Diodes as Rectifierseecs.oregonstate.edu/~traylor/ece112/beamer_lectures/diode... · Diodes as...
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Diodes as Rectifiers As previously mentioned, diodes can be used to convert alternating current (AC) to direct current (DC). Shown below is a representative schematic of a simple DC power supply similar to a au- tomobile battery charger. The way in which the diode rectifier is used results in what is called a half-wave rectifier. Wallplug T1 R1 resistive load D1 1N4001 110VAC 0V 167V pp 0V 35.6V pp 120 to 12.6VAC stepdown transformer 0V 17.1V p negative half-wave is removed by diode Figure 1: Half-Wave Rectifier Schematic The input transformer steps the input voltage down from 110VAC(rms) to 12.6VAC(rms). The diode converts the AC voltage to DC by removing the negative going part of the input sine wave. The result is a pulsating DC output waveform which is not ideal except for simple applications such as battery chargers as the voltage goes to zero for one have of every cycle. What we would like is a DC output that is more consistent; a waveform more like a battery what we have here. We need a way to use the negative half-cycle of the sine wave to to fill in between the pulses created by the positive half-waves. This would give us a more consistent output voltage. Below is a circuit that does just that. When the diodes are connected in this manner this circuit would be called a full-wave rectifier. Wallplug T1 D1 110VAC 120 to 12.6VAC stepdown transformer D4 D2 D3 R1 resistive load 0V 17.1V p D1 and D3 D2 and D4 D1 and D3 Figure 2: Full-Wave Rectifier Schematic 1
Transcript of Diodes as Rectifierseecs.oregonstate.edu/~traylor/ece112/beamer_lectures/diode... · Diodes as...
Diodes as Rectifiers
As previously mentioned, diodes can be used to convert alternating current (AC) to direct current(DC). Shown below is a representative schematic of a simple DC power supply similar to a au-tomobile battery charger. The way in which the diode rectifier is used results in what is called ahalf-wave rectifier.
FILE: REVISION:
DRAWN BY: PAGE OF
TITLE
Wallplug
T1
R1resistive load
D1
1N4001
C1
50uF
110VAC
0V 167Vpp
0V 35.6Vpp
120 to 12.6VAC
stepdown transformer
0V17.1Vp
negative half−wave is
removed by diode
Wallplug
T1
D1
110VAC
120 to 12.6VAC
stepdown transformer D4
D2
D3
R1resistive load
0V17.1Vp
D1 and D3
D2 and D4
D1 and D3
Wallplug
T1
D1
D4
D2
D3
R11K
110VAC
120 to 12.6VAC
stepdown transformer
D1
D4
D2
D3
R
+
−
input from
transformer
(positive half−cycle)
D1
D4
D2
D3
R
+
−
input from
transformer
(negative half−cycle)
L L
v_out
1
2
Figure 1: Half-Wave Rectifier Schematic
The input transformer steps the input voltage down from 110VAC(rms) to 12.6VAC(rms). Thediode converts the AC voltage to DC by removing the negative going part of the input sine wave.The result is a pulsating DC output waveform which is not ideal except for simple applicationssuch as battery chargers as the voltage goes to zero for one have of every cycle. What we wouldlike is a DC output that is more consistent; a waveform more like a battery what we have here.
We need a way to use the negative half-cycle of the sine wave to to fill in between the pulsescreated by the positive half-waves. This would give us a more consistent output voltage. Belowis a circuit that does just that. When the diodes are connected in this manner this circuit would becalled a full-wave rectifier.
