# One resistor in series and two in parallel for LEDs

So I've read that using one resistor to limit current to two or more LED's in parallel is bad practice, so I wondered what about this kind of circuit: Each LED has it's own current limiting resistor, but the current is being limited further by a resistor in series. Are there any differences compared to just having each LED connected in parallel directly to the voltage source with a 355Ω Resistor?

Also (just to double check I am doing my calculations right as I am still a student learning this stuff), assuming each LED has a forward voltage of 2V, is each LED drawing about 20mA here (7V across resistors, Rtotal = 355Ω so 7V / 355Ω = 0.0197 Amps)? simulate this circuit – Schematic created using CircuitLab

Figure 1. Equivalent circuits.

Your circuit is fine. The 270 Ω resistors will make up for any variation in forward voltages of the LEDs and prevent the one with the lower Vf from hogging the current.

Your 355 Ω calculation is wrong, however. The effective resistance for each LED is 710 Ω. The current is given by $$\ I = \frac {V}{R} = \frac {7}{710} = 10 \ \mathrm {mA} \$$. That will be quite bright for most modern LEDs.

• Is there anywhere where i can read more on how the way the resistance works for the LED's like in the diagram in your example? I've never come across that before. Oct 19, 2018 at 19:00
• Not really. With practice of reading and analyzing circuits you get to realise that some things can be simplified or transformed into another way of looking at it. This example was fairly simple and just relied on splitting the 220 Ω into two in parallel. Have a look at star-delta transform for more advanced tricks. (There may be better tutorials.) I have written articles about LEDs on LEDnique.com to compliment my answers here and you may find some of these helpful. Oct 19, 2018 at 19:19

No, this is fine. Because you have a resistor in series with each LED this takes care of any kind of problems that might arise from matching. All LED's are not created equal, the I-V curves have a tolerance. It is probably 'slightly' better to have the LED's go directly off of the rail but this configuration works. The most important thing is to not have LED's directly in parallel.

Also (just to double check I am doing my calculations right as I am still a student learning this stuff), assuming each LED has a forward voltage of 2V, is each LED drawing about 20mA here (7V across resistors, Rtotal = 355Ω so 7V / 355Ω = 0.0197 Amps)?

The drop across the shared resistor would be 220*40mA= 8.8V Something is wrong with your calc because this would mean that you would have more than a 9V drop total.

I'd just solve for the node voltage

for the 220Ω resistor you get

$$\\frac{9V-x}{220Ω}=i_{220}\$$

Each LED would be

$$\\frac{x-2V}{270}=i_{LED}\$$

The whole equation would be

$$\\frac{9V-x}{220Ω}=\frac{x-2V}{270}+\frac{x-2V}{270}\$$

Solving for x you get 4.66. Then solving

$$\\frac{9V-4.66V}{220Ω}=0.0184A\$$

$$\\frac{x-2V}{270}=0.0985A\$$

or half of the amount you calculated through each LED.

• Yet many flashlight designs have one resistor with 4 to 10 LEDs in parallel. I think LEDs of same model and color from a quality manufacture have a fairly tight match, or the LEDs would not all have the same brightness level.
– user105652
Oct 17, 2018 at 22:22
• And those same flashlights have different brightness for each LED if you look close. Especially when the LED's age. Oct 17, 2018 at 22:35
• Difficult to tell as they leave blind spots in your eyes for a few minutes. Newer designs use many cells and LEDs with PWM. So bright you cannot even glance at them directly. Just what I always wanted...
– user105652
Oct 17, 2018 at 23:15
• Usually it becomes pretty apparent when you get a low battery Oct 17, 2018 at 23:48
• Thanks for adding in the calculations. I learnt Kirchhoff's laws last week and forgot i could have used it here. Oct 17, 2018 at 23:49

Your circuit with three resistors is less bad than using a single resistor for two LEDs, but a failure of one LED will still affect the current in the other LED.

Also, you have to remember that the 220 Ohm resistor in your circuit will be carrying the current for both LEDs. If you want 20 mA in each LED (most modern LEDs will be quite bright at 10 mA or less), then you need to plan for 40 mA in the 220 Ohm resistor - with 2 volt LEDs and 270 Ohm resistor, you'd only have 1.6 volts across the single resistor, so it would have to be 40 Ohms, rather than 220.

The calculations you made are correct and the the circuit will work. Only one error is you said 20mA , where as LED's will 10mA each. how? like as below Vcc= 9V Vf of LED = 2V Rtop = 220R Rbot= (270/2) -- This is effective resistance of the circuit.

So Total current on Rtop = (9V-2V)/(220+ (270/2) =>20mA Total, which will be shared between two LEDs, Since these are identical, they share equal amount of current i.e 10mA each. (i am ignoring tolerances of resistance, temperature variation and Vf variation in this explanation)