Full transcript of “Half wave Rectifier Explained”. Click a timestamp to open the video at that moment. Need another video? Use the YouTube transcript generator.
0:12 In the previous videos of the diode, we learned that the diode allows the flow of current only in one direction. and it blocks the current in the reverse direction. Now, this property of the diode is very useful in certain applications. And one such application in which this diode can be used is AC to DC conversion. Now, the process of converting the AC voltage signal into the DC voltage is known as the rectification. And the circuit which is used for the rectification is known as the rectifier. Now, broadly this rectifiers can be classified into two categories. One is known as the half-wave rectifier and the second one is known as the full wave rectifier. So, in this video, we will learn about the half wave rectifier.
1:00 Now, whenever the sinusoidal signal is applied as an input to this half wave rectifier then this rectifier only allows the one half of the sine wave and it rejects the other half of the sine wave. Like here as you can see, it only allows the positive half cycle and it completely rejects the negative half cycle. And similarly, it is also possible to get this kind of waveform. Where this rectifier completely blocks the positive half cycle and it only passes the negative half cycle. So, in this way, this half wave rectifier converts the AC signal into the pulsating DC signal. Now, let's see the circuit of this half wave rectifier. So, as you can see over here, the input signal is applied to the load through this diode.
1:48 So, whenever sine wave is applied as an input then at the output you will get this kind of waveform. So, here this circuit only passes the positive half cycle of the sinewave and completely rejects the negative half cycle. so, now let's understand how this circuit works. And for the analysis purpose for a moment let's assume that this diode is an ideal diode. So, during the positive half cycle, the voltage which appears across the anode and cathode of the diode will be positive. It means that this diode will get forward biased. And simply we can replace this diode by the close switch. So, this entire voltage will appear across the load. so, during the positive half cycle, from 0 to T/2, the complete input wave appear across
2:37 the load. Now, during the negative half cycle, the voltage which appears across the anode and cathode will be negative. So, this diode will get reversed biased. And simply it will act as an open switch. so, there will not be any flow of current through the circuit and at the output, we will get zero voltage. So, during the negative half cycle, we will get zero voltage. And in this way whenever the continuous sine wave is applied as an input to this rectifier, then at the output we will get either positive or negative pulse trains. Now, so far in our analysis, we have assumed that the diode which is connected to the rectifier circuit is an ideal diode. But now let's also consider the forward voltage drop across the diode.
3:24 So, now the diode will get forward biased only when the applied input voltage is more than the threshold voltage. And here we are assuming that the connected diode is silicon diode. It means that it has a threshold voltage of 0.7V. So, the diode will get forward biased only when the input applied voltage is more than 0.7 V. So, now the circuit will start conducting only during this period. So, during the positive half, if you see the output waveform, it will look like this. And if you see the peak voltage at the output waveform, it will be slightly less than Vm. Because now we are also considering the forward voltage drop across the diode. So, instead of Vm, it will be equal to Vm - 0.7.
4:10 Now, during the negative half cycle, the voltage which appears across the diode will be negative. And simply it will act as an open circuit. And the output across the load will be equal to zero. So, during the negative half cycle, we will get zero voltage. Now, in most of the cases, during this rectification process, the peak amplitude of the applied input voltage used to be a much larger than the diode voltage. So, the output waveform of the rectifier will be much closer to the ideal waveform. And in that case, we can simply neglect the drop across the diode. Now, that is being said, let's see, few performance related parameters for this half wave rectifier. And the first parameter which is defined for this half wave rectifier is the average value.
4:59 Now if you see the waveform of the half wave rectifier then mathematically it can be represented like this. So during the first half from 0 to T/2, the waveform can be expressed as Vm*sin (wt) While for the other half from T/2 to T, it will be equal to zero. So, the average value can be given as 1/T, integration 0 to T, v(t)dt. And if we put the value of this V(t), then we can write the average voltage as 1/T, integration 0 to T/2, Vm sin (wt) dt. And if we solve this expression, then the average value for this half-wave rectifier
5:48 will come out as Vm/π. Now, we have already derived this expression in the earlier video of RMS and average value. So, I will give the link in the description for that video so you can check that video for more information. Similarly, the next parameter which is defined for the half-wave rectifier is the RMS value. so, the RMS value of any given signal is given by this expression. That is the square-root of, 1/T, integration 0 to T, V^2(t) dt. And if we put the value of this v(t) in this expression and if we evaluate the value then RMS value for the half-wave rectifier will come out as Vm/2. Now, if you are wondering that from this integration how we have arrived at this expression, then
6:38 don't worry we will derive this expression in the next video. Now, so far in our discussion of this half-wave rectifier, we have only discussed about the output waveform of the voltage. But whenever this diode is conducting at that time finite amount of current will also flow through the circuit. So, if Vm is the peak amplitude of the sinusoidal wave, the peak amplitude of the current can be given by the expression Vm/RL. And if you see the output waveform of the current then it looks very similar to the voltage waveform. And mathematically, it can be represented like this. So, if we find the average value of the current, then it can be given as Im/π.
