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  Filtering cell with input self         Footer  

Leveling Filter and Snoring Voltage :


We have seen that by inserting an electrochemical capacitor, between the output terminals of the power supply, one could considerably reduce the "hollows" between two current pulses, that is to say level the shape of the DC voltage.

The capacitor used therefore behaves like a "reservoir", absorbing voltage peaks and restoring them at the desired moment, that is to say when the rectifier is blocked, during negative alternations.

However, despite this capacitor, the output voltage is still too wavy, to power an electronic assembly.

With the simple alternating circuit, the frequency of the "hollows", that is to say the frequency of the ripple is 50 Hz (the assembly only rectifies an alternation per period and the frequency of the voltage is 50 Hz).

With full-wave editing, the ripple frequency is 100 Hz.

This fact is highlighted in Figure 17, where for simplicity, the four periods of the voltage are represented.


The regime of successive charges and discharges being faster in the case of double-wave mounting, it follows that leveling is easier.

More technically, it can be said that for a given value of the capacitor, the capacitive reactance, that is to say the impedance presented by the capacitor to the AC component (ripple due to current pulses) is all lower than the frequency of the component to be eliminated is high.

We have indeed :

XC  = 1 / cw,   with :

XC = capacitive reactance

c     = capacitor value in farad

w   = 2  x n x F (with F in Hertz) = 6,28 x F.

After disposing a capacitor at the output terminals of the power supply, the current in the load has the appearance shown in Figure 18-a for a single-wave circuit, and Figure 18-b for a full-wave circuit.


Also, in order to level the voltage as much as possible, filtering cells are generally used at the outlet of the rectification circuits.

These are known as PASS-LOW Filters.

A low-pass filter is a device that attenuates as much as possible the frequencies above a given value.

In other words, a low-pass filter passes all the frequencies between zero (DC) and a frequency f0, called the stop frequency, and blocks all the frequencies greater than f0.

This filter, in the case of a power supply, is composed of two chemical capacitors and a coil or a resistor (Figure 19).


The operation of the whole is very easy to understand.

We have seen the effect of the first capacitor C1. It charges during current pulses and discharges when the voltage disappears or tends to decrease.

With the filter cell, the current must now pass through the winding before reaching the load.

Now, this winding has the property of combating the variations of the current. This results in an improvement in the shape of the US output voltage.

On the other hand, the capacitor C2 connected to the output of L (or R) behaves in the same way as C1 and therefore further reduces the AC component.

In summary, the AC component at the output of the rectifier meets two elements : C1 and L (which, let us repeat it perhaps replaced by a resistor).

However, for this alternative component, C1 has a capacitive reactance lower than the inductive reactance of L or the resistance of R.

Indeed, by performing the calculation on the basis of C1 = 50 µF (value generally adopted), L = 4 H and F = 50 Hz, we obtain :


By replacing L = 4 H with a resistance of 1256 Ω, one could expect to obtain the same result from the point of view of attenuation of the AC component.

In practice, the result is less good because the coil has the property of opposing current variations, which is not the case of a resistor. However, RC cells are sufficiently effective in practice.

Using the same example, we see that in the double-wave array, where the AC component has a frequency of 100 Hz, the result is considerably improved.


Of course, the capacitor C2 has the same effect, with respect to the load of the power supply, as the capacitor C1 with respect to L or R.

For the calculation of the attenuation of the AC component (the latter causing humming in an amplifier BF, also called snoring voltage), the value of the DC output voltage US and also of the current delivered must be taken into account. by the power supply and absorbed by the load.

It is obvious, indeed, that the AC component depends on the DC output voltage, because the higher it is, the higher the AC voltage required, is of significant amplitude.

As for the current delivered by the power supply, it is equally easy to understand that the smaller the power, the more capacitor (when discharged) is able to maintain a constant flow rate.

The theoretical calculation of the attenuation, which is very complex, must take into account the charging and discharging times of the capacitor, as a function of the voltage and current parameters.

We will not discuss the details of these calculations outside the scope of this lesson, especially since the results obtained are often incorrect in practice, depending on the nature of the load that can be partially resistive, inductive and capacitive.

However, to have a rough idea, perfectly valid in practice, one can simply calculate the snoring voltage using the formulas below.

a) - Single alternating circuit, with filter having only one capacitor :

Vr = (4 x I / C1) x racine.gif     with :

  • Vr    = Snoring voltage

  • I       = Current output in mA load

  • C1  = Capacitive value of the capacitor in µF

  • racine.gif2 = 1,41

Example : A power supply with a current of 60 mA. The filter capacitor has a value of 50 µF.

Vr = (4 x I / C1) x racine.gif2 = (4 x 60 / 50) x 1,41 = 6,7 Volts.

b) - Full-wave circuit, with filter having only one capacitor :

Vr = (4 x I / C1) 

Example : Same data as in the previous example.

Vr = (4 x I / C1) = 4 x 60 / 50 = 4,8 Volts

c) - Single alternating circuit, with filter comprising two capacitors and a coil (see Figure 19).

Vr = (4 x I / C1 x C2 x L) x racine.gif2    with :

  • I en mA

  • C en µF

  • L en H (henry).

Example : Same data as in the previous example, plus C2 = 50 µF et L = 2 H.

