Diodes Aplications, Half Wave Rectifier Theory and Study

This chapter consists of the items below. If you want to go straight to an item, click on it.

2 - Rectifiers Half Wave click here!

3 - Filtration Capacitor click here!

3.1 - Ripple Voltage - ΔV_r click here!

3.2 - Ripple Voltage RMS [V_r (rms)] click here!

3.3 - Filter Capacitor Value click here!

3.3.1 - For Ripple in Peak to Peak Values click here!

3.3.2 - For Ripple with Values in RMS click here!

4 - Rectifiers of Half Wave with Real Diode click here!

5 - Reverse Voltage on Diode click here!

6 - Average and Peak Current in the Diode click here!

One of the biggest applications of diodes is its use in so called rectifier circuits. What is a "rectifier circuits"? As we have already studied, there are the circuits AC and the circuits DC. The majority of electro-electric equipment work on DC power. However, for consumers, electrification companies provide energy of the type AC. So, for the electro-electronic equipment to work properly, we must transform the energy AC in energy DC. That's where the rectifiers come in.That is, they transform energy AC into energy DC.

So, what we are going to study in this chapter is how to rectify a sinusoidal type wave, supplied to the consumers, and turn it into an energy of continuous type.

2. Rectifiers Half Wave

Basically, we have two types of rectifiers: half wave rectifiers and full wave rectifiers. Initially, let's study the half wave rectifier. In the Figure 62-01 we have represented a rectifier circuit in its simpler way.

The circuit consists of a AC voltage source, passing through a diode and feeding a load R_L. Remember that the diode only allows electric current to pass through in one direction only.

See the Figure 62-02 for the graph of the input voltage (upper part of the figure), and at the bottom of the figure, the output voltage V_o on the load R_L. Note that the negative semicircle of the sine is eliminated by the diode. So we just got the positive semicircle of the sine.

How does this happen? When the input voltage starts from the origin and its value starts to increase, there is a direct polarization of the diode and it goes into conduction. In general, conducting diode means that the diode acts as a short-circuit. Then, the entire voltage of the source appears on the load until it reaches the maximum value. From this point, the voltage begins to decrease, reaching zero. Following, the sine has negative values and this causes the diode to enter the cutoff area. That is, no there is more current flowing through the circuit, because the diode acts as an open circuit. This is it, generates at the output a voltage zero. This lasts through the entire negative cycle. When the positive cycle returns, the process is repeated.

Although the rectified voltage is not a continuous or filtered voltage, it has a DC component and a specific ripple. For a rectified half-wave signal, the voltage DC, in relation to the peak voltage V_p, is worth:

The rms value of this ripple voltage, for a half wave rectification, is:

3. Filtration Capacitor

Our challenge now is how to fill the spaces "empty" between the two positive peaks, so that we can have a constant output voltage. We have to add some element that provides energy to the load between these peaks. Nothing more natural than using a ... CAPACITOR ... in parallel with the load. Let's look at how the circuit works by adding a capacitor.

See the Figure 62-03 the modified circuit where we include the capacitor in parallel with the load. When we include the capacitor, there is a difference in circuit operation. As soon as the of the source starts to grow from zero, the diode enters conduction transferring charge to the capacitor. This situation holds until the sine reaches its maximum value. At this time, the capacitor also got the maximum voltage. At that time, which makes the first 90° of the sine wave, the source supplied current to the capacitor and to the load. For clarity, in the circuit the diode is represented by a short-circuit. And the circuit, we easily conclude that I = i_C + i_L.

From that moment, the voltage of the source begins to decrease. However, the capacitor does not decrease in the same ratio. This is due to the fact that the voltage of the has a value less than the capacitor voltage. Then the diode enters the cutoff area, acting as an open circuit. See the Figure 62-04 the diode being represented by an open key. In this way, the source stops supplying current to the capacitor and to the load. Therefore, who supply current to the load is the capacitor, indicated by the green arrow and represented in the circuit by i_L. Note that when the diode is in the cut zone the current in the capacitor obeys the equation - i _C = i _L . This is evident watching the Figure 62-03 and Figure 62-04.

From the Figure 62-05 we find that when the sine wave varies between 90° and 450°- φ (zone highlighted in yellow), is the capacitor that provides current to the load. Here, φ_c represents the conduction angle of the diode. Thus, this small time interval that the diode conducts is sufficient to again raise the voltage in the capacitor up to the maximum value of the input voltage. This repeats cycle by cycle. In the figure we are representing the first cycle immediately after the circuit is powered by the voltage source.

