Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 1 (Q.No. 7)
7.
In figure v1 = 8 V and v2 = 4 V. Which diode will conduct?
Answer: Option
Explanation:
D1 will conduct and the output voltage will be about 7 V. Therefore D2 will be reverse biased and will not conduct.
Discussion:
32 comments Page 2 of 4.
Shivanand said:
9 years ago
The diode which experiences the maximum potential differences between the two nodes, that diode will conduct. As in this case between v1 and ground Diode D1 experiences the potential difference of 8v but D2 experiences only 4v. So the diode D1 conducts.
Dhana said:
9 years ago
Suppose I will put same voltage at two nodes.
Two diodes are conduct or not please clarify.
Two diodes are conduct or not please clarify.
Vijay said:
9 years ago
According to me both diodes will conduct.
Raushan said:
9 years ago
Assume both diodes are in forward bias, then after doing node analysis, the node at the intersection of 3 resistors, has the potential of 5 V (assuming cut off voltage 0.7V), which contradicts, so only one diode is in forward bias.
Sharma said:
9 years ago
Could anyone explain clearly how diode (D1) alone conducts and not the other?
Ramkamal said:
9 years ago
V1and V2 is not mentioned they are AC/DC. Secondly practically diode voltage drop 1v then both are forward biased.
Lorlor said:
8 years ago
The answer is B but the Vo is equal to 5V only.
Kiran said:
8 years ago
@All.
First check current through both the diodes and ACC to the direction of current you can identify which diode is working.
First check current through both the diodes and ACC to the direction of current you can identify which diode is working.
Surya said:
8 years ago
Both the diodes are conducting because p>n
p:positive terminal.
n:negative terminal.
This concept from pulse and digital circuits.
p:positive terminal.
n:negative terminal.
This concept from pulse and digital circuits.
Nikkie said:
8 years ago
First take: I(1) + I(2) = I.
where I(1) = (V-8)/1k ohm.
similarly I(2) = (V-4)/1k ohm.
And I = V/10k ohm.
Next solve for 'V' ( USING KCL AT NODE V).
You will get V = 40/7 which is 5.7 something.
Hence, d1 will conduct and d2 will be reverse bias and won't conduct.
where I(1) = (V-8)/1k ohm.
similarly I(2) = (V-4)/1k ohm.
And I = V/10k ohm.
Next solve for 'V' ( USING KCL AT NODE V).
You will get V = 40/7 which is 5.7 something.
Hence, d1 will conduct and d2 will be reverse bias and won't conduct.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers