Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Save energy using LED in series

Status
Not open for further replies.

nepdeep

New Member
Hi I have the LEDs set up in circuit as follows: Very simple series and parallel. Now what I assumed is that the LEDs need 20mA current to give max light.
My question is .....if they are in series the two leds consume same current to give maximum brightness....if they are in parallel the two leds consume twice as much the current in series....so my analogy is that the battery lasts for longer time in series LEDs but the amount of brightness..for both the circuit is the same...pleas throw me some light if I am missing something.. IMG_0705 (Medium).JPG
 
Hi,

How is that a double post? That is in another forum isnt it?

But anyway, the efficiency is highest when the combined LED voltage is closest to the power supply voltage. But there are practical problems that come in because usually the source voltage varies a little so there has to be some headroom for the voltage in case it does down. So the idea is to get as close to the voltage level as possible and still have enough headroom for whatever the source and LEDs might do.

For an example, if you had an LED that was exactly 1.5v and stayed like that and you connected it to a 1.5v source, you would get max efficiency. If you had a 3v source available however, then you would need two LEDs in series to get that same efficiency as their combined voltage would be 3v. But in practice we cant get that close because the LED voltage varies a little and sometimes the source does too, so we might have to go with a 3.5v source and two 1.5v LEDs plus a series resistor to make it work right.

So if the voltage is high enough to allow more LEDs in series then yes you save power when using a source that is not changeable. However, if you use 3.5v and two 1.5v LEDs and a series resistor set for 20ma, you could instead use two sources of 1.75v and two smaller series resistors with two LEDs and you'd get the same efficiency. So it really depends on what kind of source you already have to work with or what source you can construct.
 
Hi I have the LEDs set up in circuit as follows: Very simple series and parallel. Now what I assumed is that the LEDs need 20mA current to give max light.
My question is .....if they are in series the two leds consume same current to give maximum brightness....if they are in parallel the two leds consume twice as much the current in series....so my analogy is that the battery lasts for longer time in series LEDs but the amount of brightness..for both the circuit is the same...pleas throw me some light if I am missing something..View attachment 74095

Hello nepdeep,

I try to keep it as simple as possible.

Well, I think you should consider the term "Power" as well (Power= Voltage X current) when you talk about the life of the battery here.
The useful life of a battery when connected to a Load is dependent on the power the load tries to consume from the source.
So say you have two LEDs and each LED needs say 1V and 20mA to be completely lighten (the lEDs mostly do not works by 1V though, and in real life you need a Joule Thief circuit!).
Ok because we now know that the life of a battery is related to the POWER the load attracts from it, and again we know that "Power = Volatge X current" then we can play with the 2 LEDS and the source if needed to see what it comes up:
1:LEDs in series:
The power captured from the battery by the LED's is: 2V X 20mA =40mW

2:LEDs in parallel:
the power captured from the battery:
1V X 40mA = 40mW.

So when you have two LED's and you want to experience a simallar brightness then the power consumed by them is simillar whether they are in series or in parallel. If the LED's are in parallel then you must feed them by a battery having the same voltage as EACH LED, more or less (but the current taken from the battery is DOUBLED).
If the LED's are in series then you must feed them with a battery having a voltage of say DOUBLE of EACH LED needs (here the voltage must be doubled so that each LED takes its needed voltage). So the whole power consumed by the loads form the source in both cases is simillar. And the battery life would not change so much!

Hope that helps:)
 
Last edited:
Nobody makes a 1V visible LED and nobody makes a 5V battery.

Instead think of 3.5V white LEDs and a 9V alkaline battery.
Two 3.5V LEDs in series use 7.0V and the current-limiting resistor has a 2V voltage drop. But the battery voltage drops to 7V in about 4 hours and the LEDs will slowly dim for the 4 hours then they will turn off.

If two LEDs and their current-limiting resistors are connected to the 9V alkaline battery then the current is doubled but the battery drops to 3.5V in 19 hours when the LEDs will turn off. 19 hours is much longer than 4 hours.
 
Okey, I am sorry but I am still a bit confused. Here is my calculation.(and yes in reality the voltage drop across the LED is more than 1. I just assumed for the sake of simplicity of calculation)
I assume the 5V power supply in both cases
When the LEDs are in series:
Vin=5V
Vled=2 X1[V]=2[V]
Vresistor=3[V]
Current=20[mA]
R=(3/20)*1000=150[ohm]
Now, Power consumed by LEDs= 20 *2 = 40[mW]
Power consumed by resistor = 3 * 20[mA]=60[mW]
Total power consumed in series = 100[mW]


When the leds are connected in parallel:
Vin=5V
Vled_1=1[V] , Vled_2=1[V] ,
Vresistor_1=4[V] , Vresistor_2=4[V]
Current=20[mA] each
R=(4/20)*1000=200[ohm]
Now, Power consumed by LED_1= 20 *1 = 20[mW]
Power consumed by resistor_1 = 4 * 20[mA]=80[mW]
Now, Power consumed by LED_2= 20 *1 = 20[mW]
Power consumed by resistor_2 = 4 * 20[mA]=80[mW]
Total power consumed in parallel (for same amount of light) = 200[mW]

I just want to know if my assumption is correct or not. Uncle scrooge made it quite clear about the life time of the battery but he didnot say direct yes to my question. I want the answer YES or NO please. Till now I rely on uncle scroog's explanation and assume that he supports my arguement. :) have nice weekends
 
Last edited:
thanks Marl.
You made it clear about the efficincy and few other factors. You slightly support my hypothesis but never said it directly. Is my assumption correct in ideal case?

What 'hypothesis'? - it's standard technique to place LED's in series, you get less loss from a higher supply voltage (essentially because there's only one series resistor, instead of a separate one for each LED in parallel). In either case the actual design depends on the number and type of LED's and the supply voltage available - essentially LED's are current driven, and a constant current source is the best method of feeding them (and again, LED's in series are best for that).
 
OK.....(essentially because there's only one series resistor, instead of a separate one for each LED in parallel)...this explains something....it must be as simple as piece of cake...but just eating my brains.... :)
 
OK.....(essentially because there's only one series resistor, instead of a separate one for each LED in parallel)...this explains something....it must be as simple as piece of cake...but just eating my brains.... :)

As with many (most?) 'problems' the correct application of ohms law clears everything up.
 
If you use a power supply that holds up its voltage pretty well then when you connect the LEDs in series the total power used is low.
But if a battery is used then the LEDs soon turn off as the battery voltage drops.

If you use a battery that drops its voltage as it runs down then connect each LED with its own current-limiting resistor so that the LEDs continue to produce light as the battery voltage runs down.
 
Your math is correct.

Think about it like this: 5 X .04 and 5 X .02.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top