7:27 That is equal to Vm/(π*RL). So, in this way, we can find the average value of the current. So, the next parameter for the rectifier is the peak inverse voltage. And it defines the maximum reverse voltage which appears across the diode during the operation. So, here if Vm is the peak amplitude of the applied sine wave, then during the negative half cycle the maximum voltage which appears across the diode in reverse condition will be equal to -Vm. So, we can say that for this half wave rectifier, the peak inverse voltage or the peak reverse voltage will be equal to Vm. Now, whenever you are selecting the diode for the rectifier, then you should need to
8:12 check the PIV rating of the diode. So, the peak amplitude of the applied input voltage should be less than this PIV rating. So, now let's move to the next parameter. So far we have seen that whenever the sine wave is applied to this half wave rectifier, then at the output we will get pulsating DC voltage. It means that it has some finite ripple. Now, this ripples defines the periodic variation in the output DC voltage because of the imperfect suppression of the AC signal. So, to define how much ripple is there in the output waveform, this ripple factor is used. And this ripple factor is defined as the ratio of RMS value of the ripple in the output waveform to the total average value of the output waveform.
9:00 And for the half-wave rectifier, the value of this ripple factor is equal to 1.21. So, we will derive this expression for the ripple factor in the next video. Now, the ripple in the output waveform can be reduced by using the filter circuit. So, just by connecting the capacitor across this load, we can reduce the ripple in the output waveform. So, let's understand how this circuit will reduce the ripple. And for a moment, let's assume that the diode is an ideal diode. Now, during the positive half cycle whenever the applied input voltage across the diode is positive at that time this diode will get forward biased. And simply it will act as close switch. So, through this diode, this capacitor will get charged.
9:47 And it will get charged up to the peak voltage of Vm. Now. once this capacitor will get charged up to this voltage Vm, then the diode will become reversed biased. Because at that point, both the anode and cathode will have the same voltage. So, at that time this diode will simply act as an open switch. So, after this time, the charge across the capacitor will get discharged through this resistor RL. So, this yellow line which you are seeing in the waveform is the discharging of the capacitor. Now, here we are assuming that the RC time constant of the circuit is much larger than the time period of the waveform. So, if this condition is true then we can assume that the capacitor is discharging linearly.
10:33 Now, once again, whenever the capacitor reaches at this point, at that time the input voltage is greater than the capacitor voltage. And once again, the diode will become forward biased. So during this portion, the capacitor will get charged through this diode. And once again, it will get charged up to the peak voltage Vm. And then after again it will get discharged through this resistor. So, in this way, because of the charging and discharging of the capacitor, we will get this kind of waveform. So, here this time t1 is the charging period of the capacitor. And this time t2 is the discharging time of the capacitor. Now, like I said before, here we are assuming that the RC time constant of this filter is
11:21 much larger than the time-period of the waveform. And if this condition is not satisfied then the capacitor will get discharged rapidly. And to avoid that problem we should always check this condition. So, by properly selecting the RC time constant of the filter, we can reduce the discharging of the capacitor. And hence we can also reduce the ripple in the output waveform. So, as you have seen, using this filter circuit we can reduce the ripple in the output waveform. But still, if you see the output waveform, it has some finite ripple. And the peak to peak ripple can be given by this expression. That is Idc/(f*C) Where Idc is the average current which is flowing in the given circuit.
12:06 While f is the output frequency of the half wave rectifier and C is the filter capacitance. Then the next parameter which is used for this rectifier is the efficiency. So, this parameter defines how efficiently the AC input power is converted into the DC output power. And for this half rectifier, the efficiency is equal to 40.6 %. It means that from the given input power only 40.6 %of power is getting converted into the DC power. And as you can see, for this half wave rectifier, the efficiency is less than 50%. So, these are the few performance parameters for the half wave rectifier. And these parameters can be improved by using the next type of rectifier circuit.
12:52 That is the full wave rectifier circuit. And we will see about it in the separate video. Now, before ending this video, let's see few applications of this half wave rectifier. And one of the most obvious applications of this half wave rectifier is in AC to DC conversion. So, here the block diagram of the linear power supply has been shown. So, as you can see from the block diagram, at the first stage the input voltage is stepped down using the step-down transformer. And then it is applied to the rectifier circuit. So, at the output of the rectifier circuit, you will get pulsating DC voltage. And the ripple in the output waveform is reduced using the filter circuit. Now, although this filter circuit will reduce the ripple in the output waveform, still some
13:37 variation will be present in the output waveform. And that can be removed using this regulator circuit. So, any variation which is present at the input of this regulator will be taken care of by this regulator circuit. So, in this way, using this half wave rectifier circuit, it is possible to convert the AC signal into the DC signal. But in practical cases, this half-wave rectifier circuit is barely used. Becuase it has very poor efficiency as well as the large ripple. So, in practical cases, full wave rectifier circuits are more preferred over the half wave rectifier circuits. then the next application of this half wave rectifier is in demodulation of the signal. So, if we apply this modulated signal to this half wave rectifier circuit, then it is possible to recover the envelope from the carrier frequency.
14:24 So, these are the few applications in which this half wave rectifier circuit can be used. And in summary, here is the list of different parameters which is defined for the half wave rectifier. Now, among these parameters, we will derive few parameters in the next video. So, I hope in this video, you understood about the half wave rectifier circuit. If you have any question or suggestion do let me know in the comment section below. If you like this video, hit the like button and subscribe to the channel for more such videos.
