Diodes Applications, Full 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 - Full Wave Rectifiers click here!

2.1 - Bridge Rectifier click here!

2.2 - Rectifier with Central Tap click here!

3 - Effect of the Capacitor in the Circuit click here!

3.1 - Ripple Voltage - ΔV_r click here!

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

3.3 - Filter Capacitor Value click here!

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

3.2.2 - For Ripple with Values in RMS click here!

4 - Real Diode Rectifiers click here!

5 - Reverse Voltage on Diode click here!

6 - Average and Peak Current on Diode click here!

We saw in the previous chapter the operation of the half-wave rectifier. It became evident in this type of rectifier that the capacitor supplies current to the load for a very long time long. In fact, almost the entire cycle. This requires the capacitor to have a capacitance very high so that we have a ripple voltage of reasonable value.

2. Full Wave Rectifiers

To work around the problem discussed above, we have the so called full-wave rectifiers. Basically we have two types of full wave rectifiers: the rectifier in bridge and the rectifier with central tap using transformer. Let's look at how they work.

2.1 Bridge Rectifier

As its name suggests, this circuit is characterized by using four diodes in a connection type named bridge. This circuit can be seen in the Figure 63-01.

Note that the alternating voltage of the source is connected to the ends of the diode bridge. There are two distinct cases to look at.

The first when the positive semicycle of the sine is at the top of the bridge. We can see the representation of this situation in the Figure 63-02.

In the figure to the side we can follow the path of the electric current in the circuit indicated by the green arrows. Note that we have two conducting diodes and two diodes in the cutoff zone. The diodes D₁ and D₃ are conducting and the diodes D₂ and D₄ are in the cutoff zone.

In this way, the current coming out of the voltage source and reaching the point a must pass through D₁, because D₂ is cut off. Upon reaching the point c, the current can not return to the source by D₄, because it is cut off. Therefore, the current must pass through the load R_L arriving at the point b. As D₂ is cut off, the current must pass through D₃ to return to the voltage source and close the circuit. The green arrows indicate the path of the electric current in the circuit.

Now let's look at the second case, when the sine is in the negative half-cycle on the top of the bridge.

In the Figure 63-03 we can follow how is the circuit for this case. Notice that now are the other two diodes that are conducting, that is, D₂ and D₄. Meanwhile, D₁ and D₃ are in the cut zone. Thus, the current coming from the bottom of the voltage source and reaching the point d has to pass through D₄ until it reaches point c, because D₃ is cut off.

When arriving at the point c, the current can not return to the voltage source by D₁. Therefore, it has to go through R_L. And when it reaches point b the only alternative to return to the voltage source is to pass through D₂. With that, we have a closed circuit. The important thing is to realize that in the negative half cycle there was no reversal in the direction of the electric current when crossing the load R_L. This means that in any of the half-cycles, we always have a positive voltage on the load R_L. The graph that appears in the Figure 63-04 shows the effect of the bridge rectifier on the load.

Therefore, it does not matter the polarity of the voltage source, because on the load we will always have a undulated voltage with a positive voltage. It should be noted that if you want to have a negative voltage on the load, just reverse the polarity of the four diodes. That simple.

2.2 Rectifier with Central Tap

Another configuration for a full-wave rectifier is shown in the Figure 63-05. In this configuration we use a transformer with central tap. Transformer with central tap is characterized by having two identical secondary windings completely independent of one another. Subsequently, we join the end tip of the first winding with the initial tip of the second winding. This connection is the central tap of the transformer. The other two tips are where we have the voltage going to the diode anode. The most important thing is that with this artifice we get two senoids with a phase difference of 180° between each other. For simplicity, in the figure below, we represent the transformer secondary by two voltage sources. Note the polarity of the same.

While in the anode of D₁ we have the positive half-cycle of the voltage source V₁, in the anode of D₂ we have the negative half-cycle of the voltage source V₂.

In the Figure 63-06 we show the circuit in this configuration. In this way, the current leaves the positive pole of V₁, goes through the diode D₁ reaching the point a and how D₂ is cut off, current flows through the load R_L and closes the circuit by the source V₁ when it reaches the point g.

The green arrows indicate the path of the electric current in the circuit. The above description happens in the first 180° of the sine. Thereafter, in the range of 180° to 360°, there is a reversal in the polarity of the sources. We show this situation in the Figure 63-07.

Now is D₁ that is cut off. In this way, the current leaves the positive pole of V₂, goes through the diode D₂ reaching the point a and how D₁ is cut, current flows through the load R_Land closes the circuit by the source V₂ when it reaches the point g. The green arrow indicates the path of the electric current in the circuit.

As a final result, we get the same waveform over the load when we use the bridge circuit. Below, Figure 63-08, we reproduce this waveform again.

The analyzed cases provide a positive voltage at the output of the circuit. In case an negative voltage output is needed, is enough inverter the position of all the diodes that make up the circuit. If the capacitor used is of the polarized type, as is the case of the electrolytic capacitor, it must also be inverted.

