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Devices and Applications Ctec 201. Diode Rectifier Supplement Prepared by Mike Crompton. (Rev. 7 July 2003) Half and Full Wave Rectifiers. If we now combine what we know about transformers, diodes and capacitors we can create circuits that are extremely useful. Rectifier circuits that convert the line supply AC voltage to almost any level of DC voltage. 10 : 1 V Pk = 16.26V - 0.7V = 15.56V +V D1 11.5V RMS 60Hz 16.26V Pk 115V AC 60Hz Vout 1/2 Wave rectified (Positve 1/2 cycles only) -V Fig. 1 Fig. 1 shows the simplest of these circuits known as the ½ Wave Rectifier, because it rectifies only the +ve half cycle of the applied (and stepped down) AC voltage. The 10:1 step down transformer will produce an 11.5 VRMS or 16.26 VPK sine wave at the secondary. The diode, D1, will only conduct on the +ve half cycle of this sine wave giving a ‘pulsating DC voltage’ at VOUT equal to the peak voltage of the sine wave, minus the 0.7V drop across the diode, or 15.56V. Measured with a DC Voltmeter this would register as only 4.95V (call it 5V) because (remember) the DC AVG Voltage of half a sine wave is only VPK / π or VPK * 0.3183. It is obvious that this pulsating DC is of no real use as the supply voltage for any device. It is never constant and it’s DCAVG is at a very low level compared to the 15.56 peak voltage we started with. However if we add a large filter capacitor (C1) as shown in fig 2, we can change the VOUT to a constant DC of almost 15.56V. C1 will charge to VPK on the first +ve half cycle when D1 conducts. The charge time (C * R) will be very short as the only resistance in circuit will be that of the transformer secondary and D1, both of which will be very low. When the diode cuts off on the –ve half cycle, there is nothing connected to C1, as the diode is an ‘open switch’, and there is no discharge path. C1 will 15.56V 10 : 1 +V D1 115V AC 60Hz 16.26V Pk 60Hz C1 Vout C1 Charges to V Pk (15.56V) on 1st positive 1/2 cycle and remains at that level. -V Fig. 2 2 therefore remain fully charged at approximately 15.56V, giving a steady DC output of more than three times the previous (5V) level. There is one problem with this arrangement, and that is there is no load or device connected to the circuit at VOUT. As soon as we connect anything, it’s resistance (RL) is now in parallel with C1 and will cause C1 to discharge during the –ve half cycles of the sine wave. Recall that discharge time is also C * R, and so the amount that C1 will discharge will be dependant on the size of R L. A low value of RL means C1 will discharge a lot, and a large value of RL means it will only discharge a little. C1 will of course charge up to VPK again on the next +ve half cycle, but any change in the charge level of C1 is in fact a change in the DC O/P voltage level, see fig 3. This variation is known as the Ripple Voltage, and although undesirable, it will always be present as soon as a load is connected. Ideally the peak to peak ripple voltage should be as small as possible, this will not only produce a smoother DC but also a higher voltage level, since the level will be at the average point (or middle) of the ripple (VPK - ½VRIPPLE PK PK). Ripple Pk to Pk +V DC Avg Low value of load resistor will reduce CxR and allow C1 to discharge by a large amount giving a large amount of Ripple. +V DC Avg Ripple Pk to Pk To minimize ripple the size of the filter High value of load resistor will increase CxR capacitor can be and allow C1 to discharge by only a small amount Fig. 3 giving a small amount of Ripple increased but only to the point where it is still able to charge to the peak voltage in half the period time of the AC, (one half cycle). This you will recall takes 5 Time Constants. Keeping the load resistance as high as possible will also reduce ripple, but often this is beyond your control, being determined by the internal resistance of the device being connected. The frequency of the ripple produced by a ½ wave rectifier is always the same as the applied AC Voltage. In this case it will be 60Hz. The ½ wave rectifier was popular in the days of tubes, when the cost of one diode tube would be a large percentage of the total cost. These days when semi-conductor diodes are extremely cheap and come in prepackaged chips, the full wave and full wave bridge rectifiers are much more popular. The greatest majority being, by far, the bridge rectifier which uses 4 diodes. The regular full wave rectifier uses 2 diodes, but requires a center tapped transformer, which today would be the most expensive item. The following pages contain descriptions of both. 3 The Full Wave Bridge Rectifier. As the name implies, a full wave rectifier is able to produce a +ve half cycle for both the +ve and –ve half cycles (full wave) of the applied AC Voltage. The bridge rectifier accomplishes this by using 4 diodes in a ‘bridge’ configuration. See fig 4. 