Connecting LEDs in series vs parallel
Basically there are two ways in which you can connect LEDs in a circuit. Series or Parallel.
Connecting them in Series means daisy-chaning
You start with one, then add a second one, connecting its (+) to the (-) of the previous one, then a third one, connecting its (+) to the minus of the previous one, and so on.
If you look at this image, you can notice that the voltage of the supply is much higher that the common 5V or 9V.
Well, assuming that all leds that are depicted here have a forward voltage of 2V, it is easy to understand why a 20V supply is necessary for the 9 leds in the image.
When connecting them in series, the total voltage drop between the (+) of the first one and the (-) of the last one is the sum of voltage drops for all leds. You can use various types of leds with different forward voltages, but the total voltage that is needed is the sum of all forward voltages.
Considering this, 2Volt x 9LED = 18V total necessary voltage for the leds to power up. However you still have to limit the current going through them, so you have to add a limiting resitor. Choose a small voltage drop to have over the resistor, for example 2V and calculate the required resistance, 2/0.02 = 100 Ohm.
That’s mostly it. To sum this up:
When connecting leds in series:
- Total Voltage: Sum of voltages of all leds.
- Total Current: The same current passes through all LEDs, so whether you have 1 LED or 100, it’s still 20mA.
Limitation: You shouldn’t connect in series LEDs with different current ratings. If you have 20mA passing through the first one, you’ll have the same 20mA passing through all, so if the second one has a maximum rating of 5mA, it will burn.
Workaround: use a resistor in parallel with each of the the lower rated LEDs to divert some of the current through it.
In the image, LEDs 1 1 and 3 are rated at 20mA, but LED2 is rated at 5mA
The main limiting resistor (right of circuit) is sized to allow for the largest rated current in the group, which is 20mA and has a value of 15o Ohms.
Without anything else, LED2 will just burn out because 20mA are passing through it.
To solve this issue, you have to add another resistor, in parallel with the lower rated LED.
This resistor can be easily calculated because we know the voltage that is applied to it, which is 2V (the same as it is on the LED that it connects to). The current that we want passing through it is current through circuit – led rating which in this case is
20mA – 5mA = 15mA
Therefore, the resistance of this resistor needs to be: R = U/I , meaning R = 2/0.015 = 133 (approx)
So now, by using a 133Ohm resistor in parallel with this lower rated LED you have effectively limited the current going through this led to around 5mA.
Connecting LEDs in parallel is another story. This way you can use a smaller voltage source, such as a regular battery to light up as many leds as you want, being limited only by the current that the battery can provide.
You saw earlier in this article that if you have a higher voltage battery, you can connect many leds in series and only require a small amount of current.
However, in parallel the voltage stays the same across all of them, so you only need a few Volt off your battery, but the current that you need is much higher because the total current that passes through parallel connected LEDs is the sum of currents needed for each of them.
This means that if you have 10 leds, each rated at 2V and 20mA, you can power them with a 2v battery capable of providing 200mA.
Practically, you can use a small 4.2V battery to power them using a limiting resistor of 110 Ohms, and if the battery’s capacity is 200mAh, you’ll have enough juice in it to power your leds for a whole hour.
However this will only work if the battery is able to provide 200mA. For example, a standard 9V battery can provide a maximum discharge current of 1000mA. http://data.energizer.com/pdfs/la522.pdf
In the case of this 9V battery, you could power 50 LEDs in parallel at their max current rating of 20mA, or 100LEDs in parallel at half current each (10mA)
Different parallel configurations
If you take another look at the above image, you’ll notice that it doesn’t matter how many resistors you use or how you connect them to the LEDs, as long as the current going through each LED is 20mA.
- Configuration 1, easiest to build: A single 30 Ohm resitor that drops 3V, limiting the overall current to 20mA x 5Leds = 100mA and the voltage applied to all LEDs to 2V
- Configuration 2: Each LED has its own limiting 150Ohm resistor that drops 3V off the 5VCC and limits the current to 20mA
- Configuration 3: A block of 5 150 Ohm resistors connected in parallel that is effectively the same as Configuration 1 because their total resistance is 30 Ohms when connected in parallel.
