Hi all I have a string of led lights which ran off a now deceased solar panel. I use them for camping so have a 12v supply I've connected them up to a 3.3v power supply and they draw 160mA What value of resistor do I need to run them off a 12v supply and how much current (or power) will the lights and resistor draw together? Many thanks
Nick
Assuming that you're happy with the 160mA current draw, then your series resistor would need to drop 12-3.3 = 8.7V @ 160mA and would therefore need to be (from Ohm's law) 8.7/0.16 = 54.4 Ohm.
The power dissipation in the resistor would be VxI i.e. 8.7 x 0.16 = 1.39W.
The total power dissipation would be 12 x 0.16 = 1.92W
Therefore, the LEDs would dissipate 1.92 - 1.39 = 0.53W
So, A 56 Ohm 3 Watt series resistor would be the order of the day.
You are, of course wasting the power dissipated in the resistor, so you should consider wiring the LEDs in a more efficient manner or using a switching voltage regulator to get the 12V down to 3.3V or a bit closer anyway!
contains a suitable circuit. These can be found for £1-2 if you keep an eye out, and used for this purpose. I've done exactly this.
Did the old solar panel have betteries in, so it would charge in the day, and then use the batteries at night?
If so, how many batteries did it have inside?
3.3v seems a strange figure - I would have though it would be more like 3.6 (three AA Rechargable batteries) or mabe 2.4 (2 rechargable batteries)
It did have batteries but I'd connected the lights to an available power supply which was putting out 3.3v at the time
So where was the current limit resistor in the original setup? In the LED string you still have or in the solar panel?
What was the current limit of the 3.3V power supply you tested with?
If you connected 3.3V directly to the LEDS with no resistor (if there isn't one in the LED string) then the current reading may not be correct.
MBQ
So where was the current limit resistor in the original setup? In the LED string you still have or in the solar panel?
What was the current limit of the 3.3V power supply you tested with?
If you connected 3.3V directly to the LEDS with no resistor (if there isn't one in the LED string) then the current reading may not be correct.
MBQ
Can anyone point me to an article (for novices) that will explain how to provide power to an LED (from mains and battery sources.) I Can find applications - but I am put off by lack of knowledge.
Right, so that is not necessarily the correct power is it - how many batteries did it have, and what voltage or type were they (If they were normal AA size, then they will almost certainly be 1.2v each)?
Toby...
It is a very straightforward calculation.
Firstly, we need to look at the datasheet for the LED.
Have a look at this example: -
From Electrical/Optical characteristics, you need to check out Vf. This is the forward voltage drop of the diode. In this case a typical Vf = 2V. Now it says that Vf will be 2V +- 0.1V with a maximum of 2.5V, but for the purpose of our calculations here, we can assume a Vf of 2V.
Note that this value of Vf varies according to the type, brand and technology used for the particular LED, so we should always look it up if possible.
Next you need to think about what current you want to run the LED at. Various values of current are mentioned in the data sheet, including
10mA, 20mA, 30mA and 160mA.Under absolute maximum ratings, it shows a maximum continuous forward current of 30mA continuous.
In practice, the LED will start to glow at about 0.5mA and will be at maximum brightness at 160mA. Above 160mA we can expect the diode to be permanently damaged in very short order. Also, you couldn't allow a
160mA current continuously because that would exceed the maximum power dissipation of 75mW. But we are only concerning ourselves with continuous current here, so we should be thinking about a current of between about 5mA and 30mA. Nominally, in electronics school, we would thing about using a continuous current of 20mA for this exercise.So what about this forward voltage, then? What this means is that the forward voltage drop at nominal continuous current (say 10-30mA) will be approximately 2V and is a constant value.
Now, the design principle is that when a reasonable current is flowing through the diode (say between 2mA and 30mA), the diode will "drop" a voltage equal to Vf. If our supply is greater than Vf, then we need a resistor to limit the current flowing in the circuit.
In its most basic form, you have a battery, an LED and a current limiting resistor wired in a series circuit:
The basic principle is:
R _______ -----------|______|--------- | | | | | ____ _____ \ / ___ / \/ LED | -- ---- | Battery / | | V | _____ | ___ | | | | | | | -----------------------------
The absolute key to unlocking this problem is that we choose the current we want to flow in the circuit and this choice informs the value of the resistor we need.
Let's assume we want a current of 20mA to flow.
Kirchoff's Second Law:- "The directed sum of the electrical potential differences (voltage) around any closed circuit must be zero"[1].
This means that the voltage drop across the resistor and the voltage drop across the LED must add up to and exactly equal the voltage across the battery.
Let's say that the battery comprises 2 AA cells each with a voltage of 1.5V.
So the battery voltage is 1.5V + 1.5V = 3V
The forward voltage of the LED at 20mA is 2V
Therefore, the voltage across the resistor is 3V - 2V = 1V.
Now, the value of resistor we choose is going to determine the current flowing in the circuit and, as we know from Kirchoff's First Law[1], the current flowing in the resistor is exactly the same as the current flowing in the LED.
So, all we need to do is work out the value of the resistor from Ohm's law[2].
R = V/I
In this case, the resistor will drop 1V and the current we have chosen is 20mA, so the resistor value we need is 1V/0.02A = 50 Ohms.