3. Effect of the Capacitor in the Circuit

As we have already studied in the chapter on the rectifier of half-wave, in order to the rectifier output voltage as constant as possible we must add a capacitor in parallel with the load. The Figure 63-09 shows this situation. Notice that we represent the four diodes in the cut off situation. This happens after the output voltage of the rectifier reaches the value maximum. Thus, when the input voltage exceeds 90°, that is, its value is decreasing, the capacitor voltage is greater than the output voltage of the rectifier by placing the four diodes in cut off. It is in this situation that the capacitor fulfills its role of supplying current to the load when the diodes are in the cut off.

From the moment the capacitor starts to supply current to the load, the voltage between its terminals decreases gradually. When the output voltage of the rectifier becomes greater than that of the capacitor, the diodes D₁ and D₃ are in conduction and restarts a new recharge cycle of the capacitor, raising its voltage up to the maximum output value of the rectifier. In the Figure 63-10 we can see the graphical representation of this situation.

Observe in the graph the angle of conduction φ of the diodes, represented by green. At this time, the voltage variation in the capacitor is ΔV_r. The period between peaks is T / 2. The part that is in yellow represents the situation of the four diodes in cut as described above. The voltage variation on the capacitor at this moment is also ΔV_r, but the voltage decay is determined by the time constant RC.

It is important to note that in the case of full wave rectification, the rectified wave has a frequency that is twice that of the half wave rectification.

So, with the presence of the capacitor, we will have an average voltage DC at the output, which we will call of V_DC, and which can be approximated by eq. 63-01, considering that we can represent the ripple voltage by a triangular wave.

eq. 63-01

From this average voltage DC we can define an average current in the load that can be calculated by the equation eq. 63-02 and let's name it I_DC.

eq. 63-02

3.1 Ripple Voltage - ΔV_r

Usually in problems presented in books or tests, the ripple voltage can be specified as a numerical value or, as is more common, a percentage referring to the voltage of circuit output.

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 that we must implement is that the voltage in the capacitor, between the two voltage peaks, must be equal to V_C = V_C_max - ΔV. For this, we must have a time constant (τ = C R_L) of adequate value. As the value of R_L is known, then we must calculate the value of the capacitor C that allows 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 in 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 for the value of C, let's make two simplifications, based on the fact that in this case we have τ = R_L 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. But in the case of full wave rectification, the wave period is half the period when half wave rectification occurs. This means that we have the double the frequency on the rectified signal. Substituting these values into eq. 62-02b, combining with eq. 62-02a, and making τ = R_L C , let's find:

ΔV_r = V_C_max - V_C_max (1 - (T/2 R_L C)) = (V_C_max / R_L ) (T/2 C)

Remembering that f = 1 / T, the above equation can be written as shown in eq. 63-03.

eq. 63-03

So, knowing the values of f, R_L and C, we can calculate the value of the voltage undulation. And so, we can define the index or percentage rate of peak-to-peak ripple represented by the letter r_pp, of according to the equation eq. 63-04.

eq. 63-04

Note that using the eq. 63-04 and eq. 63-03 we can write to eq. 63-05.

eq. 63-05

Note that based on eq. 63-05, we can rewrite eq. 63-03 as follows:

eq. 63-06

Working algebraically at eq. 63-01 and eq. 63-06, we can write to eq. 63-07:

eq. 63-07

Attention

It should be noted that so far 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, for example, for values below 10% in the ripple index, assumes V_DC ≅ V_C_max and therefore we can write I_DC = V_C_max / R_L. With that in mind, we can approximate the eq. 63-03 by the following expression:

eq. 63-08

3.2 RMS Ripple Voltage[V_r (rms)]

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

eq. 63-09

Now, we can define the RMS ripple index represented by the letter r, according to the equation eq. 63-10.

eq. 63-10

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. 63-11.

eq. 63-11

3.3 Filter Capacitor Value

As 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 in the problem the value of ripple is stated in RMS or peak-to-peak values.

3.3.1 For Ripple in Peak to Peak Values

In case the ripple is given in peak to peak values, that is, we are working with ΔV_r, then algebraically manipulating the eq. 63-03, we can find the capacitor value for the specified ripple, or

eq. 63-12

It is also possible to calculate the value of C knowing the values of f, R_L and r_pp using the eq. 63-13.

eq. 63-13

Note that comparing the above equations with those of the half-wave rectifier, the difference appears in the denominator, by the factor 2. This means that for the same conditions imposed on the rectifier, in the full wave rectifier we need a capacitor with half the value as in the case of the half wave rectifier.

Attention

Note that according to the observation made at the end of item 3.1, where we assume it was valid approximation V_DC ≅ V_C_max, it is possible to rewrite the eq. 63-12 as:

eq. 63-14

For the case where r_pp < 10% a eq. 63-14 can be used, as the error made is less than 3% on average. As electrical components, in general, can have a tolerance greater 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 63-5, or click here.

3.3.2 For Ripple with Values in RMS

For the case that ripple is given in value RMS, we must consider the correction factor given by eq. 63-09. Using this correction factor in eq. 63-03, we found the value of C, or:

eq. 63-15

Replacing the eq. 63-11 in eq. 63-15, we get:

eq. 63-16

Note that from eq. 63-16, after an algebraic arrangement, we can calculate the value of R_L, if the problem is given the value of f, C and r. Thus, we obtain:

eq. 63-16a

Attention

Just as we did in the previous item, where we assume V_DC ≅ V_C_max, it is possible to rewrite the eq. 63-15 like:

eq. 63-17

In this case, the considerations of the previous item regarding the error caused by this approximation also apply.

4. Real Diode Rectifiers

So far we have studied the full-wave rectifier assuming an ideal diode, that is, there was no voltage drop on it. But actually, we know that when a diode meets in the conduction zone there is a potential difference over it of 0.7 volts. Soon, to find the maximum voltage across the capacitor, ie V_C_max, we must know if the rectifier is of type bridge or with central tap.

If the rectifier is center tap, we must subtract 0.7V from the maximum or peak voltage (V_p) of the secondary of the transformer, because in this case we only have one diode conducting in each half cycle of the sinusoid.

However, in the case of bridge rectifier type, we must subtract 1.4 V from the maximum or peak voltage (V_p) of the secondary of the transformer, because in this case we always have two diodes conducting in each semi-cycle of the sinusoid.

So, when solving the problems, we must pay attention to the type of rectifier in use.

eq. 63-18

eq. 63-19

5. Reverse Voltage on Diode

All diodes come with a very important specification called reverse voltage. What does that mean? Note, in the previous explanations, that when the sine enters the negative peak, the P-N junction of the diode is subjected to a reverse bias. In order not to damage the diode, it must be able to withstand this reverse voltage. Therefore, when selecting a diode for a project be sure to verify that it supports the reverse voltage in the circuit. As an example, we can mention the series 1N 4000, where the diode 1N 4001 supports a reverse voltage of 100 volts, the 1N 4002 supports 200 volts of reverse voltage, going to the 1N 4007 that supports 1 000 volts. Thus, there are several types of diodes with different reverse voltage values. So, for a good project it is important to pay attention to details.

For a full wave rectifier, the reverse voltage is approximately equal to the sum of the maximum voltage supplied by the transformer secondary (V_p ) with the maximum voltage across the capacitor (V_C_max). In this way, we can approximate the value of the reverse voltage (V_rev) over the diode according to eq. 63-20.

eq. 63-20

6. Average and Peak Current on Diode

So far, we have verified that the full wave rectifier can produce the same ripple with a capacitor that has half the capacitance when compared to the half wave rectifier. On the other hand, the considerations regarding the peak current drawn by the rectifier diodes during capacitor charging remain valid. Larger values for the capacitor generate a smaller ripple and, consequently, the driving time of the rectifier diodes is shorter. This means that the shorter the capacitor charging time, the greater the charging current flow through the diodes.

As studied in item 3.3.1, the average current in the diode can be calculated by transforming the eq. 63-12 and eq. 63-02 in the list presented below, eq. 63-21.

eq. 63-21

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. 63-22

Where the variables are:

I_p - Peak current on the diode.
I_DC - Average current in the diode.
T_c - Diode conduction time.
T - Period = 1/f.

Note that replacing the eq. 63-21 in eq. 63-22, and remembering that f = 1/T, we can write the eq. 63-23, or

eq. 63-23

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

However, to determine the value of I_p we need to calculate the value of T_c. Since we have a sine wave, it is possible to determine the angle at which the diode starts 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. 63-24

Where the variables are:

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

You can also rewrite the eq 63-24, remembering the relation shown in the eq. 63-06 and, adapting, we get the eq. 63-25.

eq. 63-25

When the capacitor, in the charging process, reaches its 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. 63-03, we have to 2 f C R_L = V_C_max / ΔV_r. So, we can rewrite the equation above as:

eq. 63-26

Here, you can also rewrite the eq equation. 63-26, remembering the relationship shown in the eq. 63-06 and, adapting, we get the eq. 63-27.

eq. 63-27

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

eq. 63-28

See Figure 63-9 for the indication of the angles. The part in green indicates the conduction time of the diode, representing both T_c (in seconds) and φ_c (in degrees). The yellow part corresponds to the times 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 ceases to conduct.

Therefore, the eq. 63-28 fully agrees with the Figure 63-09.

Now, in possession of the value of T_c or φ_c, we can determine the peak current in the diode transforming to eq. 63-22 in the following equality:

eq. 63-29

Therefore, it is possible to calculate I_p either by eq. 63-29 or by eq. 63-22. Note that we use the angle of 180° in the relation, as we are analyzing a rectifier of full wave that has a period of π = 180°.

Attention

"The peak voltage, I_p, calculated in the problems, can be tens of times greater than the rated current of the diode. This is not a problem, as every diode has the characteristic of withstanding repetitive surges (which happen in a very short time interval) of current much higher 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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