Circuits Basic rectifier converts an ac voltage to a pulsating dc voltage. A filter then eliminates ac components of the waveform to produce a nearly constant dc voltage output. Rectifier circuits are used in virtually all electronic devices to convert the 120 V-60 Hz ac power line source to the dc voltages required for operation of the electronic device. In rectifier circuits, the diode state changes with time and a given piecewise linear model is valid only for a certain time interval.

3 Half-Wave Rectifier Circuit with Resistive Load For positive half-cycle of input, source forces positive current through diode, diode is on, v o = v s. During negative half cycle, negative current can t exist in diode, diode is off, current in resistor is zero and v o =0. Using CVD model, during on state of diode v o =(V P sinωt)- V D0 Output voltage is zero when diode is off. Time-varying components in circuit output are removed using filter capacitor.

4 Half-Wave Rectifier Circuit with Resistive Load (cont.) vo = 0 v s < V DO R R vo = v s V DO v s V DO R + rd R + rd vo v s V DO v s V DO for rd << R V P V DO v oav = D.C component = π Maximum forward current = (V P - V D0 )/R Peak Inverse Voltage (PIV) = V P Usually the breakdown voltage is taken at least 50% larger than PIV Can this circuit used to rectify a 100 mv sinusoidal?

5 Peak Detector Circuit As input voltage rises, diode is on and capacitor (initially discharged) charges up to input voltage minus the diode voltage drop. At peak of input, diode current tries to reverse, diode cuts off, capacitor has no discharge path and retains constant voltage providing constant output voltage V dc = V P.

6 Half-Wave Rectifier Circuit with RC Load As input voltage rises during first quarter cycle, diode is on and capacitor (initially discharged) charges up to peak value of input voltage. At peak of input, diode current tries to reverse, diode cuts off, capacitor discharges exponentially through R. Discharge continues till input voltage exceeds output voltage which occurs near peak of next cycle. Process then repeats once every cycle. This circuit can be used to generate negative output voltage if the top plate of capacitor is grounded instead of bottom plate (or if the diode is flipped). In this case, V dc = -(V P - V on )

7 Half-Wave Rectifier Circuit with RC Load (cont.) Output voltage is not constant as in ideal peak detector, but has ripple voltage V r. Diode conducts for a short time ΔT called conduction interval during each cycle and its angular equivalent is called conduction angle θ c. T ΔT V T V V 1 P r P RC T R C ΔT 1 ω 2V r V P θc = ωδ T = 2V r V P

8 Half-Wave Rectifier Analysis: Example Problem: Find dc output voltage, output current, ripple voltage, conduction interval, conduction angle. Given data: secondary voltage V rms 12.6 (60 Hz), R= 15 Ω, C= 25,000 μf, V D0 = 1 V Analysis: V = V V = ( )V = 16.8V dc P D0 V V I = P D0 = 16.8V = 1.12A dc R 15Ω ( Using discharge interval T=1/60 s, V P ) V V D0 T r = 0.747V R C θ 2V c = ω ΔT = = 0.290rad = 16.6 V r 0 P θ ΔT = θ c ω = c = 0.769ms = πf π

9 Peak Diode Current For Your Information Only In rectifiers, nonzero current exists in diode for only a very small fraction of period T, yet an almost constant dc current flows out of filter capacitor to load. Total charge lost from capacitor in each cycle is replenished by diode during short conduction interval causing high peak diode currents. If repetitive current pulse is modeled as triangle of height I P and width ΔT, I = I 2 T =48.6A P dc ΔT using values from previous example.

10 Surge Current For Your Information Only Besides peak diode currents, when power supply is turned on, there is an even larger current through diode called surge current. During first quarter cycle, current through diode is approximately d i ( t) = i c ( t) C V sinωt = ωcv cosωt d dt P P Peak values of this initial surge current occurs at t = 0 + : I = ωcv =168A SC P using values from previous example. Actual values of surge current won t be as large as predicted because of neglected series resistance associated with rectifier diode as well as transformer.

11 Peak Inverse Voltage Rating Peak inverse voltage (PIV) rating of the rectifier diode gives the breakdown voltage. When diode is off, reverse-bias across diode is V dc - v s. When v s reaches negative peak, PIV V ( ) 2 dc v s min = V P V D0 V P V P PIV value corresponds to minimum value of Zener breakdown voltage for rectifier diode.

12 Full-Wave Rectifiers Full-wave rectifiers cut capacitor discharge time in half and require half the filter capacitance to achieve given ripple voltage. All specifications are same as for half-wave rectifiers. Reversing polarity of diodes gives a fullwave rectifier with negative output voltage.

13 Full-Wave Bridge Rectification Requirement for a centertapped transformer in the fullwave rectifier is eliminated through use of 2 extra diodes.all other specifications are same as for a half-wave rectifier except PIV=V P.

14 Rectifier Topology Comparison Filter capacitor is a major factor in determining cost, size and weight in design of rectifiers. For given ripple voltage, full-wave rectifier requires half the filter capacitance as that in half-wave rectifier. Reduced peak current can reduce heat dissipation in diodes. Benefits of full-wave rectification outweigh increased expenses and circuit complexity (an extra diode and center-tapped transformer). Bridge rectifier eliminates center-tapped transformer, PIV rating of diodes is reduced. Cost of extra diodes is negligible.

15 The Full-Wave Rectifier Comparison Center-Tapped Bridge Average Voltage 2.( VP VDO) 2.( VP 2. VDO ) π π PIV 2V P -V DO V P -V DO Max. Forward Current (V P -V D0 )/R (V P -2V D0 )/R Transformer Size N1:2*N2 No. of Diodes 2 4 N1:N2 Half centertapped size

16 Rectifier Design Analysis Problem: Design rectifier with the given specifications. Given data: V dc =15 V, V r < 0.15 V, I dc = 2 A, V D0 =1V Analysis: Use full-wave bridge rectifier that needs smaller value of filter capacitance, smaller diode PIV rating and no center-tapped transformer. V V + 2V P dc D V = = = V = 12V rms /2 1 1 V T C = I 2A s 0.111F R dc = = = = 7.5Ω dc V r V I dc PIV = V V D = 16V P 0

17 Three-terminal IC Voltage Regulators For Your Information Only Uses feedback with high-gain amplifier to reduce ripple voltage at output.bypass capacitors provide low-impedance path for highfrequency signals to ensure proper operation of regulator. Provides excellent line and load regulation, maintaining constant voltage even if output current changes by many orders of magnitude. Main design constraint is V REG which must not fall below a minimum specified dropout voltage (a few volts).

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