Chapter No: Chapter: Semiconductor Electronics: Materials, Devices And Simple Circuits Name the majority charge carriers in p-type semiconductors. Electronics Fundamentals, 6e. Electric Circuit Materials. • In an intrinsic semiconductor, there are relatively A Semiconductor diode consists of an n material. Semiconductor Electronics, Materials, Devices and Sample Circuits. It is the branch of science which deals with the electron flow through a vacuum, gas or.
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Semiconductor —II. • Silicon is the most common material used to build semiconductor devices. • Si is the main ingredient of sand and it is estimated that a cubic. Semiconductor Electronics: Materials, Devices and. Simple Circuits semiconductors. However, after , a few semiconductor devices using. This Semiconductor Devices: Theory and Application, by James M. Fiore is Welcome to the first edition of Semiconductor Devices, an open educational.
Class 7. If we consider the capacitance, the capacitance has already charged to the peak value Vm but after this peak point, input voltage Vi is going down but then if we consider the diode this cathode and anode, at the cathode the capacitor voltage is being applied and the anode is the input voltage. It should be sufficiently high so that we get overall almost constant DC voltage. This will be the waveform and in the negative half cycle, when this lower point is positive upper point is negative of the input voltage waveform then the green line will be showing the conduction of current because the diodes D3 and D1 will be now forward biased in the negative half cycle of the input voltage. Here in the half wave rectifier we have one disadvantage or one draw back of half-wave rectifier is that in one half of the input waveform the output voltage is totally zero.
There will be 4 diodes as arranged in this configuration as we can see here. There will be D1, D2, D3 and D4; 4 diodes. The input Vi is an AC waveform. Let us consider sinusoidal. If we consider a positive half cycle of the input waveform then the conduction of which diodes will be taking place?
The diodes which will be forward biased those diodes will conduct.
We will be applying this input voltage having Vm sin omega T. This side is positive, this side is negative. When it is positive in the positive half cycle of the input voltage waveform the red colored straight line will be showing the conduction of the diodes.
The diodes D2 and D4 as you can see here will be forward biased. This is positive connected to P; this diode is also forward biased. That is why the current part will be shown by this red line.
The current will be flowing through D2 and D4. Across this resistance R, this current will produce a voltage drop V0.
As it is a resisted drop it will be following the input voltage waveform in shape. There will be a time lag in order to start the current because of the threshold voltage of the diode. This will be the waveform and in the negative half cycle, when this lower point is positive upper point is negative of the input voltage waveform then the green line will be showing the conduction of current because the diodes D3 and D1 will be now forward biased in the negative half cycle of the input voltage.
The current will be flowing in this direction. That is through D3, through all and then through D1. The output voltage for the negative half cycle is this one. For both the positive and negative half cycles of the input voltage, we get output voltage across this resistance.
So it is better than the half-wave rectifier. For both the half cycles we are getting an output waveform. In this case if we want to find out what will be the output voltage or DC voltage again we have to do the integration. But here while doing the integration as we have output waveform for both the half cycles 1 by T by 2 integration will be from zero to T by 2. That means we can take one half of the input waveform and we can integrate it.
It will be similar for the other half also. Finally we do this integration, put down the limits and we find out that omega capital T is equal to 2 pi. Again substituting we get finally the output voltage average value is equal to 2 by pi into Vm. This value is exactly twice than the output voltage which we got in half-wave rectifier.
That is 0. Here also for doing this analysis we have assumed ideal diode approximation. If we consider a practical diode having cut in voltage of VD, then we have the average value 0. There is drop here. There is drop here also. So two voltage drops will be coming into the picture. The diode drop VD will have to be multiplied by 2. This 2VD will have to be subtracted from the peak value of the input waveform to get output voltage wave and that will have to multiplied by this 0.
For a practical diode we have this as the average value. We find out the PIV of this full wave rectifier. Peak inverse voltage of each diode you can find out when the diode is non-conducting. When one diode is non-conducting what will be the maximum voltage that is across it? That is the peak inverse voltage. What will be the voltage that will be applied across this D1?
Because D1 is non-conducting that can be found out if you consider the two ends of this D1 diode. What will be this voltage?
So this drop is equal to Vi minus this drop. That is Vi-VD0. Here also we are not nearer to a constant DC voltage even now because what we are getting is a pulsating DC voltage like this.
This is what we get. But then we want get a constant voltage. This is not obtained till now. We have to now modify the circuit or we have to add something extra to get exactly a DC voltage.
But before that I will be discussing another type of full wave rectifier that is using center tapped transformer. Transformer is a device which can transform or which can convert the voltage level; the input voltage level can be stepped down to a lower voltage or it can be stepped up depending upon the turns ratio of the primary and secondary winding of transformers.
A center tapped transformer has the secondary side tapping at the center. If this is Vs this will be also Vs and this is center which will be grounded. That is in the centre tapped transformer, tapping is at the center of the secondary.
