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  Simple Alternation Restoration        Footer   

Single Alternating Rectifier Circuits :

In this lesson, we take again in a more practical form, all the notions already exposed on the recovery of the Alternative Voltage (see P.N. junction).

On the other hand, in order to facilitate a possible search, all rectifier fixtures in common use are grouped in this same lesson, also via the links at the bottom of the page.

In electronics, whatever the installation envisaged, it must always be powered by a continuous voltage.

If the power source is a battery of batteries or accumulators, no problem exists. Indeed, the battery directly delivers the DC voltage.

This solution is not always applicable. The value of the requested voltage can be high and the current high.

In this case, the use of a battery or a series of batteries would be an inefficient and prohibitively expensive system.

The mains voltage is used, which, after straightening and filtering, is able to supply the desired current.


As you know, the mains voltage is alternative and more precisely sinusoidal.

This one increases regularly from the value zero towards a positive maximum, then decreases just as regularly of this positive maximum towards zero, then towards a negative maximum before returning to a null value.

The waveform of this voltage is shown in Figure 1.


The problem of recovery is to remove one of the alternations and most often the negative alternation.

In this case, only positive alternation remains, allowing to have a positive continuous voltage.

But, we can also keep the negative half and remove the positive half, so as to obtain a negative continuous voltage.

The active element used in this operation is called RECTIFIER.

In the most common cases, it is a vacuum or semiconductor diode for small powers, a mercury gas diode or an ignitron for high powers (use in industrial electronics).

Whatever the rectifier used, the assembly is as shown in Figure 2.



Knowing the properties of diodes (see lessons Theory and semiconductors), the principle of the rectifier assembly is simple.

a) - The diode drives during the positive alternations of the AC voltage of the sector (negligible resistance).

b) - The diode does not conduct during the negative alternations of the AC voltage of the sector (practically infinite resistance).

Between the cathode and the mass, one thus obtains a tension having the form indicated Figure 3.


In Figure 2, we see that the secondary of the transformer is connected on the one hand to the diode and on the other hand to the ground (metal part of the frame). As a result, the output voltage is collected between the cathode of the diode and the ground.

This mass link can of course be suppressed. In this case, the assembly is as shown in Figure 4.


The fact of using the mass of the chassis however facilitates the wiring by avoiding the use of a common conductor (conductor connecting the low point of secondary, also called "cold point" of the output).

The output voltage is not really continuous. It is indeed in the form of impulse (See Figure 3).

Also, to make this voltage more even, it is good to insert between the output terminals, an electrochemical capacitor of high value (from a few microfarads to several tens of microfarads in the most common case).

The rectifier assembly is then as shown in Figure 5.   (Click here : back).


This capacitor, called the filtering capacitor, is charged to the maximum value of the alternating voltage (the influence of the direct resistance of the diode can be neglected) and discharges when the voltage decreases and ceases to exist (negative alternations).

The shape of the output voltage can be further improved by using, instead of a single capacitor, a leveling filter consisting of two capacitors and a coil (or resistor).

We will study these filters in this lesson.

For the moment, we will consider only the capacitor between the output terminals of the assembly, because it intervenes for the choice of the rectifier to be used in the project of a power supply.

To understand the explanations that follow, let's take a numerical example : either 250 volts, the effective voltage delivered by the secondary of the transformer.

The DC voltage, that is the output voltage without load, will have the value :

Us = 0.45 x effective U = 250 x 0.45 = 112.50 volts.

That is approximately half of the rms value, which is normal, since only one alternation is rectified.

However, the capacitor will charge at the value of the maximum voltage, Either :

U max = U eff x racine.gif2 = 250 x 1,41 = 352 Volts approximately.

Also, during the negative half-waves, when the anode is brought to the maximum negative potential, see - 352 Volts, the cathode, through the capacitor will be at + 352 Volts (considering that the capacitor is not discharge).

We will thus have a reverse peak voltage of :

352 + 352 = 704 Volts

In normal operation, that is with a load in the power supply, the average reverse voltage will not reach this value.

Indeed, to maintain the current flow, the capacitor will begin to discharge as soon as the positive alternation of the AC voltage, having reached its maximum, will return to a zero value.

