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I am working on a project where I need to drive two LEDs on a PCB (= PCB_LED) which is 1 meter away from another PCB (= PCB_MCU).

The PCB_MCU includes a microcontroller with two LED drivers. Each LED driver is a constant current buck driver. Its output current depends on what the MCU asks it to output. The LED driver simply applies a PWM onto the cathode of the LED.

More info:

  • The PWM is around 2 kHz
  • The LED have a threshold voltage of approx. 3 V
  • I would need 3 wires into the cable: 1 PWM for each LED and common VCC

I have several questions:

  • Will it work?
  • If so, is it EMC compliant?
  • If no, it means that I need differential wires into the cable?

Do you think using an RS485 transceiver in both ends is enough? I have already done this to communicate between a master and some slaves but they were communicating using UARTs, it was not simple PWM.

Thanks :)

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    \$\begingroup\$ It will depend on the rise and fall times of the flanks of your 2 kHz signal. I don't suspect any issues. Are the cables twisted? \$\endgroup\$
    – winny
    Commented Oct 27, 2021 at 9:02
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    \$\begingroup\$ The question can't be answered since you don't mention any details. Depending on length, wire thickness and current consumption, you will always get voltage drops. To know if it will work with a voltage-modulated PWM, you have to calculate the voltage loss over copper wire depending on current consumption. Ideally use current modulation, not voltage modulation. "If so, is it EMC compliant?" EMC compliant to what? Bob's local garage EMC standard? Aerospace military EMC requirements? \$\endgroup\$
    – Lundin
    Commented Oct 27, 2021 at 9:02
  • \$\begingroup\$ You can't use RS485 as the LED driver output is a PWM'ed constant current sink for the LEDs, while RS485 receiver output would be PWM but push pull logic level output limited only by whatever resistor you put there. \$\endgroup\$
    – Justme
    Commented Oct 27, 2021 at 10:09
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    \$\begingroup\$ If you filter your PWM with an inductor, and add a flyback diode, it becomes a buck driver. Hey wait a minute! You already have a buck driver! Is that one not adjustable? It would be cool if it was adjustable. Or maybe you could throw it away, and just use filtered PWM, which is a buck driver. \$\endgroup\$
    – user20574
    Commented Oct 27, 2021 at 17:40
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    \$\begingroup\$ As @user253751 says (but in less words): Why does it need to be PWM at the LED end? \$\endgroup\$
    – D Duck
    Commented Oct 27, 2021 at 20:31

4 Answers 4

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The PWM frequency is not so important, what matters is the rise times of the switch controlling the PWM. Assuming it is some sort of MOSFET, one can assume very fast rise times of 100ns to 10ns or even less.

1 meter of wire has a self inductance of ~1.5µH to ~1.8µH for a reasonable range of wire gauges. This inductance will be some what less if you run the power and ground cables directly next to each other which will cancel some of the magnetic field and thus reduce the inductance.

Unfortunately, LEDs are still diodes and they have some pretty nasty capacitance and non-linear voltage vs current behavior, so that coupled with the likely fast rise times of the PWM switch is going to cause the cable to ring at each turn-off transition with high frequency harmonic content.

It is impossible to say what the fundamental ringing frequency will be, it will depend on the cable and the LEDs, but it will ring at nearly the full LED current then decay over some microseconds, similar to this:

enter image description here

The ringing frequency is probably going to begin at single digit MHz then have harmonic content all the way to the maximum harmonic as dictated by the rise time of the switch. The actual power at harmonics that a 1 meter cable would be an effective radiator at would probably be fairly low but given the current levels, potentially still worrying.

If you don't want to take any chances, or you need to pass FCC (don't cut corners if that is the case, that is an expensive test to fail!), then your best option is to simply ease the PWM edges and slow them down a bit. A bell rings when struck, but distribute the force over more time by hitting it with a pillow and it stays quiet. That's exactly what we want to do here. The LED's capacitance interacting with the cable inductance is the bell, so if we just avoid hitting it too hard, it won't ring.

For what I assume are currents of no more an amp or two, the best option in my opinion is the tried and true RC snubber:

enter image description here

This will dissipate the ringing as heat in the resistor, so it does have a small impact on efficiency. But the snubber pictured will dissipate tens of mW since it only dissipates power when there is a switch event. At 2kHz, this is not very often. It will eliminate the ringing almost entirely:

enter image description here

Not bad for 2 very cheap passive components!

Be sure to position them at the LED side of the 1 meter cable, if you put it on the LED driver side, it won't do any good.

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    \$\begingroup\$ Or... you could just rectify the PWM into an analog voltage early on before sending it out across the wires to the LED drivers. Though if the LED driver works with PWM too internally, then that's another story. Depending on EMC requirements, the solution to that might be to forget about the LED driver and use plain old current limiting resistors. \$\endgroup\$
    – Lundin
    Commented Oct 28, 2021 at 6:34
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    \$\begingroup\$ "Hitting a bell with a pillow" is a memorable image! \$\endgroup\$
    – Camille Goudeseune
    Commented Oct 28, 2021 at 18:54
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    \$\begingroup\$ 1.5 uH assumes ground is not routed and especially not twisted with the positive supply. \$\endgroup\$
    – winny
    Commented Oct 29, 2021 at 10:49
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This is supposed to be a comment but got too long, so I decided to put as an answer.

