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So I have two LEDs with a Forward Voltage rating of 2.65V each. Would I be okay in hooking them up to a power supply of 5v in series rather than in parallel with resistors?

I'm asking mostly because they're currently glued into a model, and if I blow them it will be very difficult to get replaced, but I also don't have any resistors at the moment.

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  • \$\begingroup\$ manufacturing variations \$\endgroup\$
    – PlasmaHH
    Commented Jul 29, 2015 at 12:03
  • \$\begingroup\$ You won't blow them by putting then in series. They may be dimmer than you'd like, but without seeing a datasheet it's impossible to tell what the brightness reduction will be -most manufacturers provide a Forward Voltage vs Current Consumption graph and a Current vs Normalised Brightness graph, so you can easily translate voltage to brightness. \$\endgroup\$
    – CharlieHanson
    Commented Jul 29, 2015 at 12:09
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    \$\begingroup\$ @CharlieHanson You don't know that. An LED with 2.65V Vf(max) could well be destroyed with 2.5V across it. Negative temperature coefficient and no series resistor is a really bad combination. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 29, 2015 at 19:17
  • \$\begingroup\$ Also if they are not the same type of LED one may drive the other to a higher voltage point. However in this case I would likely gamble if the 5V was regulated if not then very risky. Putting a diode if available in series could make it safer (and still dimmer) when living on the edge. \$\endgroup\$
    – KalleMP
    Commented Jul 29, 2015 at 22:24
  • \$\begingroup\$ The problem was that I was just given this model to wire up, and there's LEDS in it. I don't have any kind of information other than what I said. Having to wait for resistors isn't the end of the world, I just hoped I could skip a step or two. \$\endgroup\$
    – Strawhare
    Commented Jul 30, 2015 at 15:01

3 Answers 3

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No. No. No. The current you get with no resistors will be very hard to predict, and will only by the greatest of chances be what you want (and small variations in the 5V and temperature will greatly affect whatever current you happen to get).

Use one resistor per LED and be safe.

You could put them in series and use a single resistor but the supply voltage should be more like like 9V in that case (and the resistor value calculated accordingly).

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    \$\begingroup\$ I thought was much, just wanted to be sure. Cheers \$\endgroup\$
    – Strawhare
    Commented Jul 29, 2015 at 13:27
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    \$\begingroup\$ Use one resistor per series string of LEDs. \$\endgroup\$
    – Solomon Slow
    Commented Jul 29, 2015 at 16:21
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Make up a constant-current-source transistor driver. If you design for a current limit of say 20mA, the voltage should be irrelevant. It just needs to be sufficient to overcome the sum of the forward voltage drops. Take care to use the same LED type in each series chain (voltage drop varies by colour).
You can boost the LED drive voltage as necessary using an inexpensive DC/DC converter board.

Something along the lines of these:

I'm working on something like this myself for a model of a fairground engine which requires a chaser lighting effect, with 4 sets of 10 LEDs in each series chain. The driver circuit can be hidden away in the base where it is accessible later if necessary and you can use fairly fine wire between the LEDs. Take care to wire them all the same way around!

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  • \$\begingroup\$ While this is generally true, it won't help the OP as they have a PSU of 5V (which is not enough to power a current source), and there's likely no place for a current source assembly in their model anyway. \$\endgroup\$
    – Dmitry Grigoryev
    Commented Jun 23, 2017 at 7:44
  • \$\begingroup\$ @Dmitry I'm using a small low-cost DC/DC boost converter to get the higher voltage. You don't know what scale the model is but as I suggested, the circuit can often be housed in the base or stand. \$\endgroup\$
    – Ed Randall
    Commented Jun 23, 2017 at 8:38
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    \$\begingroup\$ I appreciate the help Ed, and yes, the model would have easily housed any amount of circuitry. However, the simplest option was the best for this situation, as I do not maintain it, and as such making it easy to fix in case of an accident was important. \$\endgroup\$
    – Strawhare
    Commented Jun 27, 2017 at 12:08
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You will be fine with them in series as long as you wire in a resistor at the end of the series. Since your forward voltage is 5.3 and your source is 5v you don't have to disipate any excess so use a 1 ohm resistor to draw the current. You will have a very efficient circuit using this over a parallel circuit where your combined dissipation is the sum of your input.

Hope that helps

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    \$\begingroup\$ Nonsense! If the forward drop is 5.3V and the voltage source is 5V, the LEDs won't even light! Also if the voltage rises significantly above the forward drop, 1 ohm will do nothing. \$\endgroup\$
    – DoxyLover
    Commented Jul 29, 2015 at 18:43
  • \$\begingroup\$ I seriously doubt that the min voltage of each of these would be 2.6v. LED's usually have a nice margin of tolerance and your FWDV is never your min/max, it would be crazy to think it was. I work with a lot of green 1206 lights that have a FWDV of 2.2 and a maximum operation of 2.6 and minimum of 1.8v. These will operate just fine. Further, you are operating on DC so what "spikes" in power would you expect to see? In your parallel array if there was a "significant voltage increase" the resistors are going to handle it, they are already burning off the 2.3v left from the inefficient circuit. \$\endgroup\$
    – user82092
    Commented Jul 29, 2015 at 18:53
  • \$\begingroup\$ Assuming 5v Source, 2.5 v FWD V, 20ma current, 2 diodes. This will work fine! Solution 0: 2 x 1 array uses 2 LEDs exactly +5V + -|>|- -|>|- -/\/\/\- + R = 1 ohms The wizard says: In solution 0:•each 1 ohm resistor dissipates 0.4 mW •the wizard thinks 1/4W resistors are fine for your application Help •together, all resistors dissipate 0.4 mW •together, the diodes dissipate 100 mW •total power dissipated by the array is 100.4 mW •the array draws current of 20 mA from the source. \$\endgroup\$
    – user82092
    Commented Jul 29, 2015 at 19:03
  • \$\begingroup\$ Regardless of your other arguments, a 1 ohm resistor will do absolutely nothing! You're feeding garbage data into your "wizard" and it's giving you garbage results. \$\endgroup\$
    – DoxyLover
    Commented Jul 29, 2015 at 19:40
  • \$\begingroup\$ Doxy, there is no reason to use a higher resistance. The 1ohm resistor is only there to create the load not to reduce the voltage like you seem to think. There is no need for a voltage reduction and you aren't protecting the DC circuit from "erroneous magic voltage" with a higher value. I don't think you understand... at all. It boggles my mind that I provide the data for you and you still deny it while providing no counter other than "it wont work". \$\endgroup\$
    – user82092
    Commented Jul 29, 2015 at 19:49

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