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Question
Sanya performed an experiment of obtaining characteristic curves of a junction diode. When she forward biased it, she found that beyond forward voltage V = Vk, the conductivity is very high. When she reverse biased the diode she found that a very small current (of about a few microamperes) flows in the diode. It remained constant even though she varied the voltage.
(i) In Figure 1 below, which one of the diodes is forward biased?
Figure 1
(ii) What is meant by a saturation current?
(iii) When applied voltage during forward bias is small, why does no current flow in the diode?
(iv) The circuit shown in Figure 2 below contains two diodes, each with a forward resistance of 50 Ω and with infinite resistance during reverse bias. If the battery voltage is 6V, then calculate the current through the 100 Ω resistance.
Figure 2
Solution
(i) D2 is forward biased.
(ii) It is that current which remains constant even on increasing the p.d.
(iii) Due to potential barrier or barrier p.d.
(iv) Ignoring D2 as it is reverse biased and offers infinite resistance, By Ohm's law,
`I = E/(R∗ + R∗)`
= `6/(50 + 150 + 100)`
= `6/300`
= 0.02 A
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