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dark activated circuit lamp

sahilbudhiraja95

Oct 14, 2013
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i want to make dark activated lamp by use of transistor and ldr what should i do and give better idea about biasing of transistor i am giving supply by charger of 5v and 800ma and i have to convert in to3v and 200ma give idea please
 
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duke37

Jan 9, 2011
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If the LDR does not respond to light, then it is not an LDR.

If the response is not sufficiently sensitive, then an amplifier will be needed.

Measure the resistance of the LDR in bright light and at the light level where you need the light to come on.

In the final design, the LDR should be shielded from the lamp light.

Specify what power supply is available and the lamp power requirement.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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To do this you will need more than a lamp and an LDR.

If you have a circuit, show us. If not, perhaps we can find one for you.
 

GreenGiant

Feb 9, 2012
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I would say tear apart one of those light sensative night lights and use that circuitry to drive a full lamp.
 

KrisBlueNZ

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I drew up a circuit for a dark-activated motor here: https://www.electronicspoint.com/question-mosfet-t258206.html#post1536817. The diode isn't needed because a light bulb is not an inductive load.

You can replace the motor with a light bulb. As duke37 says, you need to keep the LDR away from the light source otherwise the circuit will oscillate.

I recommend using a MOSFET with a low Vgs saturation voltage such as the Fairchild FDP8880 (see http://www.digikey.com/product-detail/en/FDP8880/FDP8880FS-ND/976840). Whatever MOSFET you use, make sure it saturates with 5V gate-source voltage.

If you need to reduce the 5V down to 3V for the light bulb, you can use a series resistor. Assuming your bulb draws 200 mA at 3V (which I think is what you're saying in your first post), the resistor needs to drop 2V with 200 mA flowing through it, so according to Ohm's Law (R = V / I), its resistance needs to be 2 / 0.2 which is 10 ohms.

The resistor will dissipate some power. From the power law, P = V I, power dissipation will be 2 * 0.2 which is 0.4 watts. I would use a 1 watt resistor and mount it off the board. A larger resistor, such as 3W or 5W, will stay cooler because it has a larger surface area.
 

sahilbudhiraja95

Oct 14, 2013
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thangs sir i have done that actually it is led lamp and i have mobile charger having rating 5v and 800ma so i have to light lamp by the help of that so sir i have search so more but could not find right circuit pleasre give me reply
 

sahilbudhiraja95

Oct 14, 2013
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projects

sir i am in second year of btech i want to make some project to start it please sir give me an idea about that its my first project so give me simle topic and knowledge thangful to u
 

KrisBlueNZ

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I have no suggestions for projects for your second year of your course.

The circuit you linked to cannot provide 200 mA of current for the LED. The BC547 transistor is only rated for 100 mA.

My circuit is a lot better - it has a quality called hysteresis (look it up on Wikipedia) which ensures that it will switch cleanly.

You can drive an LED at 200 mA if you put a resistor in series with it. This resistor needs to drop the 5V from the power supply down to the 3V that the LED needs, while passing 200 mA. The calculations are shown in post #5 on this thread. You need a 10 ohm resistor.

Please post the part number of the LED you will be using, and a link to its data sheet.
 

KrisBlueNZ

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OK, that's fine. You're using a Cree XP-E white LED that has a forward voltage of around 2.9V at 200 mA. My calculations in post #5 on this thread are applicable. Your series resistor should be about 10 ohms.
 
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