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nick67
 2 years ago
Best ResponseYou've already chosen the best response.1Hi edr1c, as a rule of thumb you may evaluate the forward diode resistance as: \[R _{F}=\Delta V_{D} / \Delta I_{D}\] taking values from the quasilinear portion of the characteristic

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0so by taking 2points above the turn on voltage for the diode, i will get the approximate resistance by taking their gradient?

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0does barrier potential comes into consideration when calculating this resistance? because the previous section asked us to estimate the barrier potential.

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0the reason im raising the question is because i obtained a very high value for the resistance. Voltage, V1/V Current, I/mA 0.6 0.92 0.64 1.71 0.68 4.04 0.7 5.81 0.72 8.81 0.74 12.72 0.76 17.86 0.77 20.89 i got around 17ohm by taking the 2nd and 3rd point

nick67
 2 years ago
Best ResponseYou've already chosen the best response.1Well, if you take points too close to the diode knee you have of course a greater resistance, because it is still going into forward bias; if you take last two points you have a resistance of about 3,3 Ohm; in this region the diode is in its linear zone

nick67
 2 years ago
Best ResponseYou've already chosen the best response.1As for the barrier potential you could apply a reverse calculus taking values from last two points; you have there a diode voltage of about 0.76 V and a diode current of about 19 mA. If you subtract the voltage on the bulk resistance from the total diode voltage you can evaluate the barrier potential:\[V _{barrier} = V _{D}  R _{bulk} I _{D}\] that is 0.76  3.3 x 0.019 = 0.697 V
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