LED Driver

Hi, We would like to drive LEDs using Ramp Waveforms using Xilinx CPLDs. Is it possible to write a driver and control the LED directly from CPLD or should an external D/A Converter be used?

Reply to
pavithra.eswaran
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you can use PWM modulation to control the LED current

Antti

Reply to
Antti

Depends on the precision, and if you need linear current, or can tolerate an average value.

Most common for human-view applications is PWM or PDM to drive the LED, but that can give EMC issues, if you have large LED currents.

If you want to avoid the EMC, you can do PWM/PDM to a DC voltage, and add a linear LED driver (now you have thermal issues :)

-jg

Reply to
Jim Granville

Jim Granville schrieb:

You can route the current in alternating loops. That should remove almost all EMC issues.

Kolja Sulimma

Reply to
Kolja Sulimma

Yes, it can mitigate some of the problems, but it also requires forethought and skill, and probably an EMC test :)

'LED Driver' covers a very wide range of possible loads, so it is not easy to advise the best solution.

Between the Digital Switch, and linear drive choices, there is also soft digital drive ( Typically MOSFETS with high value Rg drive ), gives less power loss than pure Linear, but better EMC than pure digital. If the LEDs are a matrix ( moving signs etc ) you can also ease EMC by adding a deadband, and design the drive so only one axis needs to be slow.

Mostly, I use PDM (Pulse Density Modulation) a la Rate Multipliers of old. That uses the least CPLD resource.

-jg

Reply to
Jim Granville

Hi, LED's sinking current capacity is in uA. so tie Anode of LED to port pin of CPLD & Cathode of LED to gnd. If u do this kind of connection then CPLD will be in sourcing mode.so to avoid sourcing of CPLD,u can tie Vdc to LED Anode through current limiting resistor,& cathode of LED should be directly connected to CPLD port pin.in this case CPLD will be in sinking mode. I have tried this kind of thing,it woks fine.

Reply to
mohan

CMOS outputs are fairly symmetrical, they can source or sink similar amounts of current. But perhaps the sink capability of the n-channel output is slightly higher than the source capability of the p-channel pull-up.. The difference is less than 2 : 1 Peter Alfke, Xilinx, from home.

Reply to
Peter Alfke

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