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Rectifiers and filters (Rectifying with diodes)


Demonstration of the average and effective values of a periodic signal of average wave and complete wave

The effective value of the same signal as the square root of the average of the quadratic values is defined that takes throughout the period. Of an informal form but, it is possible to be defined as as that one tension that would have a battery that would give the same power to the load. If the load were a light bulb, when changing the alternating source by the battery (with its effective value) the light bulb would illuminate the samevalues average and effective average wave


In the following connection, it appears the mathematical development to demonstrate the values average and effective of the previous signal. This development does not enter matter.

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 We begin with a signal of average wave: We begin with the average value: The signal has a value:

 Vmax sen (wt) between 0 and Π and zero from Π and 2*Π. value average sine signal  

 value average sine signal 2

What we do is to calculate the surface of the disc parking and to divide it by the length, of that form the average value is obtained. In order to calculate the surface we must resort to the integrals, so that:

For the effective tension, it is come in the same way, but in this case, the values to the squared one are taken:

First that we do is to replace the sen ² (wt) by his equivalent, since:



sen ² (wt) = (1-cos (2wt)) /2

The same can be done for the signal of complete wave.

rectificadormedia wave

Rectifiers of average wave

In the image superior, is a circuit of average wave, where a single diode is used to obtain only a cycle of the signal.


From the point of view of energy efficiency, the rectifier of average wave is not appropriate, since we are not taking advantage of half the energy of the generator.

As the diode is polarized in direct in the positive half-cycle, that cycle will happen totally to the load of the circuit, represented by a resistance of 10 k.

In the negative half-cycle, the diode is polarized in inverse, and it does not let pass the current. For that reason, during that cycle, the signal of the tension in the load (in red) is zero.

Rectifiers of complete wave

In this occasion we are going to both take advantage of energy cycles to obtain a continuous tension when coming out, with better advantage of the energy. We have two ways to obtain it.

wave is incumbent on with transformer

1º Uso of a transformer with average taking.

In this circuit, we have V1 and V2 has the same value of maximum tension, but when being listed from the central point of the secondary one, V1 is opposed to V2.

In the first case, when having V1 the positive half-cycle, D1 lead and happen to the Resistance.

fotoiIn the second half-cycle, now V2 will be positive with respect to mass, causing that D2 leads and pass that half-cycle to the resistance.

2º Uso of Graetz bridge

Very used, it uses 4 diodes leading in each half-cycle, two of them and remaining the other two in it cuts.

Based on the necessary amperage, we have a very ample range, needing some them the installation of heat dissipators

We will see an explanatory video

We already have as obtaining a continuous tension, but it is necessary to maintain enters it certain values (that it does not fluctuate much in tension.


A first step is to place a condenser of high capacity (normally electrolytic) that it allows to provide current when the source falls to values very low. In the crests of the half-cycle current is provided to the load and to the condenser and in the low tensions it is the condenser that provides the current

smoothing circuit mediates wave with condenserTension of curling

Once rectified the alternating signal, we have a positive cycle for the case of the rectifier of average wave. As we needed to have a stable tension but the possible thing (that it does not raise and it lowers too much), the first step is to put a condenser that avoids those slopes and ascents. Therefore, in the cycle that leads the diode, part of the current is used in loading condenser C.load and unloading When the diode does not drive, it will be the load stored in C orders to provide it the current to the Rl load.

We have represented the three signals of the rectifier of average wave (right figure).

First it would be the signal without sizing, the one that we have in the secondary one of the transformer.

Second it is the tension when coming out of the resistance without condenser

curling tensionThe third sample the load and unloads of the condenser. In the time marked as Don, the diode leads and the condenser quickly load. The rest of time, the condenser unloading, providing itself the current to the load


For the case of a rectifier of complete wave, we are going when coming out to have both cycles, that when connecting the condenser, those slopes and ascents of tension will be smaller. To see graph of the left

 curling tension

An important aspect of the smoothing circuits is the variation of the value of the tension that there is when coming out. That variation denominates tension of curly and comes given by the formula:


Where I am the current, f the frequency when coming out and C the capacity. curly tension

We saw that the relation between the load of the condenser and the tension comes given by


a) To mount in cocodrile a bridge of diodes with a condenser and milliammeter as one is in the figure and to verify that the theoretical tension of curling agrees with the shown tension in the simulator.

circuit to see curly tension

b) To change variables C and F to verify how it varies the curling tension



  1. The filters in electronics are vitally important since, when using them, we avoided that they pass certain frequencies and noises that can they disturb the system that we want to feed.  We have three types of filters, according to what we want to eliminate:
  2. Filter low step. If we want that they pass the low frequencies. Therefore, the discharges are eliminated
  3. High-pass filter. If we want that they pass the frequencies from a concrete value
  4. Band-pass filters. In this case, the frequencies contained in a certain band (from a F1 to a F2) are going to happen, eliminating the low frequencies and high.

