Simple 2 LED 1 Flashlight circuit w/switches

Thread Starter

EnsiGeetard

Joined Dec 10, 2017
5
Hello, don't know if this is the right part of the forum to ask this, direct me if i'm wrong. I have minimal electrical knowledge.

I am making a prop build, and for this prop I need 2 LED diodes and 1 flashlight. I want everything to run on the same power-supply and have the ability to switch both leds on/off and also the flashlight on/off. As in 2 switches, one for the 2 LED's and one for the flashlight.

The flashlight normally uses 1xAAA battery, and I measured it to consume about 0.6 amps.

The LED's have a voltage of 2.8-3.1 with a draw of 20mA.

Ideally i'd want to use a 2xAA source. I could also use a 1x18650 one but maybe that is a bit extreme and possibly dangerous messing with those cells.

Doing just the LED circuit for me is no problem, but it's when I throw the flashlight and switches into the mix I have some problems understanding how it will work on one power-supply. Can anyone direct me in the right direction? Would really want to avoid cramming 2 separate battery packs in there.
 

Wendy

Joined Mar 24, 2008
23,473
Are you wanting indepentent toggle switches for each function ?Or maybe A push on/push off for each circuit? What complexity are you wanting?
 

Reloadron

Joined Jan 15, 2015
7,573
You also may want to rethink your battery. Typical AA Alkaline batteries are rated 1800 to 2600 mAh with a 50 mA drain. With a 600 mA drain for the incandescent lamp battery life will be short. I would think about 4 X AA using a 6 volt incandescent lamp with less current draw. Also, in addition to Wendy's post do you want LEDs and Incandescent switched together or separate? You can use two SPST or one DPST sort of as pictured.

LED Switches.png

Ron
 

Thread Starter

EnsiGeetard

Joined Dec 10, 2017
5
Are you wanting indepentent toggle switches for each function ?Or maybe A push on/push off for each circuit? What complexity are you wanting?
Exactly like the circuit @Reloadron posted, one switch turns both the LED's on or off, and one turns the flashlight on/off. By on/off I mean toggling it on and off, not a momentary switch. What could also be really cool is to have a 3-way switch which goes off->LED's->LED's+flashlight to condense it into one single switch. But don't know if that last one is doable/feasible.

You also may want to rethink your battery. Typical AA Alkaline batteries are rated 1800 to 2600 mAh with a 50 mA drain. With a 600 mA drain for the incandescent lamp battery life will be short. I would think about 4 X AA using a 6 volt incandescent lamp with less current draw. Also, in addition to Wendy's post do you want LEDs and Incandescent switched together or separate? You can use two SPST or one DPST sort of as pictured.

View attachment 141387

Ron
Thanks for drawing up the circuit, may I ask what software is used? It is very similar to the one i've come up with. What I can't quite wrap my head around is that if I have both of those (LEDs+torch) turned on, if I turn off the torch, won't be LED's be overpowered/affected?

The flashlight is a LED flashlight, I put my multimeter in amp mode (10A), and connected it in series with the flashlight using the 1xAAA battery, lighting up and showing 0.6. It has a theoretical runtime of about 1 hour.

I tried fitting a 4xAA holder inside the only cavity available and it didn't fit. A 2xAA does fit, though. The LED's will primarily always be on and the flashlight just occasionally under use. Is there a way I can make this work with 2xAA? How long of a runtime am I looking at here? For the switching check out the quote above. I've been trying to research all of this the past few days and it's doing my head in haha..
 

Audioguru

Joined Dec 20, 2007
11,248
2 AA cells in series are 3V only when new and soon drop to 2V but a 3V LED will produce no light when the battery has dropped to about 2.7V in a few minutes. If your LEDs are actually 2.8V then a new 3V battery will immediately burn them out so you need a higher voltage and a resistor to limit the current.

Since your flashlight has LEDs and lights from an AAA battery then it must have a voltage boosting circuit inside like a solar garden light. We do not know if it will burn out when it is powered from a voltage higher than 1.5V.
 

