4
\$\begingroup\$

Wanted to make an LED grow light and, so, was going to follow the same hybrid parallel/series circuit used in this simple 100w quantum-style light, see picture below. Also, see below that, is a schematic in circuitlab that shows this in a diagram of 13 parallel strips of 18 diodes in series. (I had tried to make a simulation for answers yet this schematic was as far as I got).

Note that there is diversity in the diodes on this board; each 18-diode series is made up of 14-white diodes and 2-red diodes.

So I have an electrical engineer who finds fault in the circuits in this standard LED light!

Claims that in this design, there may be large differences in each of the series. And that as a result, the diodes compared series to series are a different brightness. The EE suggested, instead, use a circuit of LEDs in parallel (I think) with constant voltage PWM output driver.

Firstly, my understanding with the light pictured, is that any differences over the 18 diodes in the series gets fractionally small when averaged and would equalize the power over the parallel circuit. Is the draw from each string of LEDs so similar that there really is no need to regulate?

Secondly, what are the industry standards?

Thirdly, we are planning a mix of diodes with different voltages and currents. So, would parallel circuits be plausible? Do they work with diverse diodes populating the circuit?

My concern is about deviating from an apparent industry standard. Many people have looked at this problem before and come up with today's solutions.

enter image description here

https://www.circuitlab.com/circuit/vsez77darkyd/unnamed-circuit/ photo of quantum style LED grow light

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Yeah your EE is right. A slight variation of components leading to higher current in one string leads to that steering getting slightly hotter, which in turn makes them have lower resistance, so even higher current, and so on. We call this thermal runaway. \$\endgroup\$ Nov 26, 2023 at 2:18
  • \$\begingroup\$ Yeah, it is not a perfect circuit. Most of the time it will work well enough. \$\endgroup\$
    – jpa
    Nov 26, 2023 at 16:23
  • 1
    \$\begingroup\$ Your EE has no idea what he's talking about suggesting parallel LEDs over series and PWM over constant current. Find a better engineer. \$\endgroup\$ Nov 27, 2023 at 22:49

4 Answers 4

13
\$\begingroup\$

All in parallel would be a lot worse. From a constant voltage source (PWM or not) would be worse again.

Series diodes all have the same current flowing through them and differences will tend to average out, but more importantly the resistive component of the forward voltage drop adds up so the current will be more stable with temperature. They're mounted on an aluminum core board by the looks of it, so the temperature will tend to be similar for each LED.

If you want to be completely safe, use a separate constant-current LED driver for each series string. But as you can see, it's been found to not be necessary, at least for that class of product.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ or add a small resistor in each branch an accept a little power loss. \$\endgroup\$
    – Sascha
    Nov 26, 2023 at 21:58
2
\$\begingroup\$

"Is the draw from each string of LEDs so similar that there really is no need to regulate?" Generally, yes. 18 LEDs is a large enough number that you can pretty much count on the Law of Large Numbers https://en.wikipedia.org/wiki/Law_of_large_numbers to keep you out of trouble,and if a user gets unlucky you just ship them a replacement. Furthermore, with a large number of LED strings, a failure of one string will only increase the current through the other strings by a small amount. For two strings in parallel, if one string starts hogging current and fails at twice the current in the other, the current in the other will jump by a factor of 3, to twice the design current, and this can obviously be a problem. But if you have 13 strings, as in this case, and one string fails drawing twice its design current (about 16% rather than 8%), the current will be distributed among the other 12, for an increase of about 1 1/3 %, for an increase of about 19% in the other strings. If the unit's heatsinking is competently done, this should not cause immediate problems.

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

leds in parallel will be running at the same voltage, leds in series will be running at the same amperage. with multiple parallel rows of leds wired in series, you can combine different types of leds if you have the same number of each type in every row so that the combined voltage drop of those leds is the same for all rows. the leds also have to all be rated for the same amperage.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ ah but this is usually recommended against, because random variations in each LED can make them have different voltages \$\endgroup\$
    – user253751
    Nov 26, 2023 at 15:05
  • \$\begingroup\$ What about "tolerance stacking" which usually refers to small tolerances adding up to larger tolerances, but in a series circuit it actually has the opposite effect of adding up to smaller tolerances. \$\endgroup\$
    – Effexxess
    Nov 26, 2023 at 18:30
1
\$\begingroup\$

Generally, driving a diode with a constant voltage is a Bad Idea™. The current in any diode, including an LED, is an exponential function of voltage across it; for every extra 26mV (at room temperature, less when hot) increase in voltage, you get e (~2.7) times as much current. That's more than double! So if you overshoot your target voltage by just 50mV for a diode, you're putting 7x as much current through it. 100mV and you're putting 50x current through it, until it (very quickly) burns out.

With that in mind, I want to come back to something you helpfully pointed out: you don't have just one kind of diode. For physics reasons, white LEDs have a much higher voltage; it varies exactly by manufacturer, but white LEDs tent to be around 3V, while red LEDs are closer to 1.4V. This means that if you try and drive the white LEDs at 1.4V, they won't turn on, and if you drive the red ones at 3V, they will have e⁶⁰ times as much current through them (burn out instantly). If you split the difference and call it 2.1V, the red ones will burn out and the white ones won't turn on.

There is some random and some systematic difference between forward voltage of the same kinds of LEDs. When you add up a bunch of them in a series string, the random offsets tend to cancel out. The biggest issue with putting them in series is that if one does fail, the whole string dies. If you drive parallel strings with constant current, one string dying means that the rest of the strings have to pick up the extra current, shortening their life spans and causing them in turn to fail sooner and causing a cascading effect.

If you want to avoid that, you need a current source on each string. Assuming your total voltage on each string is about the same (i.e. the same number of each type of diode), you can use a resistor for this. To do that, you drive strings like your schematic shows, only for each string, you put a resistor in series. Then you drive all strings (now each one has a resistor) with a constant voltage. I would size it so you have a few hundred millivolts across each resistor, and make sure to use adequate power resistors for the dissipation they will see.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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