What's the Diodes Zener Applications

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

1 - Introduction click here!

2 - How Zener Diode Works click here!

3 - Features of the Zener Diode click here!

4 - Using the Zener Diode click here!

5 - Voltage Regulator with Zener Diode click here!

5.1 - Case 1 click here!

5.2 - Case 2 click here!

5.2 - Case 3 click here!

5.2 - Case 4 click here!

In the previous chapters we have studied a very useful application of the diode when it is used with direct polarization. However, there are diodes that work, under special conditions, with inverse polarization. These diodes are called ZENER diodes.

There are numerous applications for the Zener diode. One of the main ones is its use as a voltage regulator. In the figure below we can see the symbol used to represent the zener diode. Note the polarity used for it to work as a zener. If there is inversion in the polarity, it will behave like a common diode. Therefore, always be aware of polarity when working with zener diodes.

In the Figure 64-01 we see the schematic representation of a zener, as well as the polarization that we must employ so that it functions as a zener.

Note that polarity is opposite to that used for a common diode. For this polarization we will have on the zener the voltage V_z for which it was designed. In the figure we show the direction of electric current I_z on it.

2. How Zener Diode Works

As we said above, the zener diode should work with a reverse polarization. That is, at the cathode a more positive voltage is applied than at the anode. In the Figure 64-02, we represent the characteristic curve of a zener diode. Notice that by increasing the voltage on the zener, in the negative direction, the electric current is initially null. This is true to the point where the voltage approaches the ruptura voltage. At this point, there is the so-called "knee", where the electric current in the zener grows steeply keeping a stable voltage on the zener itself that there is large variations of current in the zener (evidently within certain limits).

Note, on the graph, that as soon as the voltage on the diode reaches the voltage V_z, the straight line assumes a "almost vertical" characteristic, allowing the voltage to remain stable over the diode even if the electric current changes. Notice that there are two important points in the graphic. The first is I_{z min}, which is the minimum current that the zener needs in order to maintain the voltage in V_z.

Below this point there will be no guarantee of stable tension on the zener. The second is I_{z max}, which is the maximum current that the zener diode supports without failing. The value of this current depends on the voltage of the zener and the power that it can dissipate.

Due to the sharp slope of the line appearing in Figure 64-02, in the zener region, we can say that the zener has a small resistance or, also called impedance, and that is a characteristic of each zener. Normally, this value is specified by the manufacturer.

We can not ignore what happens when we call a zener diode using direct polarization. In this case, the zener diode behaves like a common diode, that is, it conducts anode current to the cathode and there will be a voltage drop on it of 0.7 volts. In the Figure 64-03 we show this situation.

3. Features of the Zener Diode

The zener diode is characterized by two basic requirements: the zener voltage and the power that it needs to support in order to perform its function satisfactorily. Because of the maximum power and the zener voltage that it can support, we easily calculate the maximum current with which it can work. Using the equation that defines power, we have:

eq. 64-01

In case of I_{z min} (minimum current in the zener) it is usual to take its value as 10% of the maximum zener current, I_{z max}, according to eq. 64-02.

eq. 64-02

There are several commercial of zener models for various different values of voltage and power. To differentiate them, there are codes that determine the voltage and power of the device. For example, in the Table 64-01 we present some of the most common in the market. In the left column we have the voltage that the zener can regulate and the next two columns are the zener codes. The middle column is for zener which can dissipate at most 0.5 W of power and in the right column those that support up to 1 W.

Table 64-01

Voltage Zener	P = 0.5 W	P = 1.0 W
4.70 volts	1N750	1N4732
6.80 volts	1N754	1N4736
10.0 volts	1N758	1N4740
12.0 volts	1N759	1N4742
15.0 volts	1N965	1N4744
22.0 volts	1N969	1N4748
33.0 volts	1N973	1N4752

4. Using the Zener Diode

For all of the features reported above, it is more than enough reason for the zener diode to be employed as voltage regulator.

When we studied the rectifier circuits, we pointed out that one of the problems was the ripple on the output voltage. To reduce it, a capacitor was added in parallel with the load. We saw that the capacitor softened the problem, but it did not eliminate it. So we have the zener diode as the savior of this situation.

