**Full-wave rectifier **Hey friends welcome to the Kohiki. In this article, we will learn about the full-wave rectifier. In the earlier article, we have already learned about the **half-wave rectifier** circuit. And in general, we have seen that these rectifier circuits are very useful for AC to DC conversion. So, if we apply this AC signal to this rectifier circuit, then it only passes the positive half cycle and completely blocks the other half.

And Full-wave rectifier we had also seen the average value or the DC value of this half-wave rectifier. which is given by the expression, Vm/ πwhere Vm is the peak value of this input signal.

Now, in this half-wave rectifier, as it is passing only one-half of the input signal, the average value of the signal will be low. So, here somehow if we can use this other half of the input signal, then we can increase the average value of the output signal Full-wave rectifier.

## Full-Wave Rectifier.

Full-wave rectifier And that is possible using the full-wave rectifier circuit. So, if we apply the sine wave as an in putto this full-wave rectifier, then we will get this kind of waveform.

So, as you can see it completely passes the positive half cycle and inverts the negative half cycle. So, using this full-wave rectifier circuit we can get the output waveform for both half cycles.

Now, Full-wave rectifier one more thing if you observe over here, the frequency of the output waveform has been doubled. Because now, the waveform is repeating itself after the T/2 time period. So, let’s say, the new time period of this waveform is equal to T’. So, we can say that this T’ is equal to T/2. Or in terms of the frequency, we can say that the new frequency f’ will be equal to 2f. That means the output waveform frequency will be 2 times the input waveform frequency.

Now, this full-wave rectifier circuit can be designed using the two methods. The first is using the center-tapped transformer and the second method is using the diode bridge circuit. So, first, let’s see, how we can design the full-wave rectifier circuit using the center-tapped transformer. So, as you can see over here, the center-tapped transformer is used at the input side.

Full-wave rectifier Because if you observe over here, the center of this secondary winding is grounded. And here the transformer ratio which is used in the circuit is 1:2. That means, if we apply the input voltage Vin, then at the output of this secondary winding, we will get the 2 times Vin. And if we measure the output voltage between the center and the one end of the winding, it will be equal to Vin. So, this 2Vin voltage will get divided between the two parts. And here let’s assume that these diodes areD1 and D2. So, let’s understand the working of this rectifier circuit.

So, whenever we apply a sine wave as an input to this circuit, then during the positive half cycle this diode D1 will conduct and this diode D2 will remain reversed biased. And in that scenario, the current will flow like this.

So, if we assume this diode D1 as an ideal diode, then this Vin voltage will appear across this load resistance. Or we can say that this Vout is equal to Vin. So, the same input voltage will also appear across this load resistance. Similarly, during the negative half cycle, this diode D1 will remain off, while this diode D2 will conduct. And the current will flow through the circuit in this fashion.

So, once again, if we assume this diode D2as an ideal diode, then this input voltage Vin will appear across this load resistance. Now, here during the positive as well as the negative half cycle, the current which is flowing through the load resistor is in the same direction.

So, the Full-wave rectifier for the negative half cycle also we will get the positive waveform. So, now if we combine this positive as well as the negative half cycle, then the output waveform of this rectifier circuit will look like this.

So, in this way, if we apply a sine wave as an input to this rectifier circuit, then we will get this kind of** waveform at the output**. And here as we are getting the output waveform for both half-cycles, the average value of this full-wave rectifier will be more than the half-wave rectifier.

And to be precise Full-wave rectifier if I say, the average value will be equal to 2Vm /π. Which is exactly doubled compared to the half-wave rectifier circuit. But here we have assumed that this diode D1and D2 is ideal. So, if this diode D1 and D2 are not ideal, in that case, there will also be a voltage drop across these two diodes.

And in that case, the average value can be given by this expression. That is 2(Vm – 0.7)/πwhere here we have assumed that the voltage drop across each diode is equal to 0.7V. So, in general, we can say that the average value will be equal to 2 (Vm – Vt)/π where Vt is the threshold voltage for each diode.

So, in this condition, if you see the output waveform, then the output waveform will look like this. Because now, the diode will conduct only when the input voltage crosses this threshold voltage. And due to the drop across this diode, the peak value will also get reduced. So, instead of Vm, now the peak value will be equal to Vm – Vt. So, this will be the expression of the average value considering the voltage drop across the diode.

## Full-Wave Rectifier : Peak Inverse Voltage

Now, whenever we are designing the rectifier circuit, then one parameter which always needs to be considered is the Peak Inverse Voltage. And it is the maximum voltage that appears across the diode in the reverse biased condition.