FILE: REVISION:
DRAWN BY: PAGE OF
TITLE
Wallplug
T1
R1resistive load
D1
1N4001
C1
50uF
110VAC
0V 167Vpp
0V 35.6Vpp
120 to 12.6VAC
stepdown transformer
0V17.1Vp
negative half−wave is
removed by diode
Wallplug
T1
D1
110VAC
120 to 12.6VAC
stepdown transformer D4
D2
D3
R1resistive load
0V17.1Vp
D1 and D3
D2 and D4
D1 and D3
Wallplug
T1
D1
D4
D2
D3
R11K
110VAC
120 to 12.6VAC
stepdown transformer
D1
D4
D2
D3
R
+
−
input from
transformer
(positive half−cycle)
D1
D4
D2
D3
R
+
−
input from
transformer
(negative half−cycle)
L L
v_out
1
2
Figure 2: Full-Wave Rectifier Schematic
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The full wave rectifier uses the one-way action of the diodes to create the output from both positiveand negative half-waves of the AC input signal. The path through the load passes through twodiodes in this scheme. The different paths through the rectifier are shown below.
FILE: REVISION:
DRAWN BY: PAGE OF
TITLE
Wallplug
T1
R1resistive load
D1
1N4001
C1
50uF
110VAC
0V 167Vpp
0V 35.6Vpp
120 to 12.6VAC
stepdown transformer
0V17.1Vp
negative half−wave is
removed by diode
Wallplug
T1
D1
110VAC
120 to 12.6VAC
stepdown transformer D4
D2
D3
R1resistive load
0V17.1Vp
D1 and D3
D2 and D4
D1 and D3
Wallplug
T1
D1
D4
D2
D3
R11K
110VAC
120 to 12.6VAC
stepdown transformer
D1
D4
D2
D3
R
+
−
input from
transformer
(positive half−cycle)
D1
D4
D2
D3
R
+
−
input from
transformer
(negative half−cycle)
L L
v_out
1
2
Figure 3: Different Paths through the Rectifier Diodes
When the the top terminal is more positive, only diodes D1 and D3 can conduct. Because of thevoltage drop across the load, diode D2 is reverse biased. Diode D4 is reverse biased from its con-nection across the transformer. As the current passes through each diode, the voltage drops by 0.7volts. Thus, even if the load was a short circuit, diode D2 could not pass current back to the inputterminal since its anode side is at least 0.7 volts below its cathode because of diode D1. Whenthe bottom terminal is more positive, only diodes D4 and D2 can conduct. In either positive ornegative half-cycles, the positive output terminal remains positive.
Now we have an output that is uni-polar put still has ripple. We can remove most of the ripple withthe addition of a capacitor shown below. The capacitor acts as a temporary storage reservoir thatsupplies current to the load between the input pulses. The capacitor is charged on the rising edgeof the half-cycles and supplies current to the load when the input voltage falls below the voltagestored on the capacitor. Such a circuit is able to provide a fairly clean output that is suitable formost electronic devices.
FILE: REVISION:
DRAWN BY: PAGE OF
TITLE
Wallplug
T1
R1resistive load
D1
1N4001
C1
50uF
110VAC
0V 167Vpp
0V 35.6Vpp
120 to 12.6VAC
stepdown transformer
0V17.1Vp
negative half−wave is
removed by diode
Wallplug
T1
D1
110VAC
120 to 12.6VAC
stepdown transformer D4
D2
D3
R1resistive load
0V17.1Vp
D1 and D3
D2 and D4
D1 and D3
Wallplug
T1
D1
D4
D2
D3
R11K
110VAC
120 to 12.6VAC
stepdown transformer
D1
D4
D2
D3
R
+
−
input from
transformer
(positive half−cycle)
D1
D4
D2
D3
R
+
−
input from
transformer
(negative half−cycle)
L L
v_out
1
2
Figure 4: Full-Wave Rectifier with Filter Capacitor Schematic
The output from a spice simulation of the full-wave rectifier is shown in fig 5. The red trace is theinput voltage. The blue trace is the output voltage. After the input waveform begins to fall below
2
the capacitor voltage, we see the discharge of the capacitor begin. Once the inverted negative half-cycle voltage exceeds the capacitor voltage, we see the capacitor being recharged. The capacitorholds-up the voltage between the voltage peaks. With an big enough capacitor the voltage droopbetween input voltage peaks can be made arbitrarily small.
Figure 5: Spice Simulation - Filtered Full-wave Rectifier Showing Output Ripple
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