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nick67 Group TitleBest ResponseYou've already chosen the best response.1
Hi 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
 2 years ago

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

nick67 Group TitleBest ResponseYou've already chosen the best response.1
dw:1340545927975:dw
 2 years ago

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

edr1c Group TitleBest ResponseYou've already chosen the best response.0
the 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
 2 years ago

nick67 Group TitleBest ResponseYou've already chosen the best response.1
Well, 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
 2 years ago

nick67 Group TitleBest ResponseYou've already chosen the best response.1
As 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
 2 years ago
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