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Old 21st October 2009, 02:01 PM   #1
Exclamation Internal Factors?/Series Ω

Internal Factors on which Power Rating of a Semiconductor Diode/Transistor Depend on...Please Justify the Answer (What abt the Diode Drop?What does that depend on?)

Another Ques is Why Series Resistors are used in Diode Circuits(to limit current)?
Why cant the Voltage be reduced? */anyhow both of them make voltage across diode same\*

Thanks in Advance

Last edited by r.vittalkiran; 21st October 2009 at 02:08 PM.
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Old 21st October 2009, 04:02 PM   #2
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Power rating is primarily based on the ability of a system to get rid of heat. There are basically thrre methods which are conduction, convection, and radiation. Mounting a semiconductor to a heatsink is an example of conduction. A cooling fan is an example of convection. Black anodizing a heat sink is an example of radiation.

If you look at the equation of a diode you notice that the current is an exponential function of the forward voltage. Past the threshold of about 0.7V, small increases in the forward voltage produce large increases in forward current. At some value of forward current the device will proceed to get warm, hot, and vaporize. It is the vaporization that effectively limits the current without a resistor. It should not be a surprise that this thermodynamic process is not reversible. So far as I know there have been no recorded cases of molecules spontaneously forming a diode.
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Old 23rd October 2009, 03:24 AM   #3
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for second question:
When a circuit at 0V potential is first connected to power, the difference may cause an instantaneous current beyond the rating of the diode (and other devices). A proper value resistor can limit this current to a (large) safe current. In normal operation the steady current isn't infinite so the resistor's losses hopefully are acceptable.
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Old 23rd October 2009, 01:03 PM   #4
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Quote:
Originally Posted by mneary View Post
for second question:
When a circuit at 0V potential is first connected to power, the difference may cause an instantaneous current beyond the rating of the diode (and other devices). A proper value resistor can limit this current to a (large) safe current. In normal operation the steady current isn't infinite so the resistor's losses hopefully are acceptable.
Thanks...But what makes it happen so?
what happens when i connect the power supply aftersome time? i.e,. First switch it ON, and then connect!??
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Old 25th October 2009, 05:11 AM   #5
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Attaching two different potentials together causes current to flow. It continues until they are equal. If there is no resistance in the path between them, the current is infinite (in real circuits it can't be infinite, so it is large).
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Old 29th October 2009, 03:10 PM   #6
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Quote:
Originally Posted by mneary View Post
Attaching two different potentials together causes current to flow. It continues until they are equal. If there is no resistance in the path between them, the current is infinite (in real circuits it can't be infinite, so it is large).
If this is Right, then why cant i use a Low Voltage, so that its obvious that current will be less(Required Amt.), now no need of Resistor???? :P ..
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