As 50 Ohms is not a preferred value (although you can get them as terminating resistors for 50 Ohm transmission lines), you can choose to use a 47 Ohm resistor and get a slightly increased LED current or a 56 Ohm resistor and get a slightly reduced LED current.
If we decided that we only wanted a 10mA current to flow in the LED (making it correspondingly dimmer) then we would still want to drop 1V across the resistor, but now we only want 10mA flowing in the circuit.
We would therefore expect to get a higher value of resistor which resists current flow more and indeed, if we apply Ohm's Law again, we get a resistor value of:
1V/0.01A = 100 Ohms.So, to recap the basic principle is:
If you have two LEDs in series, then you add up the Vfs and then think of them as one LED with a big Vf.
So, three LED's in series will have a Vf of 2V + 2V + 2V = 6V. The same current will flow through all three LEDs (say 20mA). If your supply voltage was (say) 12V, then the resistor value would be 12V-6V = 6V divide by 20mA to give a resistance of 300 Ohms.
If you have (say) three LEDs in parallel supplied via a single resistor, then the Vf remains at 2V, but the current through the resistor is doubled to 60mA (say) of which 20mA will go through each LED. So with a supply of 12V, then the resistor would be 12V-2V = 10V divide by 60mA to give a resistance of 167 Ohms.
This principle works for all types of battery and d.c supply. If a d.c. supply is unregulated, then the voltage may be higher than the nominal value. This means that if you calculate the resistor using the nominal value and then apply a higher (i.e. unregulated) value, then the LED current will be higher and the LED correspondingly brighter.
If you use an AC supply, then can use the (nominal, RMS) value of the supply voltage to do your resistor calculation. But the LED will only glow on positive half-cycles of the a.c. supply, with correspondingly reduced brightness. You then need to consider that the LED will not conduct on the negative half cycles, so the supply will effectively be connected directly across the LED. If the peak voltage of the supply (sqrt2 * RMS voltage) exceeds the maximum reverse voltage of the LED (which you get from the data sheet) then you may well bugger the LED. Therefore, you need to connect another diode (e.g. 1N4148) in inverse parallel with the LED (or indeed connect another LED in inverse parallel) so that the reverse voltage is not exceeded.
HTH DaveyOz
[1]
A lot of responses but no info as to what current your string of LEDs needs. Give them too much and they wont last at all long. Best you show us what circuit lies between the rechargeable battery and the LEDs originally
NT
messagenews: snipped-for-privacy@j36g2000prj.googlegroups.com...
Dave has provided all the background to why things are like they are, for plug in numbers and get answers simplicity
Red ,yellow, green/yellow and amber LEDs all have forward voltage of around 2,3V, it can go a bit above or below but so can usually V supply , its a rule of thumb.
True green, blue, UV and white LEDs have a higher Vf of around 3.3V for thumbs rule, green is the most failure prone colour and are very sensitive to static discharge when handling.
Cheers Adam
Unfortunately the solar panel stopped working due to physical damage and it, and the accompanying circuit board have been binned. I do know that it had two AA sized batteries in it so it's probably safe to assume that it was running off 2.4 volts.
I'ts probably also safe to assume that 160mA is way too much, so redoing the calculations with 20mA gives:-
12-2.4=9.6v 9.6v/0.02A = 480 OhmsPower dissipated in the resistor will be 9.6x0.02= 0.192W which is negligable enough for me to ignore
I'll try that and see what happens. There's no point is spending £15 on a voltage regulator as I can buy a new set of solar lights for less.
Thanks everybody
snipped-for-privacy@j36g2000prj.googlegroups.com...
Wiring in parallel is fine if you give each LED it's own resistor. That also allows you to mix any LEDs in parallel should you so wish.
If you don't have much voltage available then parallel is often the only way to drive multiple LEDs, but the current increases in proportion to the number of LEDs.
If you have plenty of voltage then wire in series, as many as you can with a single series resistor. Current remains the same as for a single resistor.
You can also parallel series strings with a resistor for each string.
MBQ
Just start with your chosen power supply and a few K ohm series resitor, and work downwards until you get the brightness you want, subject to a current limit of, say, 20mA.
MBQ
If you mean 15p you'd be about right.
Well, if there's no circuit data, a clear piccy of one of the LEDs may answer the big question and ensure its got right
NT
Water ingress at all?
few of panel fed strings seen recently have totally omitted sealing edge of panel
Not neccesarily, lot of solar units use a DC-DC invertor to get voltage high enough for blue and white LED.
Used to be a 40 someat as an oscillator now usually black blob on board that handles dark detection from panel brightness as well for switch on, quite neat what can retail for less than a quid.
LEDs won`t light at all until they get minimum Vf , over 3 for blue, white etc
Cheers Adam
snipped-for-privacy@j36g2000prj.googlegroups.com...
Linked LED calc shows wiring LEDs in pralell without any restor per LED.
Difficulty beingh , without resistor per LED or series string, is that one LED may turn current hog leaving other LEDs dimmer, then ends up self heating and into thermal runaway because with resistor sized to drive loads theres the current available.
As MBQ says , series or resistor per LED is much safer.
Cheers Adam
Dave - many thanks - most informative - but a little beyond me. Thanks
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