6
\$\begingroup\$

I have a 5V supply power, three 5mm red LEDs, and each LED has 47-ohm resistor.

Two LEDs are bright but soon as I connect the third one all of them become dim.

Where am I going wrong? I'm new so a simple explanation will be appreciated.

schematic

simulate this circuit – Schematic created using CircuitLab

Edit: This might be a terrible schematic but I just started to learn about these things a few days ago.

\$\endgroup\$
11
  • 4
    \$\begingroup\$ Are they really all connected in series? If so, that's your problem - for every LED you add, you need to increase the voltage. \$\endgroup\$ – Simon B Apr 5 at 22:42
  • 1
    \$\begingroup\$ the order of components in a series circuit is irrelevant ... you can lump the three resistors into one component ... one 141 ohm resistor ... that value may be too large for the circuit \$\endgroup\$ – jsotola Apr 5 at 23:20
  • 1
    \$\begingroup\$ think about this ... if one LED requires a 47 ohm resistor, why would you further restrict the current flow with an additional resistor when a second LED is introduced? \$\endgroup\$ – jsotola Apr 5 at 23:24
  • 2
    \$\begingroup\$ think of the LED circuit as a garden hose with a sprinkler ... pinch the hose (resistor) so that the sprinkler does not spray too far ... if you add another sprinker, do you pinch the hose in a second place or do you relax the first pinch? \$\endgroup\$ – jsotola Apr 5 at 23:29
  • \$\begingroup\$ #mohammad aminzadeh, Your question is a bit complicated. The following three links might help a bit: (1) LED Strip Light Internal Schematic and Voltage Information - WaveForm Lighting waveformlighting.com/pcb-designs/… (2) Right way of wiring discrete (Monochrome, white, 5050, 100mA ~350mA) LEDs to make a high power LED panel Asked 4 days ago Active 2 days ago Viewed 253 times electronics.stackexchange.com/questions/557806/…. / to continue, ... \$\endgroup\$ – tlfong01 Apr 6 at 2:20
21
\$\begingroup\$

Each red LED drops about 1.5V. So three of them takes 4.5V, leaving only 0.5V for the resistors.

0.5 / (47+47+47) = 0.0035A or 3.5mA. That's rather low.

If you were to remove two of the 47 ohm resistors, you would get

0.5 / 47 = 0.011A or 11mA. That's brighter, but still less than most LEDs can handle.

\$\endgroup\$
8
  • 5
    \$\begingroup\$ I think this answer is much better than the top voted answer, because LEDs are current-driven. It usually doesn't make much sense to talk about voltages when working with LEDs. You should calculate the current, like this answer does. +1 \$\endgroup\$ – Opifex Apr 6 at 8:13
  • 2
    \$\begingroup\$ 1.5V is a value more likely to be associated with a IR LED than a typical red LED. \$\endgroup\$ – Peter Green Apr 6 at 12:31
  • \$\begingroup\$ @OJW Yes - fixed it now. \$\endgroup\$ – Simon B Apr 6 at 14:50
  • \$\begingroup\$ Your calculation is wrong. There is no constant voltage drop of 0.5V on the resistor. The voltage drop depends on the current through the circuit, which depends on the electrical characteristics of the LEDs. Look at the datasheet to find the current through one LED. You may simplify the calculation by using only one LED and only one resistor (1 of 3 components) at a third of the voltage (1.67V). \$\endgroup\$ – Tobias Knauss Apr 6 at 19:21
  • \$\begingroup\$ @TobiasKnauss The electrical characteristics of the LEDs are such that there is 1.5 V drop across each of them, leaving 0.5 V across the resistor. The current through the resistor (which does obey V=IR, unlike the LEDs) adjusts accordingly. \$\endgroup\$ – Andrew Morton Apr 7 at 14:28
9
\$\begingroup\$

LEDs are nonlinear devices -- their current vs. voltage curve is exponential with voltage. In practical terms, it means that essentially no current flows at low voltages, and then the current very rapidly rises.

In the case of your LED, that "low voltage" is around 1.6V. If you look at the datasheet, figure 2, you'll see this.

