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Converting solar post caps from battery power to low voltage.

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richtad

New Member
Hi everyone,
I hope someone can help.
I recently purchased two solar caps that light for posts that are @ the top of my porch stairs.
They don't get enough sunlight.
I'd like to convert them to low voltage lighting.
Each one(there's two) is operated on a 1.2v, 600mAH Ni cad battery.
There's a small circuit board with the led, a resistor and it also looks like it may have some sort of a voltage regulator.
There's nothing on the led to tell me what the wattage is.
Anyway I can convert this, if so , how??
I would appreciate any advice anyone can offer,
Thanks
Rich
 

MikeMl

Well-Known Member
Most Helpful Member
Why not just gut the fixture and install a ultra-bright white LED in series with a current-limiting resistor. An old 5V 1A wall-wart type cell phone charger would run a dozen or more fixtures...
 

richtad

New Member
Hi Mike,
Thanks for the answer.
I found a supply that has exactly the specs you quoted.
That's good, the bad part is that even if I knew where to get the parts you specked out, how would I wire the circuit??
Sorry about that, I'm a mechanical guy. I've drawn a million schematics in my career, but, in most cases, I didn't know what they did. Would it be possible to get a supply to match the existing
battery output? Or is that a stupid question.
Thanks anyway,
Rich
 

MikeMl

Well-Known Member
Most Helpful Member
Ultra-bright white leds have a (nearly-constant) forward voltage drop of ~3.3V. Their brightness is more-or-less proportional to the current through them (they are a current operated device). For your application, I'm guessing that 20mA would be more than adequate.

To set the current through a LED starting from a fixed voltage (the 5V wall-wart), you put a resistor in series with each LED (it is a series circuit, it doesn't matter if the resistor is in the anode side or the cathode side). The resistor drops the difference between 5V (starting voltage) and the forward voltage of the LED (3.3V), so 5 - 3.3 = 1.7V

To get 20mA through the resistor, apply Mr Ohm: R = E/I = 1.7/0.02 = 85Ω. (buy 84.5 or 86.6Ω, standard 1% values).

The power dissipated by the resistor is P=E^2/R = 1.7*1.7/85 = 0.034W, so buy a resistor rated for at least 0.25W

Here is a simulation of what you are building. Note that the current through the LEDs in both circuits is identical...
 

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richtad

New Member
Thanks again,
Of course I have some questions.
On the schematic, I'm assuming v1 and v2 are the power supply.
Between the LED and the resistor, there a line with "a" on one and"k" on the other.
is this voltage? If so, do I tap it off of v1 and v2?
Also, where do I tie my grounds to?
Regards,
Rich
 

MikeMl

Well-Known Member
Most Helpful Member
Think of "a" and "k" as named wires. This is done so the simulator can print a voltage such as V(a) or V(k).
The ground symbol is where V(a) and V(k) are referenced. You do not need to "ground" your simple circuit.

In fact, if you use the suggested old cell phone charger, it puts out +5V measured with respect to its other wire. It is completely isolated (for safety) from the prongs that plug into the house receptacle. The LED illumination can remain isolated as long as the two wires from the cell phone charger connect as shown in place of the "voltage source" V1 or V2 (whichever version of the circuit you build).

The cell phone charger can run several Leds, as long as you connect the resistor-LED pairs in parallel. See below
 

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MikeMl

Well-Known Member
Most Helpful Member
Afterthought... This draws so little power, I would just leave the cell phone charger plugged in 24-7. It is not worth using a daylight sensor, or timer...
 

richtad

New Member
Mike,
I can wire this circuit up with no problem.
I'll stop at the local radio shack and get the parts.
I was going to hook this up to an existing 120v electric eye sensor that controls my outside lights.
I'll let you know how things go,
Thanks again,
Rich
 
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