I'm planning to drive two white leds like these (datasheet) with PAM2803 driver (datasheet) and a single cell lipo battery.

Before ordering a pcb, I'm testing on a breakout board. With a 0.13 Ohm shunt resistor, I'm expecting a 730mA current. As test load, I'm using a 2 Ohm resistor (5x10 Ohm in parallel). The result should be 1.46V on the resistor.

From the sheet, the device should be powered between 0.9v and "forward voltage of the led - 0.2v"

I'm powering the test system with 1V which is between 0.9 and 1.46-0.2 = 1.26v.

I'm using a 2.4 µH inductance (FDSD0420-H-4R7M=P3 : datasheet) and MBRA210LT3G schottky (datasheet).

The result is not the one expected: resistor voltage is 690mV thus 345mA. If I increase the input voltage, the current increases too. What's wrong?

enter image description here

Here's a picture of the board. It's not easy to see on it. Shunt resistor is soldiered just between the pins. Capacitors are on top of their tracks.

enter image description here

  • \$\begingroup\$ where did you get 1.46V from? According to that spec 700mA white is 3.3V.. And please post a schematic of what you are doing, so nobody gets confused. \$\endgroup\$ – Trevor_G Nov 16 '17 at 16:25
  • \$\begingroup\$ @Trevor: 1.46v is voltage with a 2 Ohm Resistor and 730mA curent. Schematic added to post. Sorry, it is the same as in the datasheet. I forgot to mention. \$\endgroup\$ – Julien Nov 16 '17 at 17:47
  • \$\begingroup\$ yup but it is not a good test load. 5R is closer to your LED. \$\endgroup\$ – Trevor_G