In fact, that LED calls out an operating voltage of 2V to 2.6V -- so using two in series with a 5V source is pushing it.

What happens when you put that 3rd LED is series is that the 5V supply just can't supply the minimum voltage necessary for the string to work -- and you get almost no light.

\$\endgroup\$
0
5
\$\begingroup\$

5mm LED has max current 20 mA. Drop voltage usually 1.5V. You can connect 3 LEDs in series with resistor 47 ohm. 10 mA current is enough. Leave only one resistor.

\$\endgroup\$
1
  • \$\begingroup\$ There are low-current 5mm LEDs that are run with much less than 20 mA. The typical forward voltage of LEDs are 1.8V (red), 2.0V (yellow), 2.2V (green) and about 3V to 3.5V for blue. This means, that you can run all 3 red LEDs without any resistor, if this rule applies to your LEDs. For more details, look at the datasheets, they will tell about the electrical characteristics. \$\endgroup\$ – Tobias Knauss Apr 6 at 19:07
4
\$\begingroup\$

Typically, modern red 5mm LEDs drop about 1.7V-2.1V at 20mA, depending on the chemistry and size of the die.

That means that more than 2 of them in series will mean it's difficult to predict the current.

If you use individual resistors and connect them in parallel across the 5V, you'd want (for 15mA) about 220 ohms each and the current (using my Vf values) should be nominally somewhere between 13mA and 15mA.

If you connect two in series you will have about 3.6V drop so somewhere around 91 ohms and the current should be between 17.5mA and 9mA (quite a range but probably still safe for the LED).

And for three- you won't be able to get 20mA at all.

Generally speaking, if current draw doesn't matter much, the individual parallel connection is better. Modern LEDs are pretty bright even at low current, so you may be able to use 470 ohms or 1K.

\$\endgroup\$
4
\$\begingroup\$

Most of the answers forgot to state the most important thing:

You don't talk about the correct voltage supply for LEDs, you talk about supply current. That's the reason for having the LEDs in series instead of parallel!

For manufacturing reasons LEDs have a Vf (junction drop voltage) extremely variable. Also being them diodes have an exponential current curve, so they are quite sensible to variations in supply.

In datasheets you will find usually the maximum Vf and the typical Vf (the minimum value often is not specified). If you just want to light the lamps, calculate a drop resistor using the sum of the typical Vf of all the LEDs in the chain.

However a better result can be had simply using a constant current regulator (usually in the range 10mA-20mA depending on the LED); a single transistor one would suffice, but they also make dedicated part for this purpose.

\$\endgroup\$
4
\$\begingroup\$

If you connect them in series, their forward voltages (Vf) add up.

Red LEDs have a Vf of about 2.0V at rated current. Three of them in series would yield Vf of 6V - more than your supply. So use two in series at most.

Also, you need only 1 dropping resistor. Knowing that 5mm (T-1-3/4) LEDs are usually rated for 20mA, we can set the current as follows:

  • R = (5.0 - 2.0 - 2.0) / 0.02A = 50 ohms

So just one 47 ohm for 2 LEDs should be ok.

\$\endgroup\$
3
  • \$\begingroup\$ i thought i connected them in parallel with this method, so how should i ? \$\endgroup\$ – mohammad aminzadeh Apr 5 at 22:50
  • \$\begingroup\$ Your diagram shows series connection. 2 in series is a good choice for 5V. This will give even brightness for the pair, and lowest losses in the dropping resistor. \$\endgroup\$ – hacktastical Apr 5 at 22:52
  • \$\begingroup\$ @mohammadaminzadeh You'll need to read some books on electronics to learn about the difference between parallel and series circuits. Series means there is just one path for current and parallel means there is two or more paths for current. Your diagram has just one path that goes through all 3 LEDs. \$\endgroup\$ – Mark Lakata Apr 6 at 23:03
2
\$\begingroup\$

Why not look at the LED specs in its datasheet? The absolute max allowed current is 30mA but 20mA is used for its forward voltage that is typically 2.0V but can be 2.6V max. Simple arithmatic shows that at least (2.6V x 3=) 7.8V is needed but one series resistor must limit the current to a maximum of 30mA.

