# Simple but tricky confusion on LEDs in parallel

Thread Starter

#### Xavier Pacheco Paulino

Joined Oct 21, 2015
727
Hi all,

When I want to design an optocoupler circuit, I usually find the series resistor to limit the current for the opto led. If I want to add an external indicator such a red LED, I add it in series with the opto LED and find the proper resistor. My question comes from picture shown below, which is a circuit that I decoded from a PCB. The circuit is supposed to work with either 5V or 12V pulse. But I find this design pretty bad if I'm not mistaken. What would the overall voltage drop across both LED be? In fact, both LEDs are different in nature. Red LED has about 1.8V drop and opto led about 1.2V drop. The PS2501 opto has a max current of 80 mA. If I were the designer, I would put red LED in series with R1 and optoled and design for a current of 20 mA which is safe for both LED if the pulse is 5V. If the pulse were 12V, obviously both LED would experience very high current for a 5V design. To sum up, what exactly is happening when two LED different in nature are in parallel?

#### crutschow

Joined Mar 14, 2008
23,808
You could also put the LED in series with the opto input, along with a reduction in the series resistor value to keep the current the same, but that would cause a larger variation in the input current between 5V and 12V.

#### dl324

Joined Mar 30, 2015
9,335
To sum up, what exactly is happening when two LED different in nature are in parallel?
They're going to share current based on their forward voltages.

I'd connect both LEDs to the; each with their own current limiting resistors.

#### WBahn

Joined Mar 31, 2012
24,978
Hi all,

When I want to design an optocoupler circuit, I usually find the series resistor to limit the current for the opto led. If I want to add an external indicator such a red LED, I add it in series with the opto LED and find the proper resistor. My question comes from picture shown below, which is a circuit that I decoded from a PCB. The circuit is supposed to work with either 5V or 12V pulse. But I find this design pretty bad if I'm not mistaken. What would the overall voltage drop across both LED be? In fact, both LEDs are different in nature. Red LED has about 1.8V drop and opto led about 1.2V drop. The PS2501 opto has a max current of 80 mA. If I were the designer, I would put red LED in series with R1 and optoled and design for a current of 20 mA which is safe for both LED if the pulse is 5V. If the pulse were 12V, obviously both LED would experience very high current for a 5V design. To sum up, what exactly is happening when two LED different in nature are in parallel?

Pulse-based behavior and steady-state behavior are two very different things in this case, because LEDs in parallel in steady state are dominated by unmanaged thermal issues. Short pulses don't suffer from this issue.

Depending on how short the pulse is, your response might be dominated by other factors such as capacitances and response times. Assuming that the pulse is long enough so that these are the determining factors, the 1.2 V LED will attempt to clamp the voltage at 1.2 V. But LEDs have a pretty soft characteristic so it won't take a huge increase in current to get it up to 1.8 V (see the data sheet), at which point the 1.8 V LED will start conducting and most of the additional current will dump into it for awhile.

So you will likely see considerably more, but not hugely more, current in the 1.2 V LED than you ideally wanted when you are getting the current you would like to see in the 1.8 V LED.

Safest would be to put a separate resistor in series with the indicator LED. The only problem being that now more things can fail and give you a false impression. But that's already the case here. Having them in series means that if the indicator LED is lit, you know you have current in the opto LED and, conversely, if it isn't then you don't. The failure modes that would make this not true do exist, but are unlikely to occur in practice.

As for handling the 12 V issue, that should be pretty straightforward (depending on the length of the pulse and the current sourcing capability of it). Put something like a 3.3 V zener across the whole thing using two resistors. Imagine a second resistor, R2, to the right of R1 and then a zener between the junction of R1 and R2 and ground. Size R2 based on 3.3 V on the zener and size R1 to give enough current to get the zener above the knee and feed the LEDs. Then at 12 V, most of the additional current will go through the zener. But this might be a fair amount, perhaps six times the total current going to the LEDs. But if it's a short pulse, that won't matter much provided the source can handle it.

The other option is to use a transistor as a switch so that the 5 V / 12 V side has very low current.

#### AnalogKid

Joined Aug 1, 2013
8,229
As above, you will get a smaller current variation between the 5 V and 12 V cases if you run the two LEDs in parallel with separate current limiting resistors. If you really want to run them in series ...

The PS2501 has a min CTR of 80%. If you put the two LEDs in series and size the resistor for 20 mA at 12 V, that comes to 450 ohms. Round up to 470 ohms. Now, at 5 V input that works out to only 4.2 mA LED current. Times 80% equals 3.4 mA worst case through the secondary side transistor (not counting temperature variations). Is that enough? Only you can say. What is the secondary side circuit?

ak

#### Wolframore

Joined Jan 21, 2019
1,482
Not sure if the opto would work at 5v with above example but it’s safer. I know you can pump the heck out of most IR (within reason) especially pulse

Thread Starter

#### Xavier Pacheco Paulino

Joined Oct 21, 2015
727
Thanks you all for the suggestions. I will build my own version of this circuit. I will use individual resistor for each LED. This circuit is used in a tripod turnstile and the pulses are short as stated above.

#### WBahn

Joined Mar 31, 2012
24,978
Thanks you all for the suggestions. I will build my own version of this circuit. I will use individual resistor for each LED. This circuit is used in a tripod turnstile and the pulses are short as stated above.
How short is short?

Huge difference between 100 ns, 100 μs, and 100 ms even though all three could well be called short.

#### AnalogKid

Joined Aug 1, 2013
8,229
Coupled rise and fall times add up to 8 us, with only 2 us difference between the two. A pulse width of 50 us will come through with less than 5% distortion..I can't imaging anything more critical than that coming out of a turnstile.

ak

#### MisterBill2

Joined Jan 23, 2018
4,579
I would not expect the circuit shown to perform very reliably, although if the input voltage is high enough to activate the opto-isolator the red LED should glow a bit, but not at the level it would have with 1.8 volts forward bias. and it does not seem like the pulse length should have much effect except that very short pulses may not be seen by a human eye. In addition, with that parallel circuit, if one LED fails the other will still function, and so the monitoring aspect of the red LED is unreliable. A series connection is called for and will provide reliable indication. What was the source of the parallel circuit?

Thread Starter

#### Xavier Pacheco Paulino

Joined Oct 21, 2015
727
I honestly do not have much information on the entire circuit. What I know is that the 5V signal comes from a computer. Basically, there is a fingersprint scanner which gives the signal to the computer, and the computer then gives the signal to the optocoupler to activate something in the other side of the optocoupler. I don't know when the 12V case comes in (maybe for another application). As @MisterBill2 says, if one LED fails the other still functions, so the monitoring aspect is not reliable. That's why I thought it was a bad design.

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