Diodes As Rectifiers +

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.

0V 35.6Vpp

167V 0V pp 17.1Vp 0V T1 D1 Wallplug 1N4001 negative half−wave is 110VAC R1 resistive load removed by diode

D1 D1 120 to 12.6VAC − stepdown transformer + D2 D2 input from 0V 35.6V RL input from RL pp transformer transformer (positive half−cycle) Figure 1: Half-Wave Rectifier Schematic (negative half−cycle) 167V 0V pp − 17.1Vp D3 + D3 0V The input transformer steps the input voltageT1 downD1 from 110VAC(rms) to 12.6VAC(rms). The D4 D4 Wallplug diode converts the AC voltage to DC by removing1N4001 theD1 and negative D3 goingD1 and D3 part of the input sine wave. negative half−wave is The result is a pulsating110VAC DC output waveform which isR1 not ideal except for17.1V simplep applications 0V resistive load such as battery chargers as the voltage goes to zero for oneD2 have and D4 of every removed cycle. by diode What we would D1 D1 v_out like is a DC output that is more consistent;T1 a waveform more like a battery what we have here. T1 D11 D1 120 to 12.6VACD2 D2 C1 Wallplug R1 Wallplug + R1 − stepdown transformer resistive load D2 50uF 1K D2 We need a way to use the negative half-cycle of the sine wave to to fill in between the pulses input from RL input from RL 110VAC 110VAC transformer transformer created by the positive half-waves. This would give us a more consistent output voltage.(positive Below half−cycle) (negative half−cycle) is a circuit that does just that. When the diodesD3 are connected in this manner this circuit would be D3 2 called a full-wave rectifier. 120 to 12.6VAC 120 to −12.6VAC D3 + D3 stepdown transformer D4 stepdown transformer D4 D4 D4 D1 and D3 D1 and D3

17.1Vp 0V D2 and D4

D1 D1 v_out T1 T1 1 D2 TITLE D2 C1 Wallplug R1 Wallplug R1 resistive load FILE: 50uFREVISION:1K 110VAC 110VAC PAGE OF DRAWN BY:

D3 D3 2 120 to 12.6VAC 120 to 12.6VAC stepdown transformer D4 stepdown transformer D4

Figure 2: Full-Wave Rectiﬁer Schematic

TITLE 1 FILE: REVISION: PAGE OF DRAWN BY: 0V 35.6Vpp

167V 0V pp 17.1Vp 0V T1 D1 Wallplug The full wave rectiﬁer uses the one-way action of the diodes to create the output from both positive 1N4001 negative half−waveand is negative half-waves of the AC input signal. The path through the load passes through two 110VAC R1 resistive load removed by diodediodes in this scheme. The different paths through the rectiﬁer are shown below.

D1 D1 120 to 12.6VAC − stepdown transformer + D2 D2 input from RL input from RL transformer transformer (positive half−cycle) (negative half−cycle)

− D3 + D3

D4 D4 D1 and D3 D1 and D3

17.1Vp Figure 3: Different Paths through the Rectiﬁer Diodes 0V D2 and D4 D1 0V 35.6Vpp D1 When the the top terminal is more positive, onlyv_out diodes D1 and D3 can conduct. Because of the T1 T1 1 D2 167V voltage drop across the load, diodeD2 is reverseC1 biased. Diode D4 is reverse biased from its con- Wallplug 0V R1 pp Wallplug R1 nection across the transformer.17.1Vp As the current50uF passes through each diode, the voltage drops by 0.7 resistive load D1 0V 1K T1 volts. Thus, even if the load was a short circuit, diode D2 could not pass current back to the input 110VAC Wallplug 110VAC 1N4001 terminal since its anode side is at least 0.7 volts below its cathode because of diode D1. When negative half−wave is 110VAC theR1 bottom terminal is more positive, only diodes D4 and D2 can conduct. In either positive or D3 resistive load removed by diode D3 negative half-cycles, the positive2 output terminal remains positive. 120 to 12.6VAC 120 to 12.6VAC D4 D4 D1 D1 stepdown transformer 120 to 12.6VAC stepdown transformer − stepdown transformer Now we have an output that is uni-polar+ put stillD2 has ripple. We can remove most ofD2 the ripple with input from the addition of a capacitor shown below. The capacitor acts asR aL temporary input from storage reservoir thatRL transformer transformer (positive half−cycle) supplies current to the load between the input pulses. The capacitor(negative is charged half−cycle) on the rising edge of the half-cycles and supplies current to− the load when the input voltage falls below the voltage stored on the capacitor. Such a circuit is able toD3 provide a fairly clean output+ thatD3 is suitable for most electronic devices. D4 D4 D1 and D3 D1TITLE and D3

FILE: 17.1Vp REVISION: 0V D2 and D4 PAGE OF DRAWN BY: D1 D1 v_out T1 T1 1 D2 D2 C1 Wallplug R1 Wallplug R1 resistive load 50uF 1K 110VAC 110VAC

D3 D3 2 120 to 12.6VAC 120 to 12.6VAC stepdown transformer D4 stepdown transformer D4

Figure 4: Full-Wave Rectiﬁer with Filter Capacitor Schematic

The output from a spice simulation ofTITLE the full-wave rectiﬁer is shown in ﬁg 5. The red trace is the

input voltage. The blue trace is the outputFILE: voltage. After the inputREVISION: waveform begins to fall below

PAGE OF DRAWN BY: 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 capacitor holds-up the voltage between the voltage peaks. With an big enough capacitor the voltage droop between input voltage peaks can be made arbitrarily small.

Figure 5: Spice Simulation - Filtered Full-wave Rectiﬁer Showing Output Ripple