\$\endgroup\$
1
\$\begingroup\$

If driving a LED with a series resistor, you need a voltage that is above the forward voltage of the LED at your desired operating current by a sufficient margin to account for LED variability.

Unfortunately component manufacturers often under-specify their products. In your case they give a graph for the typical forward voltage at different currents, but they only specify a maximum forward voltage at one particular current level and they don't specify a minimum forward voltage at all.

So lets try some designs. Lets say we want to aim for a 10mA forward current. That means a forward voltage of around 1.82V.

With one LED on a 5V supply, that gives us 3.18V for the resistor, and a desired resistor value of 318 ohms. Round that to the closest common value of 330 ohms.

This will be a stable but inefficient configuration, more voltage is dropped over the resistor than the LED, so variation in LED forward voltage will have little impact.

Now lets try two LEDs in series. Now 3.64V is taken by the LEDs, so that leaves only 1.36V for the resistors. Now we only need a 136 ohm resistor to limit the current.

This system will be more sensitive to variability in the LEDs and in the supply voltage, but it will probably still be ok in practice. At least for indicator LEDs there is a pretty wide range of currents that will result in acceptable brightness.

With three LEDs in series we have a problem. 1.82*3 = 5.46. So with LEDs that perform typically we won't achieve our goal of 10ma, even with no resistor at all! The graph becomes hard to read at that level but it looks to me like the current will maybe be around 1mA at that voltage. This is likely to be very variable with slight changes in supply voltage and temperature and between different LEDs of the same model.

\$\endgroup\$
0
\$\begingroup\$

LEDs are "semiconductors" which mean they don't act like resistors. An LED is a diode, and a diode has a complicated formula for predicting the current given some voltage. The complicated formula can be simplified using an exponential function.

You can get even simpler to view the relationship as a "hockey stick" line, sort of like the global temperature curve for global warming. As in global warming, the temperature does nothing until the 20th century and then shoots up like crazy, in a diode the current does nothing for low voltages (close to unmeasurable) until it gets to the "Forward Voltage", and then the current goes up like crazy for just a little more voltage. Unfortunately diodes don't have any internal protection and this current will destroy it immediately. You can not just connect an LED to 5V and hope it works. It will get really bright for a moment and burn out.

So you need to put in a resistor to limit the current. The easiest approximation for the circuit is to pretend that the diode has a fixed voltage on it (say 2.0V in this case). Because the total voltage has to be 5V, that means the resistor has to have the remaining 3V on it. And then you can use Ohm's law to figure out what resistor to pick if you have 3 volts and you want 20 mA. Why 20 mA? Well, that's what the data sheet tells you for a typical forward voltage. Let's see, V=IR, 3V = 0.02A x R, R = 3.0/.02 = 150 ohms. If you don't have 150 ohms in your kit, then go bigger, say 200 or 300 or 500 ohms. The LED will just get a little dimmer. DON'T go smaller, because you'll probably burn out your LED.

Now if you put 3 LEDs in series, then you need to pretend that each LED has 2.0 voltages, so that is a total of 6V. But you only have 5V. So you have a few choices:

  1. Don't put them in series. Make 3 separate circuits. Each LED should have 150 ohms. Then each circuit will have 20 mA.

  2. Put them in series and don't use a resistor at all. If the datasheet is correct and the supply voltage is correct (5V), the LEDs should all light up slightly dimmer than max. You'll be putting 5V divided by 3, or 1.7V on each LED. If you look at the current curve (I pasted it below from the datasheet), at 1.7V you'll get about 2 mA of current. It will light the LEDs but it will be very dim.

  3. Put 2 of them in series and with a resistor. In this case, the LEDs have 2V each, so that means the resistor must have 1V. Use Ohm's law: V=IR, 1V=0.02 x R, R = 50 ohms.

How do you get the data sheet? You go to google and search for the part number and the word "datasheet"

https://www.google.com/search?q=ltl-307ee+datasheet

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.