Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 16)
16.
Determine ID2.
Discussion:
14 comments Page 2 of 2.
Dipen said:
8 years ago
As diode resistance is very small which allows maximum current to pass through it.
So (20-0.7-0.7)/5.6=3.3214mA.
So (20-0.7-0.7)/5.6=3.3214mA.
(1)
D.P.Singh said:
8 years ago
ID2= (20-0.7-0.3)/5600 - 0.3/3300
=19/5600 - 0.3/3300
=0.003392857 - 0.0000909090
= 0.003301948 A .
i.e. 3.302 mA. The answer above has not taken into consideration the current flowing through the resistance 3K3 due to 0.3 V forward bias V of the diode.
=19/5600 - 0.3/3300
=0.003392857 - 0.0000909090
= 0.003301948 A .
i.e. 3.302 mA. The answer above has not taken into consideration the current flowing through the resistance 3K3 due to 0.3 V forward bias V of the diode.
(1)
Choi said:
7 years ago
ID2= (20-VD1-VD2)/5600 - VD2/3300.
=19V/5.6kOhm - 0.3V/3.3kOhm.
=3.393mA - 0.091.
So, 3.302mA.
=19V/5.6kOhm - 0.3V/3.3kOhm.
=3.393mA - 0.091.
So, 3.302mA.
(2)
Habib said:
4 years ago
@D.P. Singh @Choi.
You both are right. Thanks.
You both are right. Thanks.
(1)
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