Circuit board for flashlight - need help

Hello to all board members

May name is Simon. I would like to built a head-flashlight witch such bulbs:

https://www.conrad.pl/Żarówka-wskaź...mm,-przejrzysta.htm?websale7=conrad&pi=720183

Supply 3,5V, 200mA. I would like power it through 3 AAA batteries (3x1,2V), connected in parallel. The batteries would be rechargable ones, so I think, there is a need for additional components to stabilize the power supply. I want to apply two red and two green lights, so there must be a possibility to switch between the two groups of lights, turning the whole thing off and on. I have no idea of electronics, but know some fundamental laws. Could someone please help me to design a circuit for such a flashlight?

hmm,,, i think you are going to have problems. As you stated, those bulbs are 3.5V bulbs. the voltage of a battery is 1.2, you are going to have serious under-voltage. Why use the batteries in parallel instead of series? possibly to increase your current capacity? I figure with 3 in series you will get about 2 hours of on time before they become dead. if you make a 3x 2 array of them then youd get 4 hours of on time roughly which is pretty good(but the weight of these on your head might not be so comfy). It might be more comfortable to have the lights on your head and the battery pack and circuit board on the waist. as far as the circuit, you need to know the on resistance of the light, its foreign language to me so i cant read. Id expect the 3.5V gives 200mA but i wouldn't want to make that assumption. if you find the on resistance simply divide the 3.6V by that resistance and it tells you the current, then add resistance in series to get the current you want. for the 2 lights, just put the lights on separate wires controlled by a 3 way switch(common input to A, common input to B, common input to float) make sure the switch can handle the power (V*I) . Also include the switch on resistance when calculating the resistance needed for your desired current draw.

-Pb

Hello

I thought of something similar to that:

**broken link removed**

So I have to connect 3x AA 1,2V batteries in series. The resistance of the light is 200mA, indeed. 3,6V/0,2A yields 18 Ohm, is that right? 18 Ohm x 2 yields 36 Ohm (group of 2 green lights), then another 36 Ohm for another group of 2 (red) lights. What resistors should I put before the lights? How chould I connect the two groups of lights together? How to stabilize the current from rechargeable batteries? (So many questions

You need the 2 lights of same color in parallel not series. and 200mA in each, so 400mA for max brightness. I am not convinced on the 18 ohms, i would try to find an on resistance in a spec, for hobbying this, you could use variable resistor, and decrease the value until you get the brightness you wish. I might be missing something but whats the purpose of the crossing bjt's? the circuit i described above was simply to gave the bulbs in parallel to ground, sharing a load resistor (variable until you determine the value you need) and a switch at the top between the resistor and the batteries.maybe I do not understand the purpose of the two bulbs of same color, should they always be on together? do they blink or alternate? are the red and green ever on at the same time? as far as the rechargeable batteries, i was just going to treat them as normal batteries(like most battery powered devices).

-Pb

OK, I have made a (very) preliminary design of the circuit:

**broken link removed**

I wanted two bulbs of the same color be on together, no blinking, just on, off and switch between the green and the red group. How should I design the power supply? In series, but will 3 AA's suffice for the task? The red and green are not on the same time, there should be a choice between red and green. Can I put a shared resistor for each group?

I read somewhere, that rechargeable have another discharging characteristics than ordinary batteries, not so stable.

Could you please revise my circuit and make a suggestion, how to combine the power supply and the switch?

call the top of each pair node a and node b, connect a and b to separate terminals of a 3 way switch, leave the 3rd pos open(thats your off), connect the common node switched to a variable resistor, and the other side of the variable resistor to the batteries. using 2 bulbs in parallel will sink about 400mA, thats about 1 hour of use before batteries are changed. if you want more usage, you can A) decrease the current input by increasing the resistor, or B) add another bank of 3 AAA batteries in parallel, or C)use bigger batteries like AA, C, or D. the bottom of your simpel schematic would be ground and connected to the ground of the battery array.

a possible switch(just make sure it can handle your current and power requirement)
https://www.mouser.com/ProductDetai...G40-RO/?qs=sGAEpiMZZMvrYgwxMvI/G3SLL%2b1NlgHk

Something like that?

