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I don't understand how the diffusion capacitance is formed in a forward Biased diode

All websites give the equations but no qualitative description of the reason why There's a capacitance in a forward biased diode.

What I know is capacitance is related to charge storage so How charge could be Stored if there is a forward current through the diode?

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  • \$\begingroup\$ Think about it. All diodes have a Vdrop, with capacitive leakage. If reversed biased, it still has capacitive leakage. The model of a diode includes resistors, capacitors and inductors. \$\endgroup\$ – user105652 Apr 18 '18 at 21:30
  • \$\begingroup\$ @Harry Svensson I think it looks good now \$\endgroup\$ – Mohammed Hisham Apr 18 '18 at 22:13
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Charge transport (current) through a forward-biased pn-juntion is markedly different when compared to charge transport through a conductor. One major difference is that current through a pn-junction is due to two different types of charged particles, one positive (holes) and the other negative (electrons). This is opposed to current through a conductor, which is (typically) due to the flow of electrons only. The other huge difference is the mechanism by which current flows. In a forward-biased pn-junction, current is due primarily to diffusion of both holes and electrons through n-type and p-type materials, respectively. This is in contrast to a conductor in which current is due to an electric field which causes a potential difference to appear across the ends of the conductor.

I like to think of a forward-biased pn-junction like an army storming a city. When the gates of the city come down (a forward voltage is applied), the defenses are down and the troops (holes or electrons) come storming in. However, the holes (or electrons) are in foreign territory and at any moment they can be eliminated by the enemy (recombination). At the gates of the city (right at the barrier of the pn-junction), you can expect to find a large concentration of troops. However, the deeper into the city you go, the less troops you find because only a few of the soldiers manage to make it a significant distance before getting eliminated. All of this to say that there exists a concentration gradient of charged particles as you move from the barrier of the pn-junction to the end of the semiconductor material. This concentration gradient forces a current to flow that is NOT the result of an electric field but rather a result of a random thermal walk of like particles that is not unique to holes and electrons in a semiconductor (diffusion current).

It's very important to note that the concentration gradient is caused by real, physical charged particles that exist within the semiconductor material. The concentration gradient is determined by the applied voltage so that a particular voltage uniquely maps to a concentration gradient. The physical amount of stored charge is associated with the concentration gradient by integration. When a greater forward bias voltage is applied, a greater concentration of charged particles come pouring across the barrier. A new concentration gradient must be established, which suggests that a greater amount of charge must be stored in the material. The point here is that changing the applied voltage changes the concentration gradient and thus the amount of physical charge that is stored in the semiconductor material. Notice that a voltage is associated with a stored charge. This is exactly the definition of a capacitor with the factor of association being the capacitance, C = Q/V.

Note that charge is not actually being 'stored' in the semiconductor material. It is continuously and simultaneously being eliminated by recombination and replaced by charge diffusing across the junction from the other side. Therefore, for all intents and purposes, it appears as if there's constant excess charge hanging around inside the semiconductor material, which is the source of the diffusion capacitance.

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As you know, when a current flows through a forward biased diode, there is a voltage drop, typically in the range of 0.3-0.8V, depending on the type of a diode and the current level. (For the LED's, the voltage drop is on the order of 1-2v).

This voltage drop implies the presence of some sort of barrier in the diode: the forward current would not flow if the voltage applied to the diode is below that barrier level.

But for the voltage to build up across the barrier, the concentration of electrons have to increase on one side of the barrier and decrease on the other. So, before the current can flow through the forward biased diode, the diode acts as a capacitor that is being charged to a certain voltage. The effective capacitance of a diode is roughly proportional to the area of its junction and will naturally be larger for more powerful diodes.

Once the voltage on that capacitor reaches a certain level, the current will start to flow, but the charges accumulated on the two sides of the junction/barrier will remain there to maintain the required forward voltage or, we can say, that the charges will be stored on the junction capacitor.

Naturally, charging of that capacitor before the current can start to flow takes some time. That is why large diodes tend to be slow.

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