When we use a zener diode as a voltage regulator, we usually use the voltage output of the rectifier, and use a resistor between the capacitor and zener as a limiter of current. In the Figure 64-04 we can appreciate a typical voltage regulator circuit using a zener diode.

Important: For this circuit to work properly, is very important that the relation below to be satisfied.

eq. 64-03

Note that in this equation, V_DC - (ΔV/2) represents the minor voltage at the output of the rectifier circuit. As there will be a voltage drop in the resistor R_s, we have to leave a minimum clearance of 2.50 volts (this may be more), otherwise the zener diode will not be able to guarantee stable output voltage.

Some authors, and even teachers, accept values of the order of 0.5 to 1 volt. In theory, it may be. But if it is a design to be put into practice, never leave less than 2.5 volts or there will be great possibilities your design will not work.

Attention

Another important situation in projects with zener diodes is the possibility of the load suffering some kind of problem, such as a rupture thereof and become an open circuit. In this case, the current that had previously circulated through the load must now be absorbed by the zener diode. If zener does not support this increase in current, it will be damaged as it will dissipate a power above its capacity.

To avoid this possibility, we must verify if the equation below is satisfied by the circuit, not forgetting that I_S = I_Z + I_L.

eq. 64-04

Another possibility is that the load is not an open circuit, but a short circuit. In this case the zener diode will not be damaged. However, the resistor R_s shall withstand the excess current flowing through it. Then, in the design, the equation below must be satisfied.

eq. 64-05

When the input voltage is constant, then V_i_max = V_i. And when the input voltage is not constant then V_i_max = V_i + (Δ V/2)

5. Voltage Regulator with Zener Diode

For the circuit design of a voltage regulator using a zener diode we need some data, such as the output voltage, the maximum electric current that will circulate through the load, and also the output voltage of the rectifier plus the ripple, ΔV. In the problems and evidence of this area, we find some variations of data provided, but there should be a minimum of data that is sufficient to elaborate the project.

Another relevant fact is that the value of R_s can assume minimum and maximum values, and will depend on the minimum and maximum voltage at the rectifier output. So we can establish two basic principles for the project.

1st Principle - In order for the zener diode to maintain the stable voltage at the output, it must operate with a current greater than the calculated minimum current, I_Zmin. In other words: I_Z > I_Zmin. As a consequence, this limits the minimum value of the input voltage in the circuit, given by V_i(min) = V_DC - (ΔV/2), as well as the maximum value of the series resistance of the circuit, represented by R_Smax. These limitations allow us to write the equation below.

eq. 64-06

2nd Principle - If the current passing through the zener diode is greater than the maximum calculated current, I_Zmax, certainly the zener diode will be damaged. As a consequence, this limits the maximum value of the input voltage in the circuit, given by V_i(max) = V_DC + (ΔV/2), as well as the minimum value of the series resistor of the circuit, represented by R_Smin. From this, we conclude that we should have I_Z < I_Zmax. These limitations allow us to write the equation below.

eq. 64-07

Based on these two principles we can analyze four cases that may occur.

5.1 Case 1

Constant Load (R_L = cte) and Constant Input Voltage

In this case, the constant load means that the current in the load depends only on the voltage V_Z. On the other hand, since the input voltage is also constant (it may be a battery), then zener is being used as a reductor of the input voltage. This circuit can be used when we need a voltage less than that supplied by the rectifier. In the Figure 64-05 we can appreciate a typical circuit used for this purpose, where V_i represents the source with constant voltage.

Based on the circuit, we easily conclude that:

I_S = I_Z + I_L

Since we assume that the electric current in the load is constant, then if we work with the minimum current for the zener diode, we will be establishing the maximum value that the resistance R_S can assume. On the other hand, if we work with the maximum current for the zener diode, we will be establishing the minimum value that the resistor R_S can assume. So, depending on the choice, we can work with the two equations below.

eq. 64-08

eq. 64-09

Realize that to find R_Smax we use I_Zmin and for to calculate R_Smin we use I_Zmax. In general, in the tests, these variables are the most requested in the problem solving. And to find the practical value of R_S, we can do the arithmetic mean of the calculated values. If you are prompted by the problem, we can adjust the value of the R_S found for the nearest commercial value.

Example 1

As an example, we will use the above circuit, assuming that V_i = 14 volts, V_Z = 10 volts and a load current equal to 100 mA. Let's use the 1N4740 zener that can dissipate up to 1 watt of power.

Solution

Since we know the voltage and power of the zener, we can calculate I_Zmax and I_Zmin, or:

I_{Z max} = P_Z / V_Z = 1 / 10 = 0.1 A = 100 mA

Now we can calculate I_{Z min}, or:

I_{Z min} = 10% x I_{Z max} = 10% x 100 mA = 10 mA

With these two values we will calculate R_{S min} and R_Smax using the equations that we developed at the beginning of this case. Like this:

R_{S min} = (14 - 10) / (0.1 + 0.1) = 20 Ω

And in turn:

R_{S max} = (14 - 10) / (0.01 + 0.1) = 36.67 Ω

In this way, we can calculate the arithmetic mean of R_Smin and R_Smax to find a convenient value for our project. Soon:

R_S = (R_{S min} + R_{S max}) / 2 = 56.67 /2 = 28.33 Ω

Therefore, a commercial resistance value closer to the value found is:

R_S = 27 Ω

Of course, we could have chosen a value as 22 ohms or 33 ohms, because they would be within acceptable values for R_S, since R_Smin < R_S < R_Smax .

We must determine the power that the resistor R_S will dissipate in the circuit. The current passing through R_S is given by:

I_S = (V_i - V_Z) /R_S = 4 / 27 = 148.15 mA

Then the power dissipated is:

P_S = (V_i - V_Z) I_S = 4 x 0.14815 = 0.5926 W

Therefore, we can use a resistance of 1 W, or rather one of 2 watts. So we got a bigger slack.

To conclude, we can calculate the power dissipated by the zener.

P_Z = V_Z. (I_S - I_L) = 10 x 0.04815 = 0.4815 W

Some considerations

With the chosen value of R_S we get a security factor greater than (1 / 0.4815)> 2 for the zener diode. If R_S = 33 Ω then P_Z = 0.21 W. The zener works with low power dissipation. And if R_S = 22 Ω then P_Z = 0.82 W. In this case, the zener will work very close to its power limit.

Note how the chosen value of R_S influences the power dissipated by the zener.

Another situation: let us suppose that the load suffers a malfunction, such as an interruption in the circulation of electric current through it. In this case, the zener will have to absorb the excess current. In this example, if this happens, the zener will be damaged, because the current that will circulate through it is above the maximum value that it supports. This happens for any value of R_S calculated. Check out !!!

These are just a few considerations that should be taken into account in the project.

5.2 Case 2

Variable Load and Constant Input Voltage

Now let's look at the case where we hold the voltage V of the source and change the load, that is, I_L can range from a minimum value to a maximum value, which will certainly be explained in the problem statement. This case is a particularity of the previous. The zener continues to be used as a reductor of voltage, but the current in the load may vary. An example would be to feed a small amplifier that amplifies the signal coming from a microphone. When no one speaks to the microphone, the current consumption of the amplifier is minimal. However, when someone speaks to the microphone, the current consumption increases considerably.

Let's think with logic: if zener is able to maintain a stable (constant) voltage and voltage V_i is also constant, obviously the electric current through the resistor R_S will also be. This implies that if the current in the load changes, then the current in the zener should also vary. And in the same amount as the change in load. With this in mind, for the circuit to work properly, we must fulfill two requirements:

a) When load is consuming the maximum current , we have to ensure that the current in zener is at least equal to the minimum current of the zener , that is, we have to satisfy I_Z ≥ I_Zmin, so that the zener can maintain a stable voltage at the output.

b) When the load is consuming the minimum current, then the electric current in zener, under no hypothesis, may be greater than the maximum current of zener, that is, I_Z ≤ I_Zmax. Otherwise, the zener will be damaged as it will dissipate a power above its specifications.

Based on this information we can write the equations that allow to find the maximum and minimum values of R_S.

eq. 64-10

eq. 64-11

Realize that to find R_{S max} we use I_Zmin together with I_Lmax and to calculate R_Smin we use I_Zmax together with I_Lmin. As in case 1, to find the practical value of R_S, we can do the arithmetic mean of the calculated values. If you are prompted by the problem, we can adjust the R_S found for the nearest commercial value.