Vr = (4 x I / C1 x C2 x L) x racine.gif2 = (4 x 60 / 50 x 50 x 2) x 1,41 = 67 mV

d - Full-wave circuit, with filter comprising two capacitors and a coil (see Figure 19) :

Vr = 4 x I / C1 x C2 x L

Example : Same data as in the previous example :

Vr = 4 x I / C1 x C2 x L = 4 x 60 / 50 x 50 x 2 = 48 mV

In the last two formulas, replace L (value in Henry of the coil) by R (value of the resistance in place of the coil).

In this case, we then express :

  • I in mA

  • C in µF

  • R in kW

Example : Same example as above, but with R = 1,200 Ω instead of L = 2 H.

Vr = 4 x I / C1 x C2 x R = 4 x 60 / 50 x 50 x 1,2 = 80 mV

This value of 80 mV, representing the AC voltage, superimposed on the DC output voltage is perfectly acceptable in the vast majority of cases.

Also, the fact of using a coil rather than a resistance depends mainly on the permissible voltage drop, caused by the filter cell.

Looking at Figure 19, it can be seen that the entire current absorbed by the load (which will be connected to the output terminals US) passes through the coil L or the resistor R.

This obviously results in a voltage drop (U = R x I) and therefore the DC voltage US available, is less important.

Let's take an example with common values.

A filter coil of 4 Henrys for example, of excellent quality, has an ohmic resistance (due to the winding constituting the winding) of the order of 40 Ω.

Instead of this coil, it can take a resistance of about 1500 Ω.

If the current output is 60 mA for example, in the first case, the voltage drop will be :

U = R x I = 40 x 0,06 = 2,4 V

Thus, with a DC output voltage of 250 Volts, the load voltage US will be :

250 - 2,4 = 247,60 Volts about

With the resistance of 1500 Ω, and under the same conditions, the voltage drop will be :

U = R x I = 1500 x 0,06 = 90 Volts

The US voltage under charge will then have value :

250 - 90 = 160 Volts

If the current required was even more important (for example 100 mA instead of 60 mA), the voltage drop would be prohibitive (1500 x 0.1 = 150 Volts).

In conclusion, a filtering resistor (inexpensive and space-saving element) is used when the current that the supply must supply is relatively small.

In the opposite case, a filtering coil is adopted.

In practice, there are power supplies with CRC cells, for maximum currents of the order of 60 to 80 mA.

Beyond these values, there are CLC cells.

In both cases, when the power supply must supply current to several loads, several filter cells in series can be encountered.

This avoids the influence of one load on the other (this is called decoupling circuits).

The assembly is as shown in Figure 20.


In this example, if :

  • L = ohmic resistance = 40 W

  • R1 = 1000 W 

  • R2 = 500 W

and if the currents requested are :

  • HT1 = 100 mA

  • HT2 = 50 mA

  • HT3 = 20 mA

We will have, in the case of a DC output voltage of 250 volts :

  • HT1 = 250 - (40 x 0,1) = 246 Volts

  • HT2 = 246 - (1000 x 0,05) = 196 Volts

  • HT3 = 196 - (500 x 0,02) = 186 Volts

Of course, each cell improves filtering and adding effects ; the snoring voltage on HT3 will be lower than on HT2, itself lower than on HT1.

Finally, since there are circuits that can be powered with a poorly filtered voltage, it can be taken directly on the cathodes, across C1 (HT0).

By neglecting the internal resistance of the rectifiers, the voltage drop will be zero, therefore : HT0 = 250 Volts.


In the cells we examined, the first element was always capacitor C1.

For this reason, these cells are called capacitive input cells.

In the vast majority of cases, these are the ones found on power supplies.

However, on some montages, SELF is placed as an input element.

The diagram of this kind of circuit is represented in Figure 21.


This type of cell is adopted for gas and mercury vapor tubes (Industrial Electronics).

With these, low internal resistance, the adoption of a capacitor of high capacity, would give rise when it is discharged at an excessive charge current, which can destroy these tubes.

The SELF filter in the HEAD, decreases the AVAL ripple rate (after the coil) but increases the ripple rate AMONT.

The residual hum at the filter outlet, is calculated as above, using the formula :

Vr = 4 x I / L x C

To have a constant DC output voltage, with a low load, the input choke must have a minimum value in henries, much larger than that required for a normal load.

However, a high value of L means that the magnetic circuit is bulky (a lot of iron), and that the winding (a lot of copper) has a significant ohmic resistance.

Since it is necessary to reduce this resistance to avoid unnecessary voltage drop and power dissipation, a slip self is usually used, that is, a coil with a reduced, or even zero, gap.

Such a self has the required value when the flow is low, but falls to about one-third or more, when the flow is normal.

This phenomenon is due to the partial saturation of the iron, in which the magnetic flux remaining almost constant can not induce a counter-electromotive force in the coil, which thus loses its power to oppose the variations of current, that is to say to say, loses its inductive value.

It should be noted that in both types of cells (capacity or self at the top), the coil must be provided for the maximum current required.

Otherwise, there will be a heating (or even destruction) of the winding.

In the filter cells with capacity at the head, when L is replaced by a resistor, it must be able to dissipate heat, the power resulting from the resistance value of R and the current (P = R x I²).

Thus, in the case of a power supply 80 mA, it will be necessary to choose a filter choke capable of supporting this intensity.

If the choke is replaced by a resistance of 1500 Ω for example, it must be able to dissipate a power of :

P = R x I² = 1500 x 0,08² = 9,6 Watts see 10 Watts in practice.

We will now see other power circuits, commonly used in different applications of electronics.


Nombre de pages vues, à partir de cette date : le 27 Décembre 2019

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