In this full wave rectifier this transformer which is center tapped is used to bring the input voltage to its desired value. Here there are two diodes D1 and D2. Circuit is like this. We are getting the output voltage across this resistance which is V0.
The conduction will be taking place when this is positive, upper point is positive with respective to the center. Upper one diode will conduct and when the center is positive with respect to this that means this lower one if we consider if this is higher at a higher potential than this, then this way the diode will conduct.
But the resistance through which the current will conduct in both the cases it will be in the same direction so that the polarity of the voltage which you get across this resistance will be in the same way. Input voltage is Vi and output voltage which is obtained across this resistance will be pulsating DC as we are getting here. For each half cycle we will be getting the voltage across this resistance and what will be this peak inverse voltage across each diode in this case?
When this diode is non- conducting that is in the case when this upper diode D1 is conducting then the voltage across these two ends of this diode will be how much; the total voltage across secondary minus this drop.
If one half is VS and peak value above one half is Vm then total voltage, peak value will be 2 times Vm.
But here we are employing 2 diodes instead of 4 diodes in the bridge rectifier. That is a capacitor filter will be applied to smoothen out the DC voltage which we are getting without capacitor. But till now whatever we have got in half-wave rectifier and full wave rectifier is unipolar voltage, no doubt. But it is not purely DC. To get an ideal DC voltage we will be employing a filter circuit using a capacitor. What will be achieved using a capacitor that we will have to see in the circuit.
Let us first consider half-wave rectifier. Half-wave rectifier with capacitor filter will be this circuit where a capacitor is included across this resistance RL, load resistance. Here we will consider ideal diode for simplicity because we are concerned about the analysis.
Let us consider an ideal diode where the cut in voltage is zero as well as the average resistance is also zero.
In the positive half cycle when this upper point is positive with respect to the lower point for the input voltage Vi, the diode will conduct as it is forward biased and it will be charging these capacitors. The capacitance will be charged and this diode is having zero resistance.
The time required for charging the capacitance is almost zero. Because instantly the capacitor will be charged there will be no time lag. The capacitor will be charging to the peak value of the input voltage instantly and VC will be the output voltage.
This voltage across RL is nothing but the voltage across this capacitance. If the input voltage is Vi output voltage if we draw for a particular resistance RL we are considering, then it will be first of all charging. The capacitor will be charging to the peak value of the input voltage. When you notice the input voltage it is sinusoidal. It is varying with time.
From the peak value it is falling off. If we consider the capacitance, the capacitance has already charged to the peak value Vm but after this peak point, input voltage Vi is going down but then if we consider the diode this cathode and anode, at the cathode the capacitor voltage is being applied and the anode is the input voltage. If we consider this point for example here the capacitor voltage is V0 which is equal to Vm. But at this point the input voltage is falling down.
This diode will be reverse biased and it will not be conducting. So this circuit is off. Now what will happen? The capacitor voltage will be discharging through this resistance RL. This discharging which is taking place is shown by this green line. This capacitor is discharging through this resistance RL. This discharging rate will be dependent upon the time constant which is RL into C.
If we have a bigger time constant it will take more and more time to discharge. If we have low resistance, RL is a small value then the time constant will be small. It will be falling of very soon. That means rate of discharging will be very fast. Here the capacitor is discharging and discharging but you see that input voltage is now at this point increasing here and after this point input voltage becomes higher than the capacitor voltage.
Input voltage is higher means the diode will be again forward biased. From this point it will be again forward biasing the diode since Vi is greater than VC and the conduction of the diode will be taking place and therefore the capacitor will be charged.
Charging will be taking place. Again after this point input voltage is lower than the capacitor voltage. So discharging will be taking place. Charging and discharging will be going on, finally giving a voltage which is having a ripple. It is a not constant voltage but it will be having a ripple.
So we will get this type of voltage. It is not true. We cannot get an infinitely high voltage but we will be getting more and more towards DC if you go on increasing the RL resistance, load resistance. Capacitor is not changing; capacitor we are fixing. That is time constant should be such that it should not decay appreciably when you consider the time period of the input voltage. If the input voltage has time period capital T then this time tau which is the time constant for the discharging circuit should be sufficiently large compared with this time period of the input voltage.
Then only we do not see much falling of the output voltage and output will be almost constant like a DC.
So you have to choose this resistance RL value sufficiently larger to have a larger time constant. If we consider the diode conduction how much time the diode will be conducting?
The diode is conducting for a very small time, only for this portion. The diode current is such that it will flow for a very small time and then it will become zero. It is called a surge current. If we consider this half-wave rectifier and see the conduction of the diode, the diode current will be like a surge. Suddenly it is increasing and then within a very short time it will be falling off also to zero.
That is why it is called the surge current. For example if we consider the whole time period, only for a small portion T1 it is conducting. Electronics Engineering. Engineering Electronics.
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