Wendy

Joined Mar 24, 2008
23,473
Actually AG is right About the battery life There is a gadget called a joule thief that will solve the problem. I've designed one myself using a 555 IC. What you need is a simple single transistor version, as for the toggle switch you need a 3PST(Three Pole Pole Single Throw Switch . They are very common and cheap,You can also find DPST (Double Pole,AKA 2PST) with Center off with a little research. as for the base schematic. How about 3X AA batteries or 4X AAA Batteries?

tenp1.png

I'll look some link s What I talking about and post them later.
 
Last edited:

Reloadron

Joined Jan 15, 2015
7,573
Thanks for drawing up the circuit, may I ask what software is used? It is very similar to the one i've come up with. What I can't quite wrap my head around is that if I have both of those (LEDs+torch) turned on, if I turn off the torch, won't be LED's be overpowered/affected?

The flashlight is a LED flashlight, I put my multimeter in amp mode (10A), and connected it in series with the flashlight using the 1xAAA battery, lighting up and showing 0.6. It has a theoretical runtime of about 1 hour.
There is no software used, not sure what you are asking? The basic circuit(s) just consist of discreet components, no software. Each device, be it an incandescent bulb. or LED will draw its rated current for a rated or given voltage. Adding loads (lamps) will simply change the demand on the battery or batteries or supply.The switch or switches depending on how you choose to wire them will simply give you different options. Such as one on or the other on or both on and both off.

What is important is that your power source (voltage and current) be able to support your given loads for the amount of time you require.

Ron
 

Audioguru

Joined Dec 20, 2007
11,248
What I can't quite wrap my head around is that if I have both of those (LEDs+torch) turned on, if I turn off the torch, won't be LED's be overpowered/affected?
Overpowered? Yeah sure.
A car battery can produce 600A (!) to turn the starter motor in winter when it is VERY cold. But doesn't the 600A blow up the clock in the car that uses the same 12V but draws only 0.01A?

The electricity In my home can produce 200A. But my 0.008A clock radio is connected to it and doesn't blow up. My stove turns on and off to set its temperature but the clock radio is not affected.

The schematic in post #6 shows the polarity of the LEDs backwards.
 

Thread Starter

EnsiGeetard

Joined Dec 10, 2017
5
Hi all, super thanks for all the help and sorry for the late reply. I decided to go read more to try to understand everything better. I've come up with a circuit that works using 1x18650 battery, 6 LED's in parallel, and 1 variable switched voltage regulator (set to 1.5V) connected to the flashlight assembly. Does anyone see a problem with this?

As you said, @Audioguru, there was an additional circuitboard inside the light. I don't think my 1/4w resistors could handle a high-power chip (CREE), so I played it safe and got the voltage regulator hooked up to the entire assembly. All 6 LED's got their own 100ohm resistor, which I think should be enough to handle 4.2V (fully charged 18650). How will the decrease in voltage (as the battery is being drained) affect the LED's if at all?

@Reloadron sorry if I was unclear. I was just asking about what you used to make the picture of that circuit. Circuit designer of some kind?
 

Audioguru

Joined Dec 20, 2007
11,248
Your voltage regulator has no part number so we do not know its efficiency. It might be wasting 4.2V - 1.5V x the current = a lot of wasted power.
We know nothing about the circuit that is boosting the 1.5V to an unknown voltage that is powering the 6 LEDs. This voltage is pulsing so you cannot simply measure its DC voltage. The LEDs will not dim as the 4.2V battery runs down until its voltage is the minimum input of the unknown voltage regulator you are using.
 

Reloadron

Joined Jan 15, 2015
7,573
@Reloadron sorry if I was unclear. I was just asking about what you used to make the picture of that circuit. Circuit designer of some kind?
The drawing I posted was done in ORCAD Capture but any number of free schematic drawing software options would do about the same. :)

Also my apologies as I somehow lost this thread.

