Question on Resistors for use with 5050 LED Strips

Question

I bought a set of red and yellow 5050 waterproof LED strips for my motorcycle. I'm looking to add them to my topcase so it has running, brake and directional lights on it. Right now, there are no lights on the topcase and I feel like it's hard for people to see the bike's lights below the topcase.

These are the ones I ordered - they are the typical strips with 3 LEDs per segment, that you can cut every 3 LEds.

http://www.amazon.com/SUPERNIGHT-16-4ft-Waterproof-Flexible-Multifunctional/dp/B00BMHP960/ref=pd_bia_nav_t_3?ie=UTF8&refRID=0T7N95KYJ4TASS7Y4CC4

I want to connect about 54 LEDs in 6 separate strips to my topcase (see attached). Each side of the topcase will have a set of 3 strips: 1x12 LEDs, 1x9 LEDs, and 1x6 Leds.

I was thinking of wiring them up via a serial-parallel framework, using resistors to both my brake light and running light so the LED strips will be dim for running light purposes, and then go very bright when the brake is engaged.

For each strip (+), I was planning to have it connected both:

• the brake line with a directional diode
• the running light with a resistor and directional diode

I've been researching LEDs for a couple days now but am finding most of the stuff on the internet is for individual LED diodes - not for strips. I'm confused because I believe the strip segments have resistors built into them already - hence why they can take a full 12v input. If they didn't have resistors, they would overheat, no?

My big question is - what resistance resistor should I use to dim the LED strips. From looking around the net, it seems like 1/4 W would maybe work - but I have no math to back that up.

Furthermore, I'm confused by how LEDs react at their forward voltage rates. For example, I believe the forward voltage rate for red 5050 LEDs is 2.0v. If supplied that voltage, does the LED light up but just barely? IE: Is that the dimmest? And is that for one LED or for the one segment of the strip (3 LEds)?

Any help on this front would be appreciated. THank you all.

Answer 1

Yes, LED strips have resistors built in.

Resistors are primarily rated by ohms (Ω), and secondarily by power dissipation. Standard component resistors are rated at 1/4W dissipation. That just means that you can't put more power through them without burning them out.

To figure out what to put in the resistor block of your circuit we'll just use Ohm's law (V = I R):

1. You should measure with an ammeter to be sure, but 5050 LED strips are typically assembled with resistors to draw 1 amp per meter at 12V, which means that the strips have built-in resistance of 12Ω.
2. Decide how much "dimming" you want. If you want them at quarter brightness, and noting that luminosity is proportional to current, then quadruple the resistance by adding a 36Ω resistor, which will bring total resistance to 48Ω and reduce current to .25A/m.
3. But you can't just pop any 36Ω resistor in front of the load. Suppose you have 1 meter of strips on your helmet: That would mean that you have to dissipate .25A*12V ≈ 3W of power. Therefore, either find a ≥ 36 Ω 3W resistor, or else use twelve standard 3 Ω 1/4W resistors in series, or twelve standard 432 Ω 1/4W resistors in parallel, in order to dissipate the power used by a meter of those LEDs.

Answer 2

Data provided does not say how much current per LED or group of 3. 5050 LEDs usually have 3 dies and these are usually run in parallel. They MAY have got clever - you can check.
You can measure current per group of 3 - I will assume 60 mA and you can adjust.
I will assume 6V across 3 LEDS and 6V across a resistor - you can measure Vresistpr and adjust.

1 die = 20 m max usually.
3 in parallel = 60 mA

1 red LED = 2V 3 inparallel = 2V still
3 x 5050in series = 6V.
At 12V, 609 mA R = V/_ = (12-6)/60 mA = 6/0.06 = 100 Ohms = 100R

To get half current you need R = V_res/I = 6V / 0.03 = 200 R

So add 100R / N for half brightness for N x 3 LED strip.

In general Rseries effective
=(Vsupply - V_3_LEDS_series) / I
= (12-6)/I = 6/I

Rseries effective = 6 / (I_per_3_LEDS x N) ..... [A}
(where N = sections of 3.) Reduce current to any lesser value by increasing Vseries by adding Rnew = Rseries_effective x (1-P)/P [B] where P is the fraction of old current you want
eg for I = 40% of old current P=0.4
Rnew = Rseries_effective x (1-0.4) / 0.4 = 1.5 x Rse

So for eg 3 sections where I=180 mA and Rse = 6/.18 =~ 33R
Adding 1.5 x 33 = 49.5 ~= 47 R (std value) in series will reduce current t o 40%

You can check V_3_LEDS_series with a meter with 12v feed.
You can measure I per section of 3
You can then design as desired using [A] & [B] above

==

Answer 3

Furthermore, I'm confused by how LEDs react at their forward voltage rates. For example, I believe the forward voltage rate for red 5050 LEDs is 2.0v. If supplied that voltage, does the LED light up but just barely? IE: Is that the dimmest? And is that for one LED or for the one segment of the strip (3 LEds)?

Here's a typical normal red LED forward voltage-current characteristic: -

Look at the left graph first. With 1.9 volts across the LED the current will be about 30mA. Now look at the right hand graph - at 30mA the luminous intensity is 1.5 times greater than it would be with 20mA flowing (about 1.85 volts across it).

For this generic LED above you decide what luminous intensity you wish to have, see what current it needs and that tells you what voltage you can expect across the LED. Then, if your power supply is 5V you can roughly work out what resistor value you need.

Let's say you want to power it at 20mA; the forward volt drop is 1.85 volts and a series resistor connected to 5V would need to drop 3.15 volts (at 20mA).

The resistor value to choose is simply derived by ohms law 3.15/0.02 = 157.5 ohms. You'd probably choose a 150 ohm or 160 ohm resistor.

The problem with your LEDs is that they don't appear to have a data sheet and therefore aren't able to tell you what you fully need to know. I've said this many times on many questions but I'll repeat here: -

Read the data sheet first AND if there is no data sheet try and find one and if no luck then don't buy the device or be prepared to experiment with resistor values and expect potential disappointment. I never buy any electronic component that doesn't have a sensible data sheet.

Answer 4

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A section of a typical LED light strip. Resistors in this case were for white LEDs.

The problem with resistors is that the voltage drop will vary with the number of LEDs connected. Diodes, on the other hand, give a reasonably constant voltage drop of about 0.7 V. We can use this feature.

^{simulate this circuit}

Figure 2. Using diodes to drop the voltage in tail-light mode.

• Add as many diodes in series with the tail light to get the required voltage drop.
• Add one for the brake light to prevent backfeed.
• Check your diodes are rated for the expected / measured current.

The beauty of this configuration is that the brightness of the individual LEDs will be (pretty well) independent of the number connected.

I believe the forward voltage rate for red 5050 LEDs is 2.0v. If supplied that voltage, does the LED light up but just barely? i.e., Is that the dimmest? And is that for one LED or for the one segment of the strip (3 LEDs)?

As you can see from Figure 1, to get the LEDs to glow at all you have to supply enough voltage to overcome the forward voltage of three LEDs in series and the voltage drop (at that current) of the two resistors. I would expect that you might make out a dim glow at 5 V or so.