Download Full Wave Rectifier

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.
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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.
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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.
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