What I understood is, you have two PCBs, one having only the LEDs and the other having the MCU and the respective LED drivers for LEDs. So the LED drivers' outputs go to LEDs through a 1-m-long cable. It is more preferable to see a diagram and schematic (if possible).

This cannot be simplified as "sending PWM signal through a 1-m cable", because there are switching currents flowing through the cables. You should also indicate the maximum peak value of these currents in your question's body.

A 1-m long cable can be a good quarter-wave monopole antenna for 75 MHz. Depending on the rise time of the PWM signal, the PWM signal will probably contain 75-MHz harmonics and these harmonics will be radiated. The amplitude of the radiation depends on the rise time and the peak value of the switched currents (Switching 10 A current with 75 kHz PWM with very low duty cycle may be enough to disturb your radio from 1 meter of distance). Plus, any 75 MHz signals picked up by the cable may or may not disturb the drivers' performances. It's hard to say something about radiated immunity performance without knowing what brand/model/type LED drivers will you use.

Maybe you should move the buck converters and LEDs to PCB_LED, and send low-voltage low-current PWM control signals from PCB_MCU through buffers/low-impedance drivers.

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    \$\begingroup\$ all the 75MHz stuff is not an issue if the edges of the (2kHz!) PWM signal are slowed down adequately, which is easily achieved by combination of choice of a suitable driver and/or a simple CR network (there are probably already series resistors anywat to set LED current). All the OP is trying to do here is control LED brightness, there is no need to complicate things. At 2kHz PWM, 5% duty cycle, minimum pulse width is around 25us, you could slug the rise and fall times to around 1us, and 75MHz is not going to be an issue. \$\endgroup\$
    – danmcb
    Commented Oct 27, 2021 at 9:01
  • \$\begingroup\$ @danmcb You might be right but the OP has not shared the most important details and I talked about the "possible" risks accordingly. We don't know which driver will be used and thus the switching characteristics, we don't know the power/current levels, we don't where the current sense resistor will be placed. So there "are" risks. Plus, the OP asks about if the configuration can be EMC-compliant or not. EMC compliance does include the immunity, and there's a long cable inserted directly into the feedback network. So the "75 MHz stuff" in the second half of 3rd pgph still matters. \$\endgroup\$
    – Rohat Kılıç
    Commented Oct 27, 2021 at 9:28
  • \$\begingroup\$ well, OK, but what kind of 75MHz signal would need to be in the environment to damage an LED or the output of a constant current driver? Of course, we always have to do actual tests, but I'd expect that it would need to be a very hostile signal indeed. \$\endgroup\$
    – danmcb
    Commented Oct 27, 2021 at 9:37
  • \$\begingroup\$ @danmcb Please read carefully. I'm not talking about damaging/destroying the circuit, neither completely nor partly. Again, the OP asks if the system can be EMC-compliant or not. IEC 61000-4-x (can't remember the actual ref) standards cover radiated immunity tests for different frequencies and levels (e.g. 150 kHz to 80 MHz, up to 10V/m). And the standard requires the circuit should operate as designed during and/or after the tests. Now, is it guaranteed that these level of 75 MHz signals will not affect the feedback network and force the converter into the instability? ... \$\endgroup\$
    – Rohat Kılıç
    Commented Oct 27, 2021 at 10:21
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    \$\begingroup\$ ... According to OP's explanation, the LEDs' cathodes will be connected to their respective drivers through 1-m long cables. This means that the feedback path will be directly exposed to radiation. During the tests, if one of the LEDs goes off or starts to flicker because of instability caused by radiation then the circuit will fail the test and thus will not be EMC compliant. Do we know the compliance is a requirement? No. Again, we don't know the most important details, so we can't say any certain things. All we can do is to express "possibilities". And that's what I'm trying to do. \$\endgroup\$
    – Rohat Kılıç
    Commented Oct 27, 2021 at 10:22
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Each LED driver is a constant current buck driver. Its output current depends on what the MCU asks it to output. The LED driver simply applies a PWM onto the cathode of the LED.

I don't understand how you are driving these LEDs. Either it's a constant-current buck driver, or it's PWM. It's not both at the same time.

A constant-current buck converter is really just PWM with an inductor after it. Do you have that inductor, or not?

If you have a (big enough) inductor between the LEDs and the PWM driver, you are sending (approximately) constant current through the cable, not PWM. If you don't, then you should consider adding it - not only will it decrease EMI but it should also reduce any noticeable flicker and improve the lifetime of the LEDs, as they will be driven at a constant (for example) 40% brightness, instead of alternating between 100% (or more) and 0%.

Note that depending on the PWM driver, you may also need to add a diode alongside the inductor. The inductor needs to be able to continue pushing current through the LEDs when the PWM is off. The current will circulate through the LEDs, inductor, PWM driver, and the positive rail. Some PWM drivers will be okay with having current pushed through them like that; some won't. It's impossible to say without actually knowing how you are creating this PWM signal.

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Instead of a common VCC wire, consider a separate VCC for each LED. In this case the return current in each VCC wire will exactly match the current in its associated PWM wire, but in the opposite direction of course. Now the EMI can be greatly reduced by using twisted pairs of wire. Cat5 or Cat6 communication cable is inexpensive.

The truth is, simply twisting all three wires with a common VCC will likely help reduce EMI.

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