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In this article, we will learn about the **Zener diode** and we will see how the **Zener diode **can be operated as a voltage regulator.

## What is Zener Diode

Now in the previous Articles, we have seen how this PN Junction diode can be operated in the forward as well as in the reverse direction. And we have also seen that in case of a reverse bias condition when the applied voltage is less than this breakdown voltage, then there is hardly any flow of current through the diode.

But as soon as the voltage across the diode is more than this breakdown voltage then suddenly a lot of current starts flowing in the reverse direction. So this region of operation of the diode is known as the breakdown region of operation. And for a normal diode, this region of operation should be avoided. But there are some diodes that are meant to be used in this breakdown region. And **these diodes are known as the Zener diode**.

So here is the symbol of the Zener diode and if you see the VI-characteristic of this Zener diode then it looks quite similar to the normal diode. So as you can see, in the forward bias condition, it just works similarly to the normal diode. And in the reverse direction when the applied voltage is more than the threshold voltage then suddenly the diode starts operating in the breakdown region.

so the voltage after which the Zener diode operates in this breakdown region is known as the Zener voltage and the corresponding current which is flowing through the Zener diode is known as the Zener current.

now unlike the normal diodes, this Zener breakdown voltage can be as low as two volts or as high as two hundred volts. But if you see the normal diodes, typically the breakdown voltage used to be more than20 volts. So unlike the normal diodes, these Zener diodes are heavily doped and by changing the doping concentration, we can change the Zener voltage of this Zener diode.

So for the Zener diodes where the breakdown voltage is less than4 V, the Zener effect is the predominant effect. On the other end for the Zener diodes whose breakdown voltage is more than six-volt, the Avalanche effect is the predominant effect. And we have already discussed these two effects in the previous Article.

So for more information, you can check that Article. But irrespective of the breakdown mechanism, all the diodes which are meant to be used in this breakdown region are known as the Zener diodes.

Now in this Zener diode, in the reverse bias condition when the applied voltage is more than the Zener voltage then this Zener diode will operate in the breakdown region. And if there is a current limiting resistor in series with the Zener diode then suddenly an excessive amount of current starts flowing through this Zener diode.

And in a way, because of the excessive power dissipation, this Zener diode may get damaged. So to limit this current, it is required to connect a current limiting resistor in series with this Zener diode.

And if you see the datasheet of any Zener diode then you will find that the maximum Zener current is defined for every Zener diode. it means that the maximum allowable current through the Zener diode should be less than this current. And if you see the datasheet, one more current is also defined for the Zener diode, which is known as the minimum Zener current.

so basically this current defines the minimum current which should be required to use this Zener diode in this breakdown region. so whenever the Zener diode is used in this breakdown region then the operating current of this Zener diode should be between these two limits. So for any Zener diode when the applied voltage Vs is more than this Zener voltage and the operating current is in-between this Izk and the Izm then we can say that the Zener diode is operating in this breakdown region.

So now if you see the voltage characteristic of the Zener diode then you can see that even if we go beyond this Zener voltage then also there is hardly any change in the voltage across this Zener diode, or in other words, we can say that the voltage across the Zener diode remains constant.

so in a way, this Zener diode can be operated as a voltage regulator. So in the first approximation, we will assume that whenever the voltage across the Zener diode is more than the Zener voltage in that case the Zener diode can be replaced by a constant voltage source of VZ.

But in reality, if you see this curve is not a straight line. it means that this Zener diode should have some series resistance and this series resistance is known as the Zener resistance. so because of this Zener resistance, if the current which is flowing through the Zener diode will increase then the voltage drop across the Zener resistor will increase. And due to that, the overvoltage which appears across the Zener diode will increase. But here for our analysis, we will assume that the Zener resistance of the Zener diode is zero.

Apart from that, we will also assume that the Zener current is equal to zero. That means, in the reverse bias condition the current which is required to turn on the Zener diode is also zero so under these assumptions, now let us see how those inner diodes can be operated as a voltage regulator.

## Zener Diode as a Voltage Regulator

so under this approximation, if the supply voltage Vs is more than Vz, in that case, the Zener diode will operate in the breakdown region and it will provide a constant voltage Vz across the two terminals.

So in a way, we can use the Zener diode as a voltage regulator and this is particularly useful for designing the power supplies. So let’s say we are designing the power supply using the full rectifier circuit, and for the filtration here we are using the capacitor filter. Now if you recall, even if you use this capacitor filter then also there is some finite ripple in the output waveform.