This is because:
1/R Total = 1/R1 + 1/R2 + … + 1/Rn
Which in this case is:
1/R Total = 1/150 + 1/150 + 1/150 + 1/150 + 1/150 = 5/150
therefore, R Total = 150 / 5 = 30 Ohm
Better: Multiple resistors.
Well, connecting LEDs in parallel as shown above definitely works, but it has some risks, the main one being that because not all LEDs are the same, even if taken from the same batch.
The effect of this reality is the fact that for individual leds connected in parallel or for strings of leds connected in parallel, their forward voltage will be quite different. Here’s an image that shows what I mean:
But first, a not-so-robust schematic. Multiple strings in parallel and a single resistor can cause issues.
In the image you can see two strings of LEDs that are, though maybe a bit exaggerated, like real ones taken from a batch, in the sense that not all of them have the same voltage drop at 20mA.
You can see the real voltage drop @20mA written with white text just to the left of each one.
In the same image, there is a voltmeter to measure the voltage drop over each LED, to illustrate the fact that at the same current, each LED can have a different voltage drop depending on very subtle differences in the manufacturing process.
As you can see in the previous image, there’s only one resistor that was calculated for 2 strings of LEDs using their datasheet.
There are two strings of 20mA LEDs in parallel, so a total current of 40mA is needed.
The total voltage drop on each string at this current, according to the datasheet, would be 2V x 3 LEDs = 6V.
RLimit = (9v – 6V) / 0.04A = 75Ohm.
The first thing to notice is that because of these difference in real LEDs, the current is not actually limited to 40mA, but to 39.7mA. So far so good, no harm done, as this is still under the 40mA limit. However, look to the right of the image. You’ll see that the second string has a current of 25.3mA passing through it. This is more than the 20mA limit of the LEDs, therefore they could get damaged.
The issues doesn’t stop here. Although having LEDs that emit light at different intesities is not what you want, there’s a bigger issue here, and that is thermal runaway.
What can happen in this case is that the second string gets hotter and hotter because of the higher current, and this can cause the voltage drop on the leds to decrease, which in turn will cause more current to flow through that string, and this process continues to repeat, causing the LEDs to get even more hot, and so on. This process is called a thermal runaway.
It’s better to use multiple limiting resistors, one for each LED or string of series LEDs
As you can see in the image, if instead of the 75Ohm resistor from the previous example while keeping the exact same LEDs in the same configuration as two parallel strings of 3 series leds each – so if instead this 75Ohm resistor I use a 150Ohm resistor for each string, the current is much more balanced.
Calculating this resistor is actually more simple than in the previous example because for each resistor you only need to take into account the string that you’re putting it on.
So for the leds in the image above the calculation would be:
On a string I have 3 LEDs with specs saying 2V at 20mA. I want to pass 20mA through them so I’m using the 2V to do my calculations as follows:
2V x 3Leds = 6V
9V – 6V = 3V on the resistor
3V / 0.02A = 150Ohm for the resistor.
Now you can add or remove as many strings you want without affecting the strings that are already there, because the led imbalance will be kept isolated in each string. Of course, as many as you want as long as your power source can provide the necessary current.
One led or more leds per string.
Although this example shows 2 parallel strings of 3 series LEDs each, the same applies to 2 parallel strings of 1 LED each, which is nothing more than 2 leds connected in parallel.
Therefore, you can connect either 2,3,5,7…. LEDs in parallel, each having its own limiting resistor or you can connect 2,3,5,7….. parallel strings of 3 or 4 or whatever series connected LEDs.
More advanced methods
There are more advanced methods for powering multiple LEDs. One of them is the current mirror.
The current mirror is a circuit that mirrors the current from one path in the circuit onto another path, so in our two parallel strings example, you can use a current mirror to make sure that the same current flows through both strings of series LEDs.
The current mirror will be covered in an upcoming article, so stay tuned to find all about it.
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