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21 Questions around this concept.
The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistance R connected in series with diode for obtaining maximum current?
The circuit shown in the figure contains two diodes each with a forward resistance of
and with infinite backward resistance. If the battery is 3 V, the current through the resistance is:
The saturation current of a P-N junction germanium at is . The potential required to be applied to obtain a current of 250mA in forward bias will be:
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The inverse saturation current in a P-N junction diode is ampere. The value of forward current at 0.2 volts will be:
The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with the diode, for obtaining maximum current?
Assuming that the silicon diode having resistance of , the current through the diode is (knee voltage 0.7 V)
The circuit has two oopositively connected ideal diodes in parallel. What is the current flowing in the circuit?
A potential barrier of 0.5 V exists across a p - n junction. If the width of depletion layer is , the intensity of electric field in this region will be:
At a temperature of 300 K, a p-n junction has a saturation current of 0.6 mA. Find the current, when the voltage across the diode is 1mV, 100 mV and 1V.
A p-n junction is designed to withstand currents upto a maximum of 10 mA. A resistance of is connected in series with it. When forward biased, the diode has a potential drop of 0.5 V, which is independent of current. Find the maximum voltage of the battery that can be applied to safely forward bias the diode.
P-N Junction-
A p-n junction is the basic building block of many semiconductor devices like diodes, transistors,etc.
When a P-type semiconductor is suitably joined to an N-type semiconductor, then the resulting arrangement is called P-N junction or P-N junction diode.
Formation of a p-n Junction
Let’s imagine thin p-type silicon (p-Si) semiconductor wafer. Now, we have to add a pentavalent impurity to it. This result in a small part of the p-Si wafer converting into an n-Si. Due to this two-region will get created one is the wafer containing a p-region and another is an n-region with a metallurgical junction between the two. There are two important processes that take place during the formation of a p-n Junction:
As we have learned that in an n-type semiconductor, the concentration of electrons is more than that of holes similarly in a p-type semiconductor, the concentration of holes is more than that of electrons.
When a p-n junction is being formed, diffusion of holes starts from the p-side to the n-side (p→n) while diffusion of electrons occurs from the n-side to the p-side (n→p). The reason behind this diffusion is the concentration gradient across p and n sides. Due to this, a diffusion current generates across the junction. Let’s discuss both the scenarios one by one -
Electron diffusion from n→p -
Electron diffusion leaves a positive charge ( ionized donor ) on the n-side. This positive charge is bonded to the surrounding atoms and is not moveable. As diffusion is going on, more electrons start diffusing to the p-side, and a layer of positive charge on the n-side of the junction is formed.
Hole Diffuses from p→n
Hole diffusion leaves a negative charge on the p-side. As the diffusion proceeds, holes start diffusing to the n-side, a layer of negative charge on the p-side of the junction is formed. Both the phenomena i.e., diffusion of electrons and holes across the junction depletes the region of its free charges, these space charge regions together are called the depletion region.
This process is shown in the above figure. The thickness of the depletion region is very small and its thickness is around one-tenth of a micrometre. Since there is an electric field which is directed from the p-side to the n-side of the junction. Due to this electric field electrons moves from the p-side to the n-side and holes from the n-side to the p-side. This motion of charged carriers due to the electric field is called drift. From this we can conclude that the drift current direction is opposite to the direction to the diffusion current. This is also seen in the figure given above.
The Last Stages of Formation of a p-n Junction
When the diffusion starts, the diffusion current is large as compare to the drift current. As diffusion process continues, the space-charge regions on either side of the junction start extending. Due to this the electric field gets strengthen and same with the drift current. This process will continues till diffusion current = drift current. This is how a p-n junction is formed.
Barrier Potential
In the state of equilibrium, there will be no current in a p-n junction. Due to increase in potential difference across the junction of the two regions due to the loss of electrons by the n-region and the subsequent gain by the p-region. This potential opposes the further flow of carriers to maintain the state of equilibrium. This potential is called Barrier potential.
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