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Photo sensitive circuit

Bruce Puana

May 7, 2017
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I need help with a problem I cant seem to solve.

I am attempting to create a circuit with the following characteristics. I need a light to remain illuminated, at full voltage, then turn off when it gets dark. I know that's not normally what people try to do....but I need the light on when its light and off when its dark.

I created the following circuit and it worked with one exception. As you can see I used a 9v battery but the output to the lamp was only 3 volts. I need the lamp to be at full power (9-11 volts). I would appreciate any help I can get. I didn't have the transistor listed in the schematic so I used a 2N3904.

Would you recommend different resistors?
Would you recommend a better circuit?
Maybe something with a relay?

Thanks in advance for your assistance.

circuit.PNG
 

73's de Edd

Aug 21, 2015
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Sir Bruce Puana . . . . .

Taking into consideration the max current spec of the 3904, try a 40 ohm in place of the 300 and see how close you can then come to that lamp voltage, within the limitations of your present 3904's choice of utilization.
Is a less power hungry . . . LED . . . instead of the incandescent lamp being permissible ?

Due to different ambient lighting condition changes this definitely will require more sophisticated circuitry to cover at the gradual / fuzzy sunup-sundown transitions .

A heftier long time current suppy than provided by a 9 V battery will found to be needed with the lamp..




73's de Edd
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Maybe something with a relay?
Excellent idea. The relay will provide reliable on/off operation of the lamp. Perhaps you should consider replacing the lamp with a relay having a set of Form C contacts and a 6V DC coil. Use the normally-open contacts to connect the lamp to a voltage source, turning the lamp on when the relay is energized. This will occur when the LDR is illuminated sufficiently to turn on the transistor.
 

Bruce Puana

May 7, 2017
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Excellent idea. The relay will provide reliable on/off operation of the lamp. Perhaps you should consider replacing the lamp with a relay having a set of Form C contacts and a 6V DC coil. Use the normally-open contacts to connect the lamp to a voltage source, turning the lamp on when the relay is energized. This will occur when the LDR is illuminated sufficiently to turn on the transistor.

A friend suggested this circuit. I don't have all the components to put it to the test jut yet. Do you think this would work better? The 6v would be replaced with a 9v. Is R2 330R necessary? Again, I would like the full 9 volts to make it to the LED.
dark-sensor-2.jpg
 

73's de Edd

Aug 21, 2015
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Now you are a' cookin' . . . the distinct relay pull in and drop out thresholds solve the shifting / transitioning dawn and dusk light level situations .
Increasing R3 value can increase your light threshold sensitivity if needed.
Likewise, IF required, the tailoring of the gain of the Q1-Q2 darlington arranged pair can be altered with a resistor up from ground to their shared Coll-Base junction. In which case, it will be for the lowering down of their cascaded gain.
Omit R2 . . . . OMIT R2 . . . . . WHASSAMATTAHYOU ? . . . . want to see D1 LED go off like a FLASHBULB . . .and leave a BLACK flyspeck where the light emitting junction USED to be ?

BTW what is the end result of the use of this applicatiion ?


73's de Edd
 
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Bruce Puana

May 7, 2017
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Now you are a' cookin' . . . the distinct relay pull in and drop out thresholds solve the shifting / transitioning dawn and dusk light level situations .
Increasing R3 value can increase your light threshold sensitivity if needed.
Likewise, IF required, the tailoring of the gain of the Q1-Q2 darlington arranged pair can be altered with a resistor up from ground to their shared Coll-Base junction. In which case, it will be for the lowering down of their cascaded gain.
Omit R2 . . . . OMIT R2 . . . . . WHASSAMATTAHYOU ? . . . . want to see D1 LED go off like a FLASHBULB . . .and leave a BLACK flyspeck where the light emitting junction USED to be ?

BTW what is the end result of the use of this applicatiion ?


73's de Edd

First off thank you all for your input. Please excuse my ignorance. I'm an R/C hobbyist not an electrical engineer. lol

I'm a trying to make an led anti-collision strobe light for a drone.

I'm able to remotely control the drones navigation lights. I want this anti-collision strobe to be able to be turned off with that light. This will allow me to turn off the navigation lights and the strobe for short periods of time to take photographs then turn them back on. I don't want to mess with the wiring of the drone so I want this strobe to work independently of the drones systems....the reason I want to use the ldr. I'll just attach the ldr to the face of the drones light.

When I connect the led strobe directly to the 9 volt it runs just fine. You got me thinking.....obviously this strobe has its own circuitry. When connected to the above circuit the voltage drops to 3 volts and the strobe is very dim. No longer useful as an anti-collision light.

20170507_110241.jpg
 
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hevans1944

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Jun 21, 2012
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The circuit your friend provided is overkill for what you are trying to do: turn on the strobe light if the drone navigation light(s) illuminate the LDR. However it should work as well as your original circuit. The strobe light should be connected to the NO contacts instead of the NC contacts, so the strobe turns on when the relay is energized. And, yes, you can remove the resistor R2 since the "LED" on your schematic is really a "flasher" circuit with its own current limiting.
 

73's de Edd

Aug 21, 2015
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The application info clarified the LED situation . . .up to that time it was just interpreted as a common, puny little single LED.
PLUS up to this point . . . . I was thinking of this as having daylighting being the light source.
 
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Bruce Puana

May 7, 2017
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I'm making assumptions based on my limited knowledge of circuitry. My original circuit ran through a resister before powering the light which gave it a lower voltage/current. Even if I were to simplify the circuit to just the ldr in-line with the power supply it would still reduce the current. So hopefully my last question is this.

If I remove the R2 in the relay circuit, the led would have a direct line to the battery, no resistors, once the relay completed the circuit? Is that correct? No voltage drop?
 

73's de Edd

Aug 21, 2015
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Right-o . . . . Full power through . . . . put on your shades and hope that you are not epileptic prone . . .
 
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