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Newbie question about simple LED circuit

hpops

New Member
Hi

I know very little about electronics so I apologise in advance!

I am trying to light a diorama that I have built with three led's. Two white and one slow colour changing. I have wired up the circuit and calculated the resistor value (think I got it right anyway) and powered the circuit with a 9V battery.

It worked fine, but only lasted about two weeks before dimming and then dying :(

Is there another battery that I could use to poser the circuit or any other modification I could apply that would enable it to last longer?

I am restricted on space for the battery/circuit to fit in the diorama so I want to keep it as small as possible.

Many thanks for your help :)
 

hpops

New Member
I used a Duracell 9v standard battery. I don't have a schematic of the circuit I'm afraid!!! I've just soldered two white and one slow flashing led in series with a (I think) 330ohm resistor. I put the resistor between the positive out from the battery and the first led.

Does that help? If not, I can take the wiring out of the diorama but it's all glued in atm!!

Thanks for your help :)
 

Visitor

Well-Known Member
Are the LEDs in series? In parallel in series with a single resistor?

Do you have a data sheet for the flashing LED?
 

audioguru

Well-Known Member
Most Helpful Member
The datasheet for the Duracell 9V alkaline battery shows that its small size and weight produces a low amount of power, about 700mAh at a few mA of current.
Your white LEDs are about 3V each and the red/green flashing LED is about 2V so the total of 8V leaves 1V across the 330 ohm resistor which is a current of only 3mA.

For the battery to last longer then the current must be less which makes the LEDs appear dimmer. Increasing the resistor to 1k will allow the battery to last 3 times longer.
 

Visitor

Well-Known Member
Warning: A thought experiment follows. A LiIon battery could be a decent solution....but a USB power bank might be a more practical answer.

I was curious, so I punched out some numbers.

A 9v alkaline battery, with 600 – 700 mAh capacity, has a volume of 22.5 cm³, an energy density of around 28 mAh/cm³.

A 1000mAh LiIon cell has a volume of 10 cm3, an energy density of 100 mAh/cm³.

A 2000mAh LiIon cell has a volume of 15cm³, an energy density of 133mAh/cm³.

So a 2000 mAh LiIon cell has about 3× the capacity of a 9 volt battery in about half the volume.

But the output voltage of a LiIon cell is only 3.7 volts, so the LEDs would have to be in parallel, each with its own resistor. Therefore, the total current draw would be 3× that of using a 9v battery, canceling the energy density gain.

A 4000mAh battery should be about the same size as a 9 volt battery. The net effect of increased current density vs increased current draw would be about twice the life of a 9 volt battery. Of course, the profile of the LiIon cell is different than a 9 volt battery, and a charging module would be required.
 

Diver300

Well-Known Member
Most Helpful Member
The expensive lithium thionyl chloride batteries will give you about twice the life.
https://uk.rs-online.com/web/p/9v-batteries/7781080/
However, the voltage of one of those will be more like 10.5 V when running, while the alkaline will be nearer 8.5 V when 3 mA is being taken. That will make a huge difference to the voltage across the resistor, so you will get more current, much more light but possibly even less run time.

To get the current the same, to give the same light output and a longer run time, you will have to increase the resistance to at least 1 kOhm

With a lithium thionyl chloride battery and the same current, I would expect twice the run time, and constant light output until it is exhausted, when it will go from full brightness to dim or off in an hour or so.

There is one odd characteristic that I have seen in lithium thionyl chloride batteries. I am not sure if it applies to these. The internal resistance can be higher when new than when they have been slightly used. You can see this effect in the discharge graphs here:- https://docs.rs-online.com/9168/0900766b8070972c.pdf where the votlage goes up initially at higher loads.

For your application it could mean that the LEDs will get brighter over the first 5 - 10 minutes of use of a new battery.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
There is one odd characteristic that I have seen in lithium thionyl chloride batteries. I am not sure if it applies to these. The internal resistance can be higher when new than when they have been slightly used.
Their internal impedance is very high anyway, hence the common need to have supercaps in parallel with them to provide any high current pulses that might be required. They are commonly used in GSM/GPRS logger applications, and they don't usually work without supercaps.
 

hpops

New Member
So if I were to combine two of the suggestions above and use a lithium thionyl chloride battery and increase the resistor to 1k would this be a significant improvement? If so any ideas how long it would provide power for?

I never imagined it would be so complicated to light up 3 leds!

Thanks for your help guys :)
 

Diver300

Well-Known Member
Most Helpful Member
So if I were to combine two of the suggestions above and use a lithium thionyl chloride battery and increase the resistor to 1k would this be a significant improvement? If so any ideas how long it would provide power for?

I never imagined it would be so complicated to light up 3 leds!
You should get around twice as long from the lithium thionyl chloride battery. It is rated at 1.2 Ah compared to the approx 0.6 Ah of an alkaline battery.

It's not all that complicated. You got two weeks of operation from your initial attempt. The change in resistor is needed because of the change in voltage, plus the fact that the circuit with three LEDs and a resistor will be very sensitive to voltage changes.

You could also look at lithium manganese dioxide batteries, like this:- https://uk.rs-online.com/web/p/9v-batteries/7845992/

They are similar prices and capacity to the lithium thionyl chloride ones, but slightly lower voltage so you won't have to change the resistor.
 

audioguru

Well-Known Member
Most Helpful Member
Of course it is complicated to light 3 LEDs brightly with a tiny battery for a long time.
1) If the LEDs and resistor are all in series then there is not enough voltage for the LEDs to light up for a long time. The LEDs will be dim.
2) If the LEDs are in parallel then 2/3rds of the battery voltage is wasted heating the current-limiting resistors.

I made some LED chasers that have 10 LEDs in a circle and they have a circuit that blinks each LED for a short time. Then the two or three AA battery cells last for 3 months. The batteries are AA so that they light the LEDs very brightly. I could have used smaller AAA cells and made the LEDs dimmer or have the batteries replaced more often. You can add a blinker circuit to allow your little 9V battery to last a lot longer.
 

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