It is easy to see from the Figura 62-05, that the lower the value of (φ_c), also minor will be the conduction time (φ_c) of the diode. This implies that the diode work the time necessary for the capacitor to return to its maximum voltage (V_C_max). After that, the diode re-enter in the cutoff zone.

So, with the presence of the capacitor we will have at the output an average voltage DC that we will call of V_DC , which can be approximated by eq. 62-01, considering that we can represent the ripple voltage by a triangular wave.

eq. 62-01

V_DC

eq. 62-02

I_DC

eq. 62-02

3.1 Ripple Voltage - ΔV_r

Normally in problems presented in books or tests, the ripple voltage can be specified as a numeric value or, as is more common, a percentage referring to the circuit output. In addition, there is the possibility of expressing the ripple voltage or ripple, as a RMS value (represented by V_r (rms)) or a value peak to peak (represented by ΔV_r). In this item we will study the case of the value peak to peak and in the next item the value RMS.

Taking as reference the last circuit, we see that it is simply a capacitor with an initial voltage V_C_max (immediately after the diode enters cut) in parallel with the load R_L. Over time, the capacitor will be discharged by the load. The basic idea to implement is that the voltage across the capacitor between the two voltage peaks is equal to V_C = V_C_max - ΔV_r. For this, we must have a time constant (τ = C R_L) of adequate value. As the value of R_L is known, then it remains to calculate the value of the capacitor C that allow the conditions of the problem to be satisfied.

From the above, we can write that:

ΔV_r = V_C_max - V_C

eq. 62-02a

As we know the initial voltage of the capacitor (V_C_max), after a time t a voltage in the capacitor will be:

V_C = V_C_max (1 - e^-(t/τ))

eq. 62-02b

To find an approximate equation of the value of C, let's make two simplifications, based on the fact that in this case we have τ = R C >> t. Therefore, we can write:

(1 - e^-(t/τ)) = 1 - (t/τ) and t ≅ T

In the above approximation, T is the period of the sine wave. By replacing these values in eq. 61-02b and combining this with eq. 61-02a, let's find:

ΔV_r = V_C_max - V_C_max (1 - (T/R_L C))

On the other hand, we know that the frequency is the inverse of the period. Soon, solving the above equation algebraically, we arrive at:

eq. 62-03

Therefore, knowing the values of f, R_L and C, we can calculate the value of tension of ripple. And so, we can define the index or percentage rate of the peak to peak ripple represented by the letter r_pp, of according to the equation eq. 62-04.

eq. 62-04

Note that using the eq. 62-03 we can write eq. 62-05.

eq. 62-05

Note that based on eq. 62-05, we can rewrite the eq. 62-03 as follows:

eq. 62-06

Working algebraically at eq. 62-01 and eq. 62-06, we can write the eq. 62-07:

eq. 62-07

Attention

We must emphasize that up to this moment the only approach made has been to assume the undulation of the wave rectified as a triangular waveform. This is the most accurate way to present the basic theory about rectifiers. However, in most textbooks, their authors make more approximations such as for values below 10% in the ripple index, it is assumed V_DC ≅ V_C_max and therefore we can write I_DC = V_C_max / R_L. With that in mind, we can bring to eq. 62-03 by the following expression:

eq. 63-08

3.2 Ripple Voltage RMS [V_r (rms)]

In the previous item, to study the ripple , we made an approximation of the voltage by a triangular waveform. So, we define ΔV_r as the variation in voltage peak to peak . Another way to study ripple is to consider the RMS value. For this, it is necessary to establish a correspondence between the value RMS and the value peak to peak of the triangular voltage. Using a little bit of calculation we will find that

eq. 62-09

Now, we can define the RMS ripple index represented by the letter r, according to the equation eq. 62-10.

eq. 62-10

By doing algebraic work with the equations we have already studied, we can write an equation that relates r, V_r (rms) and V_C_max, as can be seen in the eq. 62-11.

eq. 62-11

3.3 Filter Capacitor Value

Since between the peak to peak values and the RMS value there is a correction factor, let's study each case separately. Therefore, we must know in advance if the problem statement contains a value of ripple in RMS or peak to peak values. The Figure 62-06 shows the circuit of a half-wave rectifier with the filtering capacitor, C.