In practice, with a load power supply, the DC output voltage will be equal to the value of the effective AC voltage applied.

In conclusion, when planning a simple alternating power supply, with a filtering capacitor, the following three factors must be taken into account :

  • a) Reverse voltage,

  • b) Maximum current required,

  • c) Capacitor isolation voltage.

Let's take an example : Element of computation to realize a simple alternating power supply, delivering a maximum current of 400 mA under 100 Volts :

a) Since the DC output voltage is equal to the effective AC voltage, it will be necessary to use a transformer with a secondary that can deliver a current of at least 400 mA and a voltage of 100 Volts.

b) For the choice of the rectifier, we will have to take a component capable of supplying a rectified average current of 400 mA, and able to withstand a reverse voltage of two to three times the value of the effective voltage, that is to say :

100 x 2 = 200 at 100 x 3 = 300 Volts.

c) As for the insulation voltage of the capacitor, a theoretical value of 242 Volts would be necessary.

However, since the power supply delivers a current, so that the capacitor discharges, it is generally adopted (for reasons of economy) a lower value.

The peak reverse voltage is applied only intermittently and for a very short time.

It is thus possible to take a capacitor isolated at 150 volts (voltage service), able to withstand a "peak voltage" or "test voltage" much higher.

However, for safety reasons, a power supply must never run empty

Indeed, for a zero flow, the output DC voltage passes, in the example given, from 100 volts to 142 volts. During the same time, the alternating voltage on the anode can be - 142 volts, so the voltage between the anode and the cathode is 284 volts (142 + 142).

Under these conditions, it may result in the destruction of the chemical filtering capacitor.

To avoid this inconvenience, very often the output of power supplies is inserted, a protection resistor called Bleeder (Anglo-Saxon meaning : who draws power). This resistance is shown in dotted line in Figure 5.

The value of this must be sufficient to determine a flow of a few milliamps, capable of causing the discharge of the capacitor, but should not be too low, so as not to overload the power supply.

For example, a flow rate of 5 mA may be allowed.

The Bleeder resistance will then have the value : R = U / I = 100 / 0.005 = 20 kΩ.

It then remains, in consulting a manufacturer's catalog, to choose the rectifier meeting the standards determined in b, (BRY 13, 1N 334, SFR 264, etc ...).

To protect it from any overloads, it can provide a series resistance, placed before or after the rectifier, but BEFORE the LOAD ; the average value of this resistance is of the order of 5 Ω.

Since the current supplied by the power supply will circulate in it, this element must be able to dissipate a power of :

P = R x I² = 5 x 0,4² = 0,8 watts

As a precaution, it is good to provide a resistance of 1 to 2 watts.

In the case of a power supply via a transformer, it can be considered that the ohmic resistance thereof provides sufficient protection. On the other hand, in the case of a direct supply on the network, this resistance is essential.

In the preceding figures, the rectifier has always been represented with the symbol of a semiconductor diode.

Obviously, with the same assembly, it is possible to use a vacuum diode. It suffices simply to provide the transformer secondary, an additional winding, for heating the filament (see Figures 6 and 7).


It can be seen in Figure 6 (diode with indirect heating) that one end of the filament is connected to the cathode. This connection, which does not change the operation of the tube, is intended to prevent possible ignition between the filament and the cathode.

With this arrangement, it is obviously not possible to connect one end of the "heating-filament" secondary to ground and to do the same with one end of the filament (solution facilitating wiring).

In this case, indeed, the filament would be subjected to the rectified DC voltage.

Modern vacuum diodes, being designed with sufficient cathode-filament isolation, this connection is not always essential.

In the assembly of Figure 7 (diode direct heating), it is the filament that serves as a cathode.

The DC output voltage is taken between the ground and this electrode.

Again, it is not possible to connect to the earth one end of the secondary, "heating filament", and one end of the filament.

In this case, in fact, the DC voltage would be directly short-circuited with the ground.

The table in Figure 8 summarizes what has been said about the rectifier mounts.


This type of mounting is simple, but the leveling of the DC output voltage is difficult, because the negative alternations of the AC voltage are purely eliminated.


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