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voltage divider problem

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khatus

khatus

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Hello guys I have some bulbs like this.



Which takes about 3.8v,0.3A current. Now i wanted to drive 21 bulbs in parallel.But i have only 12v volatge source.So i decided to use a voltage divider . when i calculate the resistor value required for voltage divider i get the following result.



Since each bulb takes 0.3A current and voltage across each bulb is 3.8V.






But when i simulate it in proteus i have seen that increasing the load causes decreasing the output voltage across the resistor R2.i.e, the output voltage drops from 3.8V to 3.76V.But my requirement is 3.8V. My question is there anything wrong in my equation or solving method??

 

Hi,

Why do you want to drive all the bulbs in parallel and then divide down the voltage like you proposed? Efficiency wise, it's poor to do so. Why not consider connecting a number of them in parallel and then having a number of such parallel setup connected in series? That way the loop will not be broken if a bulb goes bad, and the efficiency will be way better.
 

Hi

In your simulation, the more loads/bulps you add in parallel the lower the voltage on R2.
21 bulps in parallel will use 21 x 0.3A = 6.3A.

So your R1should be (12V-3.8V)/6.3A
 

Consider 7 bulbs connected in parallel. 3 sets of such parallel setups connected in series.
 

The calculation is wrong. The voltage divider has to be corrected for the parallel connected load. But apart from this point, it's a completely unrealistic design. Did you notice that voltage divider draws more than 400 A from the battery respectively generates 5 kW losses?
 

R2 is the culprit, and it is not needed assuming that the OP much connect these lamps all in parallel. The parallel connected lamps should serve as R2 instead, and R1 should be (12-3.8)V/6.3A = 1.3 ohm. That way 6.3A current will drawn from the battery.

But efficiency is...I don't know.
 

Don't forget that a bulbs hot resistance is about 10 times its cold resistance.
 

Now i wanted to drive 21 bulbs in parallel.But i have only 12v volatge source.
Click to expand...

May I ask why you wanted to drive the 21 lamps in parallel? Any special reason? Let us know.

Best (most optimum in terms of power or energy efficiency) will be to have groups of 3 lamps in series with a series resistor: the series resistor value can be calculated as (12-3.8-3.8-3.8)/0.3=2 Ohms.

The size of this resistor (power rating) can be calculated as 0.3*0.3*2=0.18W say (round up) 0.5W. This group of 3 lamps with a 0.5W 2E resistor acts like a 12V lamp.

Now you can use 7 such lamps in parallel.

Total power used by the 21 lamps is 21*3.8*0.3=23.9W; Total power used from the battery at 12V will be: 7*12*0.3=25.2W

That means you are using 23.9W out of 25.2W supplied by the battery. Not bad!!
 
EASY FIX TO THIS:

PUT 3 BULBS IN SERIES, THIS IS 12V'S WORTH, NOW DO THIS 6 MORE TIMES, THIS IS ALL THE 21 BULBS

NOW PUT EACH STRING OF 3 ACROSS THE 12v ALL THE BULBS ARE GOING NOW - ALL 21 OF THEM - FROM A 12v SUPPLY - AT FULL BRIGHTNESS

NO VOLTAGE DIVIDER NEEDED...!!!!!!!
 

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