And this ripple can be further reduced if we use this Zener diode at the output of this capacitor filter. So in this way, this Zener diode acts as a voltage regulator and it provides a constant voltage across the two terminals. Now here even if we connect the load resistance across the Zener diode then also it will provide a constant voltage across this load resistance.

but now whenever we connect this load resistance, then first of all we need to find the Thevenin’s equivalent voltage across these two terminals. Because under the no-load condition whenever we apply the supply voltage then the entire voltage is appearing across this Zener diode.

But when we connect this load resistance RL then first of all we need to ensure that the voltage across this two-terminal is more than the Zener voltage. so to find this voltage across this Zener diode, first of all, we will find the Thevenin’s equivalent voltage across this two-terminal. And it can be given as RL divided by RL plus RS times.

Now if this Thevenin’s voltage is more than VZ in that case, this Zener diode will operate in the breakdown region. so if the Zener diode operates in the breakdown region then it will provide a constant voltage Vz to this load resistor. And in that case, the load current IL will be equal to VL divided by RL. And here VL is nothing but the VZ.

That means the load current IL will be equal to VZ divided by RL. Similarly, now let us see what is the source current or the series current which is flowing through this resistor Rs. So this source current Is can be given as Vs minus Vz divided by Rs.

so this is the source current which is supplied by this voltage source. so now we know the source current and this load current. And the remaining current will flow through this Zener diode. And this current is known as the Zener current. so if we apply the KCL at this node then we can say that this current is Is equal to Iz plus IL. Or we can say that the Zener current Iz is equal to Is minus IL.

So this is the current which is flowing through this Zener diode. And the power which is getting dissipated across the Zener diode is equal to Vz times Iz. so basically it is the product of Zener voltage and the current. so in this way, we can find the current through the diode. so let us take one simple example so that this concept will get clear to you.

### Example No.1

so in this example, we have been asked to find the Zener current and the power which is getting dissipated across this Zener diode. so first of all, we need to ensure that the voltage across the Zener diode is more than 10 volt.

### Step No.1

so for that, we will find the Thevenin’s equivalent voltage across these two terminals. so here this can be given as 1-kilo ohm divided by 1-kilo ohm plus 250-ohm times 18 volt.

### Step No.2

And if we calculate the value then it will come out as 14.4 volts. so as this Thevenin’s equivalent voltage is more than 10 volt, so this Zener diode will operate in the breakdown region and it will provide a constant voltage of 10volt across this load resistor.

### Step No.3

so from this, we can say that the load current will be equal to 10 volts divided by 1-kilo ohm, which is equal to 10 milliamperes and here the series current Is or the source current Is will be equal to18 volt minus 10 volts divided by 250 ohms and it will be equal to 32 milliamperes.

### Step No.4

So this current Is is equal to 32 milliamperes. Now we know that Zenercurrent Iz can be given as Is minus IL that is equal to 32 minus 10 milliamperes.

### Step No.5

Which is equal to 22 milliamperes so this is the Zener current which is flowing through this Zener diode. and the power which is getting dissipated can be given as Iz times Vz, which is equal to 22 milliampere times 10 volt.

### Final Answer

so the power which is getting dissipated across the Zener diode is equal to 220 milliwatts. so in this way, we can find the Zener current as well as the power dissipation.

- now whenever the Zener diode is used as a voltage regulator then depending on the application the supply voltage or the load resistance will change.
- But despite the change in these parameters, this Zener diode should provide a constant voltage across the load.
- Or in other words, it should work properly as a voltage regulator. so, in general, we can say that whenever the diode is used as a voltage regulator then there are three variables.
- one is the supply voltage and the other one is the load resistance. And the third variable is the series resistance.

so here what we will do, we will consider the two parameters constant and we will assume that the third parameter is a variable and we will find the maximum and the minimum value of this parameter such that the diode will operate in the breakdown region. Or in other words, this regulator circuit will work properly as a voltage regulator. so first of all we will assume that these supply voltage Vsand this series resistance Rs are constant.

## Zener Diode as a Voltage Regulator with variable load

Zener diode voltage regulator And the load resistance RL is variable. so we will find the minimum and the maximum value of this load resistor such that the Zener diode will work properly as a voltage regulator. Zener diode voltage regulator so first of all let us find out the minimum value of this load resistance. Zener diode voltage regulator and to find that first of all let us write the Thevenin’s equivalent voltage across this diode. so this voltage Vth can be given as RL divided by RL plus RS times Vs Zener diode voltage regulator.

Zener diode voltage regulator now from the equation if you observe as the value of this load resistance RL reduces, the Thevenin’s equivalent voltage across these two terminals will reduce but Zener diode voltage regulator if thisThevenin’s equivalent voltage goes below this voltage VZ in that case this Zener diode will come out of the breakdown region so the Thevenin’s equivalent voltage should be at least equal to Zener voltage and from this, we can find the minimum value of this load resistance.

so Zener diode voltage regulator if we equate this expression with VZ and if we rearrange the terms then the minimum value of this load resistance can be given as Rs times VZ divided by Vs minus VZ.zener diode voltage regulator so this will be the minimum value of the load resistance to ensure that the Zener diode works properly in the breakdown region.