Filter band elimination. The opposite does right who the previous filter

Activity of investigation. With the knowledge given previously along with the expression of the impedance of a condenser given by:

Part 1º Calcular the impedance that offers the condenser that is next to the load in the circuit of the figure for a frequency of 50 Hertz if the capacity of the condenser is 27 nF and in case 100 C= µF. To repeat the process for and a frequency of 5 KHZ

Part 2º Montar in Crocodile the following circuit: Filter low step The condenser has a capacity of 100 nF. Its capacity raises 1000nF. That happens. Because?

 Part 3. To mount to replace the values of C and F of the circuit by the values of the section to and calculating the tension that we have in the condenser (Value of the sounding). It is considered that the tension of the generator is 10 volts.

Part 4. To calculate the 10 volt and voltage drop that has the condenser of 1 microfarad with the same generator of 5 KHz of maximum value. To verify in cocodrile.


Band-pass filter

 it formulates band-pass filter

 In order to pass a frequency band, we needed, as it is indicated in the book, a coil and a condenser in series, but also is possible to be used a circuit with two condensers, constructing a filter step low and high step. The union of both does not allow to have a band-pass filter. We will see how we can calculate the cutoff frequencies: band-pass filter with condensers We must consider that the cutoff frequency happens when the gain falls in 3 dB. This supposes that we have a loss of -3 gain dB, or what is the same:

 band-pass filter formulates 2

 In order to calculate log decimal of a number, we must make the inverse one of the logarithm, that is to see what is the exponent that has base 10. This is:

 frequency cuts band-pass filter

In order to calculate the frequencies where it happens this, we must analyze the filter as one of low step and another high step. This simplification supposes an approach very near a complex analysis but, but it offers some very similar results. In the case of the high step, the cutoff frequency comes given by the first resistance and the first condenser:

 it formulates frequency cuts band-pass filter

The result of fc is 66314 Hertz. We come equal for the other cutoff frequency, taking the other values and have:

praciticas of filters Exercises and practices

Calcular the average and effective value of an alternating signal of 10 volts tip to tip

Calcular the average and effective value of the rectified tension (complete wave) if the value of tip is 30 volts

Calcular the average and effective value in a rectified tension of average wave. The value of tip is 60 volts

does 4º If we have in network a tension of 220 volts, what peak voltage have that tension?

Tenemos a power supply with exit of complete wave (it uses transformer with intermediate taking), that has a 1200 load Ω, a current of 6 mA and the condenser has a 20 value of µF. To calculate the curling tension, if the tension of feeding of network is 220 volts.

Hacer the same calculations if now we have a signal of average wave

Montar the circuits in cocodrile and to verify that the standard firing datas agree with the simulated ones

answer frequency Simula in proteus a filter of high step if the resistance is 1.2 K Ω and the 4 condenser is of nF. To verify that the cutoff frequency agrees with the calculated value. Important. Response” within proteus is necessary to add yn “frequency.  For it, to choose in the column of instrumentation, the component frequency. Puncturing in the zone of work, we give size to that window. For the rest of steps, to see example that we have in

answer band-pass filterthe page on use of the oscilloscope. In particular it is explained in exercise 5 of filters.

9º Montar the circuit of step band that with the values that appear in the circuit of the figure and of seeing the frequency response.  To verify that the theoretical values (cutoff frequency inferior and superior) agree with the values of the simulator. For it, db is necessary to lower 3 on the maximum gain of the filter.

To show some solutions

In an alternating signal, as the positive cycle is equal to the negative, the average value is 0

On the other hand, the effective tension comes given by the maximum value between the root of 2. Therefore we have a value of 10/1.41 = 7.09 Volts

For a complete wave of 30 volts tip, the value half no longer is zero. If we looked as the average value for a rectified wave of half cycle calculated, where we only have a half-cycle by complete cycle, the average value is Vmax/Π. In this case, as we have a complete wave, the S-value the double, therefore Vmed = 2 *Vmax/Π = 60/3.14 = 19.10 volts.

complete Wave means that the frequency is going to be the double, therefore, the frequency that we must put is 50*2 = 100 Hertz.

Vr = 6 mA/(100 * 20 µF) = 6 *10 -3/(100 * 20 * 10 -6) = 60.000/2.000 = 3 volts.

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