Electronic I Lecture 3 Diode Rectifiers. By Asst. Prof Dr. Jassim K. Hmood

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1 Electronic I Lecture 3 Diode Rectifiers By Asst. Prof Dr. Jassim K. Hmood

2 Diode Approximations 1- The Ideal Model When forward biased, act as a closed (on) switch When reverse biased, act as open (off) switch

3 This model neglects the effect of the barrier potential, the internal resistance, and other parameters Diode Approximations

4 Diode Approximations 2. The Barrier Potential Model The forward biased diode is represented as a closed switch in series with a small battery equal to the barrier potential VB (0.7 V for Si and 0.3 V for Ge) The positive end of the equivalent battery is toward the anode. This barrier potential cannot be measured by using a multimeter, but it has the effect of a battery when forward bias is applied. The reverse biased diode is represented by an open switch, because barrier potential does not affect reverse bias.

5 Diode Approximations

6 Diode Approximations 3. The Complete Diode Model The forward biased diode model with both the barrier potential and low forward (bulk) resistance ( r d )

7 HALF-WAVE RECTIFIER CIRCUITS The basic rectifier circuit converts an ac voltage to a pulsating dc voltage. A filter is then added to eliminate the ac components of the waveform and produce a nearly constant dc voltage output.

8 HALF-WAVE RECTIFIER WITH RESISTOR LOAD The simplest single-phase diode rectifier is the single-phase half-wave rectifier. The circuit consists of only one diode that is usually fed with a transformer secondary. During the positive half-cycle of the transformer secondary voltage, diode conducts. During the negative half-cycle, diode stops conducting.

9 HALF-WAVE RECTIFIER WITH RESISTOR LOAD

10 HALF-WAVE RECTIFIER WITH RESISTOR LOAD The average value of output voltage V dc is defined as: V p Vdc 0.318V The average current I dc is : I The root-mean-square (rms) value of output voltage v o is V o, which is defined as: V p V o V V o p And 2 The rectification efficiency is I o R 2R p dc V R dc V P R 2 I R dc dc 2 in o d P P I r R

11 HALF-WAVE RECTIFIER WITH RESISTOR LOAD By Substituting and simplifying, we get on: If diode resistance r d is neglected, then 0.46 r R 1 d 0.46 or 46% PIV It is the maximum voltage across the diode in the reverse direction. PIV for diode in half wave rectifier is V p

12 HALF-WAVE RECTIFIER WITH RESISTOR LOAD 1- For this case, the output voltage is one diode-drop smaller than the input voltage during the conduction interval: 2- The output voltage remains zero during the off-state interval. The input and output waveforms for the half-wave rectifier, including the effect of Von, are shown in the figure for V P = 10 V and V on = 0.7V. Half-wave rectifier output voltage with V P = 10 V and V on = 0.7 V.

13 RECTIFIER with FILTER CAPACITOR The unfiltered output of the half-wave rectifier is not suitable for operation of most electronic circuits because constant power supply voltages are required to establish proper bias for the electronic devices. A filter capacitor can be added to filter the output of the rectifier circuit to remove the time-varying components from the waveform.

14 RECTIFIER with FILTER CAPACITOR A load must be connected to the circuit as represented by the resistor R. Now there is a path available to discharge the capacitor during the time the diode is not conducting.

15 RECTIFIER with FILTER CAPACITOR The output voltage is no longer constant as in the ideal peakdetector circuit but has a ripple voltage V r. In addition, the diode only conducts for a short timet during each cycle. This time DT is called the conduction interval, and its angular equivalent is the conduction angle q C where q C = ωt. the ripple voltage: The conduction angle and conduction interval are: The PIV of diode with capacitor filter is:

16 Half-wave diode rectifier Example: Find the value of the dc output voltage, dc output current, ripple voltage, conduction interval, and conduction angle for a half-wave rectifier driven from a transformer having a secondary voltage of 12.6 V rms (60 Hz) with R = 15W and C = 25,000mF. Assume the diode on-voltage V on = 1 V. Solution The ideal dc output voltage in the absence of ripple is The nominal dc current delivered by the supply is

17 The ripple voltage is Half-wave diode rectifier The conduction angle is and the conduction interval is

18 Home work H.W4: Find the value of the dc output voltage, dc output current, ripple voltage, conduction interval, and conduction angle for a half-wave rectifier that is being supplied from a transformer having a secondary voltage of 6.3 V rms (60 Hz) with R = 0.5W and C = 500,000mF. Assume the diode on voltage V on = 1 V. Answers: 7.91 V; 15.8 A; 0.527; ms; 19.7

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