Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 19)
19.
Determine the average value of the current through the load resistor.
Discussion:
8 comments Page 1 of 1.
M G Ayag said:
2 years ago
I max= (10-0.7)/4000 =2.325 mA,
I ave = I max( 2/pi ) = 1.48 mA.
So, the right answer is D.
I ave = I max( 2/pi ) = 1.48 mA.
So, the right answer is D.
Ejaay said:
7 years ago
@Jeni.
Actually, you can analyze the current by doing some current division in the node after it gets through the diode. (10-0.7)/2K = 4.65mA.
But since the current is divided into two 2k resistors, The current is now 4.65mA/2 = 2.325mA. Then apply the formula for average full rectified current in Vo using (2*I/π) = (2*2.325)/(3.1415) = 1.479 mA.
Actually, you can analyze the current by doing some current division in the node after it gets through the diode. (10-0.7)/2K = 4.65mA.
But since the current is divided into two 2k resistors, The current is now 4.65mA/2 = 2.325mA. Then apply the formula for average full rectified current in Vo using (2*I/π) = (2*2.325)/(3.1415) = 1.479 mA.
(2)
Jeni said:
8 years ago
How to find the load resistor?
Rahul said:
1 decade ago
Iav = Ipeak/1.414 = 1.67 ans.
Litun said:
1 decade ago
pi means 3.14 it is the numeric value of pi or 22/7.
Azhar said:
1 decade ago
What is pie?
Kothapeta siva krishna said:
1 decade ago
During +ve half cycles d1(diode which is on right side)will conduct and then I throw load resistor is (10-0.7)/4k = 2.375ma.
lly during -ve half cycles d2(left side diode will conduct ).
Then current is same.finally if we observe the o/p waveform across load resistor is appears as full rectified o/p with max as 2.375. For this waveform avg value is (2*max i)/pi = (2*2.375)/pi = 1.480.
lly during -ve half cycles d2(left side diode will conduct ).
Then current is same.finally if we observe the o/p waveform across load resistor is appears as full rectified o/p with max as 2.375. For this waveform avg value is (2*max i)/pi = (2*2.375)/pi = 1.480.
Sanoj said:
1 decade ago
Avg. Current = 2I/pie (here I=2.325 m amp).
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