In the Figure 64-06 we can see the circuit that represents this situation. It should be noted that, when the value of resistor R_S is chosen, for the circuit to work satisfactorily, it is worth at any moment the relation:

I_S = I_Z + I_L

Example 2

As an example, we will use the circuit show in the Figure 64-06, assuming that the voltage input V_i = 20 volts, V_Z = 12 volts and a load current ranging from 20 to 80 mA. Let's use the zener 1N4742 which can dissipate up to 1 watt power.

Solution

Since we know the voltage and power of the zener, we can calculate I_Zmax and I_Zmin, or:

I_{Z max} = P_Z / V_Z = 1 / 12 = 83.33 mA

Now we can calculate I_Zmin, or:

I_{Z min} = 10% I_{Z max} = 10%. 83.33 mA = 8.33 mA

With these two values we will calculate R_Smin and R_Smax using the two equations we developed at the beginning of this case. Like this:

R_{S min} = (20 - 12) / (0.0833 + 0.02) = 77.44 Ω

And in turn:

R_Smax = (20 - 12) / (0.00833 + 0.08) = 90.56 Ω

As in the previous case, we will calculate the arithmetic mean of the values found:

R_Smax = (77.44 + 90.56) / 2 = 84 Ω

The nearest commercial value is:

R_S = 82 Ω

Let's calculate the current that flows through R_S.

I_S = (V_i - V_Z) /R_S = 8 / 82 = 97.56 mA

Then the power dissipated in R_S is:

P_S = (V_i - V_Z) .I_S = 8 x 0.09756 = 0.78 W

So we can use a resistor of 1 W or 2 W.

Now let's calculate the current in the zener diode as a function of the current variation in the load. Note that when the load consumes 80 mA, in the zener we will have a current of:

I_Z = I_S - I_L = 97.56 - 80 = 17.56 mA

And when the load consumes 20 mA, the current in the zener diode will be:

I_Z = I_S - I_L = 97.56 - 20 = 77.56 mA

That is, the zener diode is working to its specifications. So we can write that the current of the zener vary between the values 17.56 ≤ I_Z ≤ 77.56 mA.

5.3 Case 3

Constant load (R_L = cte) and Variable Input Voltage

In this case, where we keep the load fixed, that is, I_L is constant, there will be variations in the zener current, I_Z. This is because the input voltage is no longer fixed, so we will have a ripple . Like this, when there is variation in tension, there will be variation in the current that circulates through R_S. This variation will be reflected in the zener current.

Therefore, for the circuit to work properly we must fulfill two requirements:

a) When the input voltage is maximum, the current in the zener will also be maximum and this will determine the minimum value of R_S.

b) And when the input voltage is minimum, the current in the zener will also be minimum and this will determine the maximum value of R_S.

Based on this information we can write the equations that allow to find the maximum and minimum values of R_S.

eq. 64-12

eq. 64-13

Realize that to find R_Smin we use I_Zmax together with V_i(max) and to calculate R_Smax we use I_Zmin together with V_i(min). As in other cases, in order to find the practical value of R_S, we can do the arithmetic mean of the calculated values. If you are prompted by the problem, we can adjust the R_S found for the nearest commercial value.

See in the Figure 64-07 a typical circuit for this case.

Here!

5.4 Case 4

Variable Load and Variable Input Voltage

The latter case is characterized by all the currents and voltages involved being variables, except for the voltage on the zener, which MUST be constant. Therefore, for the circuit to function properly, we must fulfill two requirements:

a) When the input voltage is maximum, the current in the zener will also be maximum and the current in the load will be minimum . This will determine the minimum value of R_S.

b) And when the input voltage is minimum, the current in the zener will also be minimum and the current in the load will be maximum. This will determine the maximum value of R_S.

Based on this information we can write the equations that allow to find the maximum and minimum values of R_S.

eq. 64-14

eq. 64-15

Realize that to find R_Smin we use I_Zmax together with V_i(max) and I_Lmin. To calculate R_Smax we use I_Zmin together with V_i(min) and I_Lmax. As in other cases, in order to find the practical value of R_S, we can make the arithmetic mean of the calculated values. If you are prompted by the problem, we can adjust the R_S found for the nearest commercial value.

See in the Figure 64-08 a typical circuit for this case.

Here!

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