I'm not sure how the addition of the intrinsic region affects the built in potential of the diode. I also can't seem to find any information about it online. Thanks for the help.
How does the built in potential of a pin diode compare to a pn diode?
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2\$\begingroup\$ So what exactly is your question? Edit the question and don't post the question in the comments. \$\endgroup\$ – Transistor Jan 28 '16 at 19:56
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\$\begingroup\$ Have you tried googling PIN diode? or differences between PIN an PN diode? Do some research. Wait, I'll do it for you google.com/search?q=PIN+PN&ie=utf-8&oe=utf-8 \$\endgroup\$ – Voltage Spike♦ Jan 28 '16 at 20:40
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The PIN diode will have the same barrier for the material; however, it will have a different frequency behavior. The N and the P regions are highly doped, take a standard band diagram and then stretch it out. The barrier height will be the the same.
I have used these devices as attenuators, and the low frequency behavior is identical, but at high frequency, they look much like a resistor.
I believe that there is a write up on them Andy Grove's "Physics and Technology of Semiconductor Devices"; however, I have not seen them described in detail anywhere else.