So, let’s find out the peak inverse voltage for this particular circuit. And to find the peak inverse voltage, let’s assume that during the positive half cycle, this diode D1 is conducting and this diodeD2 remains off. And let’s say, in the reverse biased condition, the voltage across this diode D2 is equal to Vd2.

So, if we apply the KVL in this particular loop, then we can write it as, Vin +Vo = Vd2 Now, if we assume this diode D1 as an ideal diode, in that case, this Vout will be equal to Vin. So, we can say that this Vd2 is equal to 2*Vin. And if we assume the peak voltage of this input signal, then it is equal to VM. So, we can say that the peak inverse voltage for the diode will be equal to 2*Vm.

So, for the full-wave rectifier circuit which is designed using the center-tapped transformer, the peak inverse voltage is equal to 2*Vm. So, the PIV rating of the diode should be more than this value.

So, let’s say if we are applying an input signal which has a peak voltage of 10V then, the peak inverse voltage of the diode should be more than 20V. So, while designing the circuit for the specific voltage we should always check the maximum reverse voltage which can withstand by the diode.

## Full Wave: Bridge Rectifier

Alright so now let’s see the second method using which we can design this full-wave rectifier circuit. So, as you can see over here this full-wave bridge rectifier circuit consists of four diodes. So, if you compare this circuit with the previous circuit then does not involve any kind of transformers. And due to that, the size of the circuit will get reduce.

Alright so now let’s see the working of this full-wave bridge rectifier. So, if we apply the sine wave as an input to this circuit, then during the positive half cycle this diode D2 and D4 will conduct. While this D1 and the D3 diode will remain off.

So, during the positive half cycle, the current will flow through this path. And if we assume this diode D2 and D4 are ideal, in that case, this input voltage will appear across this load resistor. So, the same input voltage will also appear at the output terminal. Similarly, during the negative half cycle, these diodes D2 and D4 will remain OFF.

And on the other end these diodes D1 and D3will conduct. So, during the negative half cycle, the current in the circuit will flow in this way. And if we observe over here, during the negative half cycle also the current which is flowing through the load is in the same direction.

So, during the negative half cycle, we will get this kind of waveform. So, if we combine the positive as well as the negative half cycle, then we will get this kind of waveform.

Now, if we consider these four diodes as an ideal diode, in that case, the average value of the full-wave rectifier will be equal to2Vm/π But if we also consider the voltage drop across this diode, in that case, the average value can be given by this expression.

That is 2(Vm- 1.4)/π. Or in general, we can say that the average value will be equal to 2*(Vm -2Vt)/π Where Vt is the forward voltage drop across these two diodes. And here in the waveform, this T’ is the time period of the output waveform. Where T’ is equal to T/2. Alright, so now let’s find out the peak inverse voltage for this full-wave bridge rectifier circuit. And to find the peak inverse voltage,

let’s assume that during the positive half cycle, this diode D2 and D4 are conducting. While this diode D1 and D3 are in the off condition. So, in that case, as we had seen, the current will flow in this fashion. And in that condition, let’s say the reverse voltage which appears across this diode D1 and D3 are Vd1 and Vd3 respectively.

So, if we apply the KVL in this particular loop then we can say that this Vd1 is equal to Vo. And here we are assuming that all four diodes are the ideal diode. So, in that condition, the same input voltage will also appear across this load.

That means we can say that this Vd1 will be equal to the input voltage. And the peak value of the input voltage will be equal to VM. That means the peak inverse voltage which will appear across this diode D1 will be equal to VM.

And the same is true for this diode D3. So, if we apply the KVL in this loop, then we can find that this Vd3 will be equal to Vin. And the peak value of the reverse voltage will be equal to VM. So, for the full-wave bridge rectifier circuit, this peak inverse voltage is equal to VM.

## Full Wave Rectifier : center-tapped transformer

So, so far we have seen how we can design this full-wave rectifier circuit using either a center-tapped transformer or using the bridge rectifier circuit. But if you observe this output waveform, itis not completely DC voltage.

Because still there is some periodic variation in this output waveform. So, this periodic AC variation in the DC output voltage is known as the ripple. And this ripple can be reduced by using the filter circuit at the output of the rectifier.

So, as shown in the figure, by using the filter circuit, we can minimize the effect of ripple. So, with this filter, if you see the output waveform, then the output waveform will look like this. So, this blue waveform is the waveform just before the filter and this yellow waveform is the output of this filter.

