2.8 Rectifiers

Diodes are very often used in circuits that convert an alternating voltage (ac) to a voltage with constant polarity (dc). This process is called rectification. Electrical energy is generated at power plants (gas, nuclear, wind…) and transported through very high voltage lines.  Transformers convert these high voltages to lower values.  In domestic wall outlets, the output voltage in most European countries is an alternating voltage of 220 VRMS: a sinusoidal voltage with a frequency of 50 Hz – one period lasts 20 ms – and an amplitude VP of 220.√2 V ≅ 310 V.  Most electronic equipment however operates on a constant voltage, e.g. television sets, smartphone, PCs….  The purpose of a rectifier is to generate a voltage as stable as possible, starting from an alternating voltage.

The Half Wave Rectifier

The most elementary rectifier is the half wave rectifier (Figure 2.21).  The load is considered to be resistive, which means that it behaves like a resistor, but the load can be anything (the ‘remaining circuit’).  There is only one loop, and so only one current, going through all components.diode_rect_halfwave

 Figure 2.21 Half wave rectifier circuit

The voltages are related according to Kirchhoff’s law. Note that this equation is always true, whatever the state of the diode, or the value of the input voltage:

vin – vD – vout = 0 V

Figure 2.22(a) shows the input voltage as a function of time.  During the first half period (0 ms – 10 ms if the frequency is 50 Hz) the input voltage is positive, and the source wants to ‘push’ the current in the direction of the diode arrow.  The diode will conduct, and will ideally behave like a closed switch. The voltage across the diode is then 0 V (see fig. 2.22(c)) and the voltage across the load equals the input voltage (fig. 2.22(b)).  The current can be calculated by applying Ohm’s law on the load (fig. 2.22(d)):

i = vin / RL if vin > 0V

diode_rect_halfwave_signals

Figure 2.22 Voltages and currents as a function of time for a half wave rectifier: (a) input voltage, (b) output voltage, (c) diode voltage, (d) current through all components

During the second half period (10 ms – 20 ms) the input voltage is negative, and the source wants to push the current in a direction that is not allowed by the diode.  The diode will not conduct, and behaves like an open switch.  There is no current (fig. 4.14(d)):

i = 0 A if vin < 0 V

If there is no current through the resistive load, there is no voltage across it (fig. 4.14(b)).  From Kirchoff’s voltage law, it follows that the diode voltage equals the input voltage (fig. 4.14(c)).

Care should be taken that the minimum value of the diode voltage – the most negative peak, which is called the PIV or Peak Inverse Voltage – does not exceed that break-down voltage of the diode:

PIV = VP < VBD

In summary: for a half wave rectifier, the output voltage equals the input voltage if the latter is positive, and the output voltage equals 0 V if the input voltage is negative.  The input – output characteristic is depicted in Figure 2.23.

diode_rect_halfwave_IOFigure 2.23 Graphical representation of the relation between the input voltage and the output voltage for a half wave rectifier

The Half Wave Rectifier with Capacitor

The previous circuit results in an output voltage that is never negative, but it is far from constant.  By placing a capacitor in parallel with the load (see fig. 2.24), the output voltage is smoothed.  Note that a capacitor is a component across which the voltage can not change abruptly: it has to be charged or discharged in order to do so, and that takes time.

diode_rect_halfwave_withCFigure 2.24  When a capacitor is placed parallel to the load, the output voltage will stabilize.

Suppose that the capacitor is completely discharged at start-up.  When the input voltage rises, the diode is conducting, and the voltage of both the capacitor and the load equals the input voltage.  The diode current will split in a part that charges the capacitor, and a part for the load:

iD = iL + iC

The capacitor will be charged until the input voltage reaches its maximum value.  When the input voltage decreases, the capacitor can not discharge through the diode, since this would require a current in a direction that is not allowed.  The diode will be “off”, and what is left is an RC-circuit: the capacitor will discharge through the load.

iL = – iC

In practice, large capacitor values are used, so the process of discharging will be slow (large time constant t = RLC).  The output voltage decays exponentially (Figure 2.25).

diode_rect_halfwave_signals_withCFigure 2.25  The input voltage (a) and the output voltage (b) as a function of time for a half wave rectifier with smoothing capacitor.

At a certain moment, the input voltage will again be larger than the output voltage.  The diode will conduct again, and the capacitor will again be charged until the peak value.  The overall result is a rectified output voltage that fluctuates only a little bit.  The maximum fluctuation of the voltage is called the ripple voltage VR.  It’s value can be calculated from the very popular formula for a capacitor:

C.ΔV = I.Δt

It is obvious that one should know which values to fill in.  C is the value of the capacitor and ΔV is the ripple voltage VR.  During the discharge process the discharge current I equals the current through the load (iL) that remains almost constant (as long as the ripple voltage is small compared to the peak voltage vP). Δt corresponds to the discharge interval toff, the length of the time interval during which the diode is off.  As a result:

VR = iL  .toff / C

It is important to see the logic in this formula.  A larger load current iwill result in a faster discharge of the capacitor, and thus in a larger ripple voltage.  A larger toff means that the capacitor will discharge during a longer period, also resulting in a larger ripple voltage.  A larger capacitor will discharge more slowly, so that the ripple voltage remains small.  Note that the formula above is an approximation.  It is assumed that the discharge process is linear, while in reality it is exponential.  In practice, the calculation is accurate enough, since the discharge process is so slow that the exponential curve can be approximated by a straight line.  Also the values for  iL and toff are usually approximated: |iC| = iL ≅ VP / RL and toff ≅ T = 1/f.

Challenge: Try to prove yourself that the capacitor current iC during charging equals C.VR/ton and that PIV = 2.VP.

The Full Wave Rectifier

A better circuit is the full wave rectifier (Figure 2.26), which is also the most commonly used.  If the input voltage is positive, diodes D1 and D4 are conducting, and the output voltage equals the input voltage. For a negative input voltage, diodes D2 and D3 are conducting, and the output voltage is the inverse of the input voltage. The output voltage is always positive, and corresponds to the absolute value of the input voltage.  The input-output characteristic is shown in Figure 2.27. If a capacitor is placed in parallel with the load, the fluctuating output voltage will be smoothed, resulting in an almost constant output voltage (see fig. 2.28).

The advantage of a full wave rectifier, compared to a half wave rectifier, is the also the negative half period of the input signal is ‘used’. The discharge time is much shorter – it halves – resulting in a much smaller ripple voltage.  Try to prove that PIV = VP.

diode_rect_fullwave

Figure 2.26  The circuit of the full wave rectifier. Again, a capacitor can be placed in parallel with the load to smooth the voltage.

diode_rect_fullwave_IO

Figure 2.27  Input – Output characteristic of the full wave rectifier.

diode_rect_fullwave_signals

Figure 2.28  The input voltage (a) and the output voltage (b) as a function of time for a full wave rectifier. A capacitor smooths the output voltage.

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4 thoughts on “2.8 Rectifiers

    • The formula is correct, since iC equals iL (in magnitude) if the diode is off.
      According to the chosen directions of the currents, iC = -iL, indicating that the capacitor is discharging.
      Thanks for this comment. I added some elements in the formula.

  1. Why is it that we use the diode model 1 (without a knee voltage) in the half-wave rectifier ? It states that when the diode conducts, the voltage over it is 0V, shouldn’t this be 0.7 V ?

    • Diode model 1 is an ideal diode, so it’s knee voltage is 0 V.
      It is typically used when dealing with high voltages (> 100 V).
      Note that models are used to make approximate calculation.

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