16.26V Pk. A 10 : 1 115V AC 60Hz D1 16.26V Pk 60Hz D2 V SEC D4 D3 B Fig. 4 + + 14.86V Pk. (16.26 - 1.4) C1 RL V RL Vout Note that RL is still in parallel with the filter capacitor C1, and the transformer is giving the same output voltage as the ½ wave rectifier circuit on the previous pages. Also note that point A and point B are connected to the secondary, and the Ground is actually one end of the load RL, and NOT one side of the secondary winding as it is in the ½ wave. The output voltage (VRL) is shown as it would be with the filter capacitor disconnected to illustrate the fact that there is a +ve half cycle for both +ve and –ve ½ cycles of VSEC. The O/P voltage level is 14.86VPK or 16.26V minus 1.4V for the voltage drop across 2 diodes. The circuit action of the bridge rectifier is broken down into two distinctly different sets of circumstances: 1. When Point A is +ve and Point B is –ve during the +ve half cycle of VSEC as shown in Fig 5, and 2. During the –ve ½ cycle of VSEC when Point A is –ve and Point B is +ve, as shown in Fig 6. Under each of these conditions, only two of the four diodes are actually in circuit at any one time, D1 and D3 during +ve ½ cycles of VSEC, and D2 and D4 during –ve ½ cycles. A description of each follows on the next page. 4 During the +ve ½ cycle of VSEC (See Fig. 5 at right), point A is +ve and point B is –ve. D2 and D4 are reversed biased and can be considered as open switches i.e. Not in circuit. D1 and D3 are forward biased with the anode of D1 connected to the +ve end of the secondary and D3 cathode connected to the –ve end. RL is between the two. Electrons will leave the –ve end (bottom) of the secondary, flow thro’ D3, flow thro’ RL from right to left, thro’ D1 and back into the +ve end (top) of the secondary. This will produce a positive half cycle voltage waveform across RL of VSECPk minus the voltage drop across the two diodes. i.e. 16.26V - 1.4V = 14.86VPk. 10 : 1 A Positive 1/2 Cycle D1 Positive 115V AC 60Hz D3 Negative B RL Fig. 5 Electron Flow through RL. On the –ve ½ cycle of VSEC (See Fig. 6 at right), point A is –ve and point B is +ve. This means that D1 and D3 are reverse biased and can be considered as ‘not in circuit’. D2 and D4 however are forward biased, the anode of D4 connected to Point B which is now the +ve end of the secondary, and the cathode of D2 connected to Point A the –ve end. RL is between them. Electrons will flow from the top of the secondary thro’ D2, thro’ RL from right to left , thro’ D4 and back into the +ve end (bottom) of the secondary. This will produce an identical positive half cycle voltage waveform across RL of VSECPk minus the voltage drop across the two diodes. i.e. 16.26V - 1.4V = 14.86VPk. The full and continuous O/P voltage will be as shown in Fig. 4 on the preceding page. A Negative D2 10 : 1 115V AC 60Hz D4 Negative 1/2 Cycle Positive B RL Fig. 6 Electron Flow through RL +V 1/2 Wave Ripple Pk to Pk The advantages of the bridge rectifier over the ½ DC Avg wave are apparent when we look at the O/P voltage with the filter capacitor C1 in circuit and an identical load connected to both rectifiers. In Fig. 7 Full Wave Fig 7 the top half is the ½ wave O/P voltage and +V the lower is the bridge O/P voltage. The C1*RL DC Avg time constant is the same for both, but the bridge Ripple Pk to Pk rectifier re-charges C1 every ½ cycle compared to every cycle for the ½ wave. This means that C1 does not discharge as much and the peak to peak ripple voltage is considerably less. This, in turn, actually increases the O/P voltage to a slightly higher value. An additional bonus is the frequency of the ripple is double the input frequency, making it easier to filter if necessary. 5 Full Wave Rectifier The full wave rectifier, as with the ½ wave rectifier, is almost an extinct animal. Although it only uses 2 diodes, it needs a transformer with a center tapped secondary and it’s DC O/P voltage, although fully rectified, is only ½ of the O/P voltage of the bridge rectifier. See Fig 8 below. 10 : 1 +V 8.13V Pk D1 115V AC 60Hz Center Tap C1 V Pk = 8.13V - 0.7V = 7.43V Vout The output peak voltage is only 1/2 of the output peak voltage in the Bridge rectifier. RL 8.13V Pk -V D2 Fig. 8 Vout The center tapped secondary of the transformer produces two sine waves that are 180 out of phase and have a VPK of exactly ½ the normal full VSEC, in this case 8.13V compared to 16.26V for the ½ wave and bridge rectifiers. When the 0.7 volts drop across the diode is taken into account the VPK is only 7.43V. Only one of the diodes is conducting at any one time, D1 during the +ve ½ cycle of the upper sine wave and D2 during the +ve ½ cycle of the lower sine wave. This results in each half cycle producing a ½ cycle voltage pulse across RL for each complete cycle of the AC voltage (primary) applied. The waveform shown as VOUT in Fig 8 presumes that the filter capacitor C1 is not connected. This is to show the pulsating DC O/P which would normally be ‘smoothed’ by the filter action of the capacitor. Ripple voltage will once again depend on the resistance of the load and the ripple frequency will be twice that of the applied AC voltage as it is for the bridge rectifier. The only real advantage of the full wave rectifier over the ½ wave is the reduced ripple. When weighed against the disadvantage of the reduced DC O/P voltage (½) and the need for the center tapped transformer, there does not seem to be any reason to choose the full wave rectifier. When compared to the bridge rectifier, one advantage is the use of 2 diodes instead of 4. This would however be a false economy due to the increased cost of a center tapped transformer and the ridiculously low cost of diodes. Consequently, the bridge rectifier is used almost exclusively for converting AC to DC via rectification. 6