So, with this filter circuit, how well this ripple is removed, depends on the value of this load resistance RL as well as this capacitor. So, before we see that,

first of all, let’s understand how this filter circuit works. So, during the positive half cycle, this capacitor will charge up to the peak voltage. And once this capacitor gets fully charged, then this diode D1 will become reverse bias.

So, after the peak voltage, this capacitor gets discharged through this resistor RL. And due to this discharge, we will get this kind of waveform.

So, once again the capacitor starts charging whenever the voltage across the capacitor just goes below this input voltage. So, after that time once again this capacitor starts charging through this path. And once again, when the voltage across the capacitor will become Vm, then this diode D2 will become OFF. And once again, now this capacitor starts discharging through this resistor RL.

So, due to the charging and the discharging of this capacitor, we will get this kind of waveform. So, in the circuit, how well the ripple is removed, depends on the RC time constant of this filter. So, for the better rejection of this ripple, the RC time constant should be much larger than this time period.

So, suppose if this RC time constant is less than this time period T’, in that case, this capacitor will get discharged rapidly. So, due to that, we will get more ripples in the output waveform. So, avoid that this RC time constant should be much larger than this T’.

But still, if you observe, there is some ripple in the output waveform. And this ripple voltage is known as the peak to peak ripple. So, if VM is the peak voltage, and this voltage is the peak to peak ripple, then this peak to peak ripple can be given by this expression. That is Vm/ (2*f*RL*C)Where here f is the frequency of the input signal.

So, if we consider the frequency of this output waveform, then this 2f should be replaced by this frequency f’.

Now, for this waveform, suppose if we want to find the DC voltage, then this Vdc can be given by the expression, Vm – Vr (p-p)/2 That means if this voltage is the peak to peak ripple, then the DC voltage will be the subtraction of this Vm minus half of this ripple voltage. And that will be the approximate DC voltage of this output waveform. So, this will be the expression of DC voltage after the filtration.

And here if we assume that this ripple voltage is much less than the peak amplitude of the sine wave, in that case, the approximate value of this DC voltage will be equal to VM. Now, this ripple voltage can also be expressed in terms of the load current. That is Idc/ (2*f*C)So anyway, we will derive all these expressions in the separated Article.

So, using this rectifier and the filter circuit, we can reduce the ripple in the output waveform. And how well this ripple is removed from the circuit can be defined by the parameter which is knowns as the ripple factor.

**YouTube Video**

So here are a YouTube video based on a Full-wave rectifier & Bridge Rectifier, Which Was Uploaded by Neso Academy

## FAQ’s

### What is full wave bridge rectifier

### What is a full wave bridge rectifier used for

### What is bridge rectifier and its working

### What stage is the full bridge rectifier

### What are the disadvantages of half wave rectifier

**half**-cycle through per sinewave, and the other

**half**-cycle is wasted. This leads to power loss.

They produces a low output voltage.

The output current we obtain is not purely DC, and it still contains a lot of ripple

## Summary/Last Word About Full-Wave Rectifier

So, this ripple factor is a ratio of the RMS value of this ripple divided by the DC voltage. And for the full-wave rectifier, without any filtration, this ripple factor is equal to 0.43. Now, if you recall the ripple factor for the half-wave rectifier, it was equal to 1.21. So, for the full-wave rectifier circuit, the ripple factor has been improved drastically. And it can be further improved by using the filter circuit.

So, for the full-wave rectifier with the filter circuit, the ripple factor can be given by this expression. So, we will derive all these expressions in the separate Articles. Now, one more parameter which is defined for the full-wave rectifier is the efficiency. That is how well the AC input power is converted into the DC power. And for the full-wave rectifier, this efficiency is equal to 81.2 %. And if you recall for the half-wave rectifier, the efficiency was 40.6 percent.

So, using the full-wave rectifier, we can almost double the efficiency. So, in summary, here is the different parameters that we have discussed so far for the full-wave rectifier. And these parameters are also defined separately for the center-tapped as well as the bridge rectifier circuit. Now, out of all the parameters, some of the parameters we have discussed very briefly. But in the separate article, we will also derive all these expressions.

So, I hope in this Article, you understood how we can design a** full-wave rectifier** circuit using the center-tapped transformer as well as using the bridge circuit.

So, if you have any questions or suggestions, About** Full-Wave Rectifier** do let me know here in the comment section below. If you like this article, hit the like button and subscribe to the Kohiki for more such Articles.

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