Zener diode voltage regulator similarly, now let us find out the maximum value of this load resistor. now whenever this load resistance RL is maximum at that time the current which is flowing through this load is minimum or in other words we can say that the IL minimum will be equal to VZ divided byRL maximum. diode voltage regulator nowhere as the Vs and Rs are fixed so the value of current Iswill is fixed. So this current Is can be given as Vs minus VZ divided by Rs.

So Zener diode voltage regulator this current Is will remain fixed. now as we increase the value of this RL then the current which is flowing through this load resistor will reduce and at the same time, the current which is flowing through the diode will increase.

So as we keep the diode voltage regulator on increasing the value of RL then the current which is flowing through the diode will increase but we know that the current which is flowing through a diode should be less than the maximum rating or in other words, it should be less than Izm.

Zener diode voltage regulator so whenever the maximum current is flowing through the diode at that time the current which is flowing through this load RL is minimum. or mathematically we can write it as Is is equal to Izm plus IL(min). diode voltage regulator That means the IL(min) is equal to Is minus Izm. And from this, we can find the maximum value of this load resistor. So to understand it better let us take one example.

### Example No.2

Zener diode voltage regulator so in this example, we have been given this regulator circuit and we have been asked to find the minimum and the maximum value of this load resistor.

- so first of all let us find the minimum value of this load resistor and we know that the minimum value of this load can be given as RS x VZ divided by Vs minus VZ.
- And if we put the values then we can write it as hundred-ohm times five volt divided by five volts.
- That means the minimum value of this load resistor is equal to a hundred ohms. Similarly, now let us find out the maximum value.
- so for finding the maximum value first of all we will find the source current Is. And the source current Is can be given as 10 volts minus five volts divided by hundred ohms.
- That is equal to 50 milliamperes. so this is the source current which is supplied by this 10-volt source and from this we can say that the IL(min) will be equal to Is minus Izm, which is equal to 50 minus 30.
- that means 20 milliamperes. so this will be the minimum current which will flow through this load resistor and from this we can say that the maximum value of the load resistor will be equal to VZ divided by IL(min).

Zener diode voltage regulator That means five-volt divided by 20 milliamperes. so the maximum value of the load resistor will be equal to 250 ohms.

Zener diode voltage regulator so from this, we can say that for the given parameters when the value of the load resistor is in between hundred ohms and the 250 ohms then the given circuit will work properly as a voltage regulator. so now we will assume that these series resistor Rs and the load resistor RL are constant and this supply voltage Vs is variable.

## Zener Diode as a Voltage Regulator with variable source voltage

so we will find the minimum in the maximum value of the Vs such that the Zener diode will work properly in the breakdown region. (In A Peak Voltage Detector I Explained )Orin other words this circuit will work properly as a voltage regulator. so first of all, we will find the minimum value of this supply voltage and for that once again let us write down the Thevenin’s equivalent voltage which appears across the Zener diode.

that is equal to Vs times RL divided by Rs plusRL. now from the expression if you observe as we reduce the value of this Vsthen the Thevenin’s equivalent voltage which appears across the diode will reduce and if it goes below this Vz then the diode will come out of the breakdown region.

so the minimum value of this Thevenin’s equivalent voltage should be equal to VZ. And from this, we can find the minimum value of this supply voltage. So the Vs minimum will be equal to RS plus RL divided by RLtimes Vz.

so this will be the minimum value of the supply voltage which will ensure that the diode will work in the breakdown region. similarly, now let us find out the maximum value of this supply voltage. nowhere if you observers the diode is operating in the breakdown region so the voltage at this node will be equal to VZ which means the lower current IL will be equal to VZdivided by RL.

Now here as this Vs is variable so as we increase the supply voltage then the source current is will increase and here the source current can be given as Vs minus Vz divided by RS. So from the expression, we can say that as the value of Vs will increase the source current Is will increase. but here as this load current Is is fixed, so the remaining current will flow through this diode.

so as we keep on increasing the value of Vs, the current which is flowing through the **Zener diode **will increase. But we know that the maximum allowable Zener current is equal to Izm. so this will limit the maximum value of this source current. or in other words, we can say that the Is (max)will be equal to Izm plus IL. And from this Is(max) we can find the maximum value of this source voltage. that means the Is(max) will be equal to Vs(max) minus Vz divided by Rs so in this way we can also find the maximum value of the supply voltage.

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