A super simple torch
There was a time when I never had a torch handy when I needed one. Being the sort of person who has bags of LEDs lying around, why not make myself a few torches just for the fun of it so there’s always one handy?
This also can be useful for my other hobby, Astronomy, where I could make a couple of red ones so I don’t ruin my night vision: Background on this.
I can also use this post to document the two simple equations to use to make sure you don’t blow up your LEDs.
- 1 x 9V switched battery box – £2.19 from Maplins
- 2 x LEDs (your choice of colour) – I used these,
- 1 x 220Ω resistor
- Wire cutters
- 5mm drill
- Hot glue
I saw these switched 9V battery boxes in Maplins and thought they would be ideal as it already contained a switch built in and was small enough to be easily pocketable
So starting with that, I then need an LED. If you look at the LED linked in the parts, the points of interest are the forward voltage (2.4V) and the max forward current (100mA). 100mA is way too much though so I aim to run the current at around 20mA, which is mentioned on that page in the typical light output.
Running a single 2.4V LED from a 9V battery means that I’d be dropping the other 6.6V across the resistor though, so I opted to use 2 LEDs, which should keep the power dissipated by the resistor down (even though this wouldn’t be an issue here really).
This is about the simplest circuit going:
The role of the resistor in the circuit is to limit the current running through the LEDs. Without this, you will immediately blow any LED you try to connect. We have 2 LEDs in series using 2.4V each and a 9V battery, which means that the resistor will drop the remaining 4.2V across it.
We want the current to be limited to 20mA (=0.02A). Here’s where we use the first simple formula, one of the first you learn in Electronics: V = I * R (Voltage = Current x Resistance).
We are looking for the value of resistor to use, so re-arranging R = V / I. Entering the values gives our required resistance of 4.2V/0.02A = 210Ω. Resistors come in set sizes, the closest I have to this is a 220Ω one.
Working this back into the equation, I see that this should give me I = V/R = 4.2/220 = 19.1mA. Sounds about perfect to me, 220Ω it is then.
The only other thing to work out here before getting the soldering iron out is to make sure the resistor can handle the power. We shall use this equation, P = I * V (Power = Current x Voltage). So power in the resistor is 0.0191 * 4.2 = 0.08W (Watts).
Seeing as I’m using a resistor from this pack, which are 0.25W resistors, we are ok.
Note that if I was going to use a single LED, that would have given a 330Ω resistor at 0.13W so it would have still been ok.
This was very quickly thrown together. I marked up and drilled two 5mm holes in the front of the box and push the LEDs through from the back.
The LED legs were then bent towards each other (anode from one, cathode from the other), soldered together and cut to size.
The other legs were bent to the outside of the box, the anode of the first LED soldered to the resistor then to the red wire coming from the switch (a little notch was cut next to the battery connector to allow me to pull the wire through into the box), then the cathode from the last LED soldered to the black wire from the switch.
Hopefully you can see that in this photo
and here’s a shot of the LEDs poking out of the box
Everything was then stuck down with some hot glue, then covered with some electrical tape to insulate the bare wires.
Because I tucked everything as neatly as I could into the corners, you can still just fit a batter into the box, which of course was what it was designed for in the first place and here it is working!
I made this particular torch for use whilst playing with my telescope, but these LEDs are WAY too bright, oops. Next time I will get some much dimmer ones, ah well.
One last thing, how long should I expect this battery to last?
An Alkaline 9V battery typically has a capacity of 565mAH (milli Amp Hours), so let’s use that. If I divide that value by the limiting current we set of 19.1 mA, we should expect a continuous usage of 565 / 19.1 = 29 and a half hours!
Not bad at all 🙂