**broken link removed**

I don't know, if the resistors are on the correct side of the leds and where should be plus and minus... Shouldn't be there some kind of capacitor?

to save a resistor id put them on the top between the battery and the switch, but what you have works too. tell me what would the purpose of the cap be? as far as the electrical use of a rechargeable battery they are the same as a normal battery, granted the voltage during discharge is more constant along wit the lines of a LiO battery, and then drops quickly at the end of the cycle. you do not need to 'filter' battery sources since they are the cleanest voltage supplies out(no noise). ps- feel free to click the helpful button if any of my responses have helped

Hi

I have assembled a circuit like that:

**broken link removed**

I wanted to build a flashlight for reading maps and observations, so I needed green and red light. Two bulbs for stronger light. Yes, you have helped me a lot

makowka said:

Hi

I have assembled a circuit like that:

**broken link removed**

I wanted to build a flashlight for reading maps and observations, so I needed green and red light. Two bulbs for stronger light. Yes, you have helped me a lot

Click to expand...

Hi I have another small question. I have modified my circuit to include a ldo stabilizer of type LP2950. Do you know, what current and voltage this device needs for its own operation? Ist the capacitor 100 uF enough?

**broken link removed**

Im not sure why you are so concerned with the voltage being unstable. If you have headroom on your voltage, you could use a voltage regulator to prevent drop out for a time being, but you do not have the excess voltage for this. You are running off batteries so you do not need to worry about voltage noise, or dynamcis, this is a simple dc operating circuit. can you explain why you feel the need for caps and the regulator? My comment about headroom is that if you have say 4V and a 3.3V regulator, your output will be regulated at 3.3V while the 4V supply drops up until the point the supply reaches the regulated voltage, then they will both drop together. Youe supply is only at best 3.6V and so you dont really benefit any from having a regulator, if you added another row of AAA batteries to get 4.8V then I could see the justification in it. Also you are using rechargable batteries, these behave closer to Lion batteries as opposed to alkaline batteries( the voltage will stay flat and then at the end of its charge suddenly drop at a rapid rate. This further reduces your need for a regulator.

-Pb

OK, I understand, but would't the regulator give me a more stable current? If I don't use the regulator AND the resistors before the LEDs, can I use a transistor and resistors behind the Leds, like here?

Again this is a dc circuit, the combined resistance of the bulbs + the series resistor sets the current draw with a fixed voltage. you are trying to use the bjt as a current source with variable resistor at its base to control that current. but it cant sink more current then the resistor at the bottom + the resistance of the bulbs. you might as well just replace the bjt with the tapped resistor you have on its base. I just dont think this circuit requires the added complexity.as long as you have a fixed R and a fixed voltage you will have a fixed current. yes as the voltage drops the current will drop proportionally. If you are worried about dimming you could add a series current regulating circuit of some kind, but it will make the run off occur faster, in other words if you have at the end of the battery life, x time of dimming, without current regulation, and then add a current regulator, your bright life might increase a fraction of time x, but your total time lit ( bright + dim) will decrease, the dimmed time will last maybe half as long or so( think of it in power, the dimming is the voltage and current dropping, so P is dropping which means it is consumed at a lower rate, if you fix I and allow only V to drop, then your P will still be decreasing but at half the rate, so your total time of dimming will be near half).

Ok, I understand, if I add a current regulator, the leds will be "feeded" with constant current at the cost of voltage? Hence the shorter light span?

I think I will leave it the easy way, just 4 resistors before each of the leds or one resistor before each group of leds. I wanted to make something "very superb", but if it isn't necessary, why complicate.. Thanks again a lot, helped me indeed.

It has become popular to use LED's in battery-powered flashlights. They draw less power, so the batteries last a longer time than with incandescent bulbs.

I have had success powering a bright white LED from 3 half-used AA cells (which I seem to have plenty of).

Screenshot: