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Double Alternating Rectifier Circuits :



2. - DOUBLE ALTERNATION RECOVERY

To overcome the disadvantages of the previous assembly, that is to say : DC output voltage dependent on the value of the filter capacitor and difficulty in obtaining a good leveling, it was realized another type of rectifier circuit, where the two alternations of the voltage are actually used.

This new assembly is called full-wave rectifier, or current doubling rectifier (not to be confused with voltage doubler rectifiers which we will discuss in this lesson).

In this type of circuit, two rectifiers (vacuum or gas diodes or semiconductor diodes) and a transformer with secondary midpoint (Figure 9) are used.

 Redresseur_a_double_alternance.GIF

This secondary is divided into two equal parts, by the middle point, connected to the mass.

Each of these two parts, S1 and S2, provides the effective alternating voltage, to straighten.

Everything happens as if we had two transformers with secondary in parallel and in opposition.

Thus, during the positive alternation of the AC voltage, when it appears across S1, the polarities are indicated in Figure 10, across S2 these polarities are reversed.

Inversion_des_polarites_secondaires_S1_et_S2.GIF

The opposite phenomenon occurs during the negative alternation of the AC voltage (Figure 11).

Changement_de_polarites_secondaires_S1_et_S2.GIF 

It is clearly seen from Figures 10 and 11 that when the rectifier It is clearly seen from D1 is conducting, D2 is off (positive half-cycle), and that when D2 is conducting, D1 is off.

Therefore, each rectifier becomes alternately conductive, during the full period of the AC voltage.

As a result, the DC output voltage US has the form shown in Figure 12.

Forme_de_la_tension_de_sortie_redresseurs.GIF

This assembly requires a transformer with secondary delivering twice the required voltage.

Indeed, if one desires an effective voltage of 250 Volts for example, the complete secondary will have to provide 500 Volts.

After connecting the midpoint to the ground (common point), each half-secondary S1 and S2 will deliver 250 Volts, the required value.

By examining Figure 12, it is easily understood that the DC output voltage US must have a greater value than that obtained with the simple alternating circuit.

In this case, in fact, there is no interruption between each current pulse.

Without leveling capacitor, this output DC voltage has the value :

US = 0,9 x U eff.

With a capacitor placed immediately after the diode, the DC output voltage US, has, as with the previous assembly, a value of :

US : 1,41 x U eff.

In both cases, indeed, the capacitor charging at the maximum value of the AC voltage, it is normal that the value of the DC output voltage is the same.

However, with a normal load, the available DC output voltage has the value :

US = 1,2 x U eff.

Or significantly more than in the previous edit.

Above all, it must be emphasized that, with the full-wave circuit, the value of the DC output voltage does not depend on the capacitive value of the capacitor.

On the table in Figure 8, we have mentioned :

- DC output voltage with normal charge : < Ueff depending on the value of the capacitor.

It is easy to understand on this subject, since there is an interruption equal to half a period, in the DC voltage pulses, that it is the capacitor which must supply the current absorbed by the load, during these interruptions (see Figure 13).

Restitution_de_la_tension_entre_2_impulsions.GIF




It is also easy to understand that the charge of a capacitor depends on its capacitive value (Q = C x U) ; the higher it is, the higher the stored charge will be, so it will be able to supply a current, between the two pulses A and C (Figure 13)

With a full-wave power supply, the DC pulses follow each other regularly without interruption, and the capacitor does not have to compensate interruptions, but only to "level" the current (see Figure 14).

Restitution_de_la_tension_entre_2_impulsions_1.GIF

As for the reverse peak voltage supported by the two rectifiers, it is, as in the previous assembly, equal to :

2,82 x U eff.

All that has been said for the choice of elements, for a single alternating power supply, therefore remains valid with the full-time power supply.

It should be noted however that with this last assembly, each rectifier provides only half of the intensity required.

It is for this reason that this editing is often called intensity doubler.

Thus, taking the previous example, to achieve a power supply 400 mA under 100 Volts, we can use two rectifiers, capable of providing only 200 mA each.

The table in Figure 15 summarizes the characteristics of the full-wave rectifier assembly.

Tableau_des_rappels_1.GIF

With this last circuit, one can, as with the simple alternation circuit, use rectifiers of any type.

However, solid-state rectifiers are more advantageous and more flexible to use.

Let us examine the curves of Figure 16, relating to the anode characteristics of a vacuum diode and a semiconductor diode.

Diode_a_vide_et_diode_a_semi_conducteur.GIF

The Figure 16-a relates to a vacuum diode. On this one, we see that the supplied current depends a lot on the voltage applied on the anode.

Thus, when the manufacturer indicates for a vacuum diode (EY 82, for example) rectified current : 180 mA, anode voltage : 250 volts, the current of 180 mA represents the maximum value.

By increasing the anode voltage, the tube will work at saturation and the current will not increase so far. On the other hand, if the anode voltage is less than 250 volts, the current will be much lower.

Thus, with this tube, it is necessary to have an anode voltage of 250 volts, to obtain a current of 180 mA.

It is not the same with a semiconductor diode. Indeed, with this type of rectifier, the current increases very rapidly, as soon as the threshold voltage is reached (point A in Figure 16-b).

However, the threshold voltage is very low (about 0.6 volt). By exceeding the voltage limit value set by the manufacturer, there is no saturation as for the vacuum diode, but runaway and destruction of the rectifier.

Thus, the semiconductor diode SFR 164 for example can provide a current of 1 ampere and withstand an anode voltage of about 150 volts.

This diode will still operate normally with a much lower voltage 100 V - 50 V and even less, while still providing a large current.

We can therefore use this component to rectify higher voltages, although it is provided for 150 Volts, while with a vacuum tube, it is advisable to respect the voltage given by the manufacturer, the delivered current falling quickly, when the applied voltage decreases in value.

On the other hand, the semiconductor diode is capable of providing safe, instantaneous current, significantly higher than the rectified average current.

This instantaneous current can be requested by the filter capacitor during each load.

On the contrary, the vacuum diode can not, without danger, provide an instantaneous current much higher than the rectified average current.

For this reason, in order to limit the instantaneous current, the manufacturers always indicate for the vacuum diodes : maximum value of the filter capacitor.

Example : vacuum diode EY 82 - filter input capacity = 60 µF maximum.

This value is never given for solid-state rectifiers, which can precisely provide a high instantaneous current.

It is also for this reason, the fact that the capacity of the capacitor can be very high, it is advisable to mount a protection resistor (Rs), before or after the rectifier, but before the filter capacitor (see Figure 5).

The value of this resistance is generally of the order of .

To determine its value in all cases, it is assumed that it must be between 1 to 5% of the value of the load.

We must therefore define what is meant by "normal load" of food.

The pure and simple application of the law of Ohm makes it possible to define this load.

In fact, in the case of a power supply having to supply 400 mA at 100 volts, it is immediately understood that such a current at 100 volts is that which flows in a resistor having as value :

R = U / I = 100 / 0,4 = 250 Ω

This value therefore represents the normal load of the power supply.

As a result, the protection resistor may have a value between :

(250 x 1) / 100 = 2,5 Ω at (250 x 5) / 100 = 12,5 Ω

Finally, it should be noted that this protection resistor (which can also be installed in vacuum diode rectifier circuits) is essential, in the case of a direct supply on the mains network, it is a power supply without transformer.

With a transformer, the total resistance of the input circuit is generally sufficient, the value of which is :

Calcul_resistance_d_entree_d_un_transfo.GIF

With R1         =  resistance of the primary winding,

         R2        =  resistance of the secondary winding,

        N2 / N1 =  transformation ratio.

 







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