Ron
 

Thread Starter

EnsiGeetard

Joined Dec 10, 2017
5
Your voltage regulator has no part number so we do not know its efficiency. It might be wasting 4.2V - 1.5V x the current = a lot of wasted power.
We know nothing about the circuit that is boosting the 1.5V to an unknown voltage that is powering the 6 LEDs. This voltage is pulsing so you cannot simply measure its DC voltage. The LEDs will not dim as the 4.2V battery runs down until its voltage is the minimum input of the unknown voltage regulator you are using.
Very late so didn't word myself that clear hehe, maybe this crude drawing as I was building/figuring it out will help. I probably butchered some circuit drawing norms as well :oops:



According to the site of the shop I got the regulator from, it is based on the MP1584 (spec-sheet PDF LINK). States 96% efficiency. And now that I look at the specs I realize that the "optimal" input range is 4.5V to 48V. An 18650 starts at 4.2V but drops to, and mostly stays around 3.7V. Is it ok to deviate a bit like my circuit here or should I find something else?
 

Audioguru

Joined Dec 20, 2007
11,248
An 18650 is fully charged at 4.2V and drops to 3.2V when it should have its load disconnected. Then its average discharge voltage is (4.2V + 3.2V)/2= 3.7V.
When the battery is 4.2V then the current in the 2V red LED 22mA. When the battery has dropped to 3.2V then the current in the LED is with a 100 ohm resistor is (4.2V - 2V)/100 ohms= 12mA but will look dimmed only a little. Why do you have a 1.5V old incandescent light bulb?

Today I took apart a hand squeeze charging flashlight to see its super capacitor or rechargeable battery. But it is a cheating piece of junk. It has a switch for LEDs on or off and I assumed it charges the battery and lights the LEDs when it is off when it is squeezed over and over. Inside is a two position switch: one way the charger mechanics connects to the LEDs and the other way the battery connects to the LEDs. The battery is never charged because it is not rechargeable, it is three button alkaline cells.
 

Thread Starter

EnsiGeetard

Joined Dec 10, 2017
5
An 18650 is fully charged at 4.2V and drops to 3.2V when it should have its load disconnected. Then its average discharge voltage is (4.2V + 3.2V)/2= 3.7V.
When the battery is 4.2V then the current in the 2V red LED 22mA. When the battery has dropped to 3.2V then the current in the LED is with a 100 ohm resistor is (4.2V - 2V)/100 ohms= 12mA but will look dimmed only a little. Why do you have a 1.5V old incandescent light bulb?
That's good to know, thanks :) I try not to discharge my 18650's too much.

My drawings may be completely wrong but I my flashlight is an LED flashlight.

Today I took apart a hand squeeze charging flashlight to see its super capacitor or rechargeable battery. But it is a cheating piece of junk. It has a switch for LEDs on or off and I assumed it charges the battery and lights the LEDs when it is off when it is squeezed over and over. Inside is a two position switch: one way the charger mechanics connects to the LEDs and the other way the battery connects to the LEDs. The battery is never charged because it is not rechargeable, it is three button alkaline cells.
Reminds me of those "SSD"'s from china with a USB stick and bolts glued into them for weights haha. That stinks.
 

SxyWood

Joined Oct 8, 2017
25
Not much of what posted already really helps you yet. You already have the flashlight picked out, so post a picture with a ruler beside it and a little more information about it. Then we can truly help you with the parts you need and any circuit to go along with it. I will also add the following, most times the simplest way is truly the best way. So think about using the factory switch for the flashlight and adding just what’s needed for the wanted components (switch & circuitry & LED’s).
 

SxyWood

Joined Oct 8, 2017
25
Not much of what posted already really helps you yet. You already have the flashlight picked out, so post a picture with a ruler beside it and a little more information about it. Then we can truly help you with the parts you need and any circuit to go along with it. I will also add the following, most times the simplest way is truly the best way. So think about using the factory switch for the flashlight and adding just what’s needed for the wanted components (switch & circuitry & LED’s).
Also there are lithium ion cells just as capable but smaller dimensions available. For instance I have about 100 14500 cell’s and they are the same size as a standard AA but have the capability closer to a 18650 than stack of alkalines. A cheap cheap place to buy them is totallywickedeliquid.com Currently they are on clearance for about $1.50 each. Mine I purchased from them are Samsung 800 mah.
 
Top