doubt in rc circuit about charging time

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Bhuvanesh Nick

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is my above circuit correct

!my idea is to to on the led after some times..to do so how to calculate the time
 

charging equation of RC series circuit.
vc = v(1 - e-t/rc)

In this case capacitor will never charge more than 0.7 V that is LED's forward voltage drop.
 
vyomax said:
charging equation of RC series circuit.
vc = v(1 - e-t/rc)

In this case capacitor will never charge more than 0.7 V that is LED's forward voltage drop.
Click to expand...

from what i learned capacitor charges 60% of supply voltage and then it supplies for led .am i right
 

vyomax said:
charging equation of RC series circuit.
vc = v(1 - e-t/rc)

In this case capacitor will never charge more than 0.7 V that is LED's forward voltage drop.
Click to expand...

Actually that equation works only upto the point where LED current is very very low. Once the voltage reaches the diode knee, this is no longer valid.
Secondly, I know of no LED whose knee is as low as 0.7v. LED's are typically 1.6v .. 3.2v depending on the colour.
 


Actually u r right. LED produce around. 1.3v
I just considered it as a pn junction diode.

- - - Updated - - -

Bhuvanesh Nick said:
from what i learned capacitor charges 60% of supply voltage and then it supplies for led .am i right
Click to expand...



The equation is just for reference
in your circuit it is not valid for all t (time)

Capacitor will charge only upto forward drop across LED. remain there forever
It will not charge upto 60%
 


Sir,capacitor has infinite resistance.so in that way,why dont it charges first and then supply it to led
 

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