3.3.1 For Ripple in Peak to Peak Values

For the case of ripple to be given in values peak to peak, that is, we are working with ΔV_r, then manipulating the eq. 62-03 , we can find the capacitor value for the specified ripple, or

eq. 62-12

Knowing the values of f, R_L and r_pp, it is also possible calculate the value of C using the

eq. 62-13

Attention

Note that according to the observation made at the end of item 3.1, where we assume it was valid the approach V_DC ≅ V_C_max, it is possible rewrite the eq. 62-12 as:

eq. 62-14

For the case where r_pp < 10% to eq. 62-14 can be used, as the error made is less than 3% on average. As electrical components, in general, can have a greater tolerance than 5% and, in addition, they are sold with default values that do not always match the calculated values, so this error is perfectly tolerable. As an example, see Problem 62-2, or click here.

3.3.2 For Ripple with values in RMS

In case the ripple is provided in RMS value, we must consider the correction factor given by eq. 62-09. Using this correction factor in eq. 62-03, we find the value of C, or:

eq. 62-15

Replacing the eq. 62-11 in eq. 62-15, we get:

eq. 62-16

Attention

Note that according to the observation made at the end of item 3.1, where we assume it was valid the approach V_DC ≅ V_C_max, it is possible rewrite the eq. 62-15 as:

eq. 62-17

4. Rectifiers of Half Wave with Real Diode

So far we have studied the half-wave rectifier using an ideal diode, that is, there was no voltage drop on it. But in fact, we know that when a diode is in the conduction zone, there is a potential difference on it of 0.7 volts. Soon, to find the maximum voltage on the capacitor, that is, V_C_max , we must subtract 0.7 V from the maximum or peak voltage (V_p) the secondary of the transformer or other type of source used. So we can write

eq. 62-18

5. Reverse Voltage on Diode

All diodes come with a very important specification called reverse voltage. What does that mean? & nbsp; Note, in previous explanations, that when the sine enters the negative peak, the P-N junction of the diode is subjected to inverse or reverse polarization. So that there is no damage to the diode this should be able to withstand this reverse voltage. Therefore, when selecting a diode for a project be sure to check if it supports the reverse voltage in the circuit.

As an example, we can cite the 1N 4000 series, where the 1N 4001 diode supports a reverse voltage of 100 volts, whereas the diode 1N 4002 supports 200 volts of reverse voltage, going up to the 1N 4007 that supports 1 000 volts . Thus, there are several types of diodes with different values of reverse voltage. So, for a good project it is important to be attentive to the details.

For a half wave rectifier, the reverse voltage to which the diode is subjected, is equal to the sum of the maximum voltage supplied by the transformer secondary (V_p) with the maximum voltage on the capacitor (V_C_max). In this way, we can approximate the value of the reverse voltage (V_rev) over the diode according to the eq. 62-19.

eq. 62-19

6. Average and Peak Current in the Diode

From everything that has been studied so far, we can conclude that the larger the filtering capacitor, the smaller the ripple will be. With this, we will obtain a higher voltage and better filtration at the filter output. Large values for the filter capacitor, on the other hand, affect the peak current drained by the rectifier diodes during the capacitor charge. As we have seen, large values for the capacitor generate a smaller ripple and, consequently, shorter the conduction time of the rectifier diodes. This means that the shorter the charge time of the capacitor, the greater the flow of charging current through the diodes.

Look at Figure 62-07. We intentionally represent the output voltage of a rectifier circuit using a capacitor with low capacity. This results in a ripple appreciable in the output voltage. Therefore, the time T_c is also great. Bearing in mind that, in a capacitor, the relation is valid Δq = C ΔV, this tells us that the charge current of the capacitor is relatively small, as we know that I = Δq / Δt . So if Δt is large, then I is small. Note that T_c, in relation to Figure 62-05 , is exactly the angle φ_c, that is, the angle of conduction of the diode. Therefore, it is evident that the lower the capacitor value, the longer the conducting time of the diode is.

Look at Figure 62-08. Here we represent the output voltage of a rectifier circuit using a capacitor with high capacity. This results in a ripple too small in the output voltage. Therefore, the time T_c is also little. How is the relation valid Δq = C ΔV, so now the charge current of the capacitor is much higher than in the previous case, based on the fact that I = Δq / Δt . So, when Δt is small, then I must be large. We found that the higher the capacitor value, the diode conduction time is shorter. In the figures, the conduction time of the diode is shown in green.

As studied in item 3.3.1, the average current in the diode can be calculated by transforming the eq. 62-12 and eq. 62-02 in the relation below, eq. 62-20.

eq. 62-20

It should be noted that in the case of half-wave rectification, the frequency value is the nominal value of the operating frequency of the electrical system. In Brazil, the value is f = 60 Hz.

Assuming that during the charging time of the capacitor the current in the diode is constant and the average current drawn from the source is equal to the average current in the diode, we can write the equation that determines the peak current in the diode, or

eq. 62-21

Where the variables are:

I_p - Peak current in the diode.
I_DC - Average current in the diode.
T_c - Conduction time of the diode.
T - Period = 1/f.

Note that replacing the eq. 62-20 in eq. 62-21, and remembering that f = 1 / T , we can write the eq. 62-22, or

eq. 62-22

See that the product 2 C V_DC r_pp it is exactly the charge Q that must be supplied to the capacitor in a time T_c . And as by definition I = ΔQ / ΔT, then this equation is perfectly in agreement with what has been studied so far.

However, to determine the value of I_p we need to calculate the value of T_c. As we have a sine wave, it is possible to determine the angle at which the diode begins to conduct, which we will call θ₁ . So, considering the possibility of approximating the ripple waveform by a triangular wave, we can write

v = V_C_max sen θ₁ = V_C_max - ΔV_r

From this equation we easily determine the value of θ₁, or

eq. 62-23

Where the variables are:

θ₁ - Angle at which the diode begins to conduct.
ΔV_r - Peak to peak ripple voltage.
V_C_max - Maximum voltage on the capacitor.

It is also possible to rewrite the equation eq. 62-23 , remembering the relation shown in eq. 62-06 and, adapting, we obtain the eq. 62-24.

eq. 62-24

When the capacitor, in the charging process, reaches the maximum voltage, that is, V_C_max, the current in the diode drops to zero and it stops conducting. The time the diode remains conducting depends on the time constant, or τ = R_L C. In this way, we can determine the angle at which the diode stops conducting, represented here by θ₂.

θ₂ = π - tg^-1 ( 2 π f C R_L )

By the equation eq. 62-03 , we have to f R_L C = V_C_max / ΔV_r. So, we can rewrite the equation above as:

eq. 62-25

Here, it is also possible to rewrite the equation eq. 62-25, remembering the relation shown in eq. 62-06 and, adapting, we obtain the eq. 62-26.

eq. 62-26

Here we will make a small parenthesis to make it very clear what θ₁ and θ₂. The angle θ₁ represents the angle, from zero, until the moment when the diode starts to conduct. And the angle θ₂ represents the angle, since zero, until the moment when the diode ceases to conduct, that is, when the capacitor reaches its maximum voltage. It is important to note that the difference between θ₂ and θ₁ is exactly the angle of conduction of the diode , φ_c, that is

eq. 62-27

See Figure 62-09 for an indication of the angles. The green part indicates the conducting time of the diode, representing both T_c (in seconds) and φ_c (in degrees). The part in blue corresponds to the times when the diode is cut. The angle θ₁ goes from zero to the point where the diode starts to conduct. And the angle θ₂ corresponds to the angle from zero to the point where the diode stops driving.

Therefore, the eq. 62-27 fully agrees with the Figura 62-09.

Now, having the value of T_c or φ_c , we can determine the peak current in the diode transforming the eq. 62-21 on the following equality:

eq. 62-28

Therefore, it is possible to calculate I_p by both eq. 62-28 as by eq. 62-22. Note that we use the angle of 360° in the relationship, as we are analyzing a rectifier half wave. For a full wave rectifier this value is different. In the next chapter we will analyze this relationship.

Attention

"The peak voltage, I_p, calculated in the problems can be tens of times greater than the nominal current of the diode. This is not a problem, as every diode has a characteristic of withstanding repetitive current surges (which happen in a very short time) much greater than the rated current. For example, the 1N 40xx series has a rated current of 1 A, however it supports a repetitive peak current of the order of 30 A, if the peak time does not exceed 8,3 ms."

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