Electronic Devices - Diode Applications - Discussion

Discussion Forum : Diode Applications - General Questions (Q.No. 16)
16.
Determine ID2.

6.061 mA
0.7 mA
3.393 mA
3.571 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

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.
(2)

Muhammad Ali said:   1 decade ago
Current always adapt shortest path. It will never flow through 3.3 k resistance.
(1)

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.
(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.
(1)

Habib said:   4 years ago
@D.P. Singh @Choi.

You both are right. Thanks.
(1)

SREEKALA K S said:   1 decade ago
THE CURRENT WILL FLOW THROUGH THE LOW RESISTANCE PATH THAT IS THROUGH THE D2 ONLY , ID1=ID2
ID2=(20 - 0.7 - 0.3)/5.6 K

Niranjan Reddy.P said:   1 decade ago
How to find the values of 0.7 & 0.3?

Arul said:   1 decade ago
Suppose if ask ID1 how to calculate?

Brojo said:   1 decade ago
It is the minimum amount of voltage which is required to operate components like for silicon diode it is 0.7 V while for Germanium Diode its 0.3 V.

Suprith said:   10 years ago
ID1 is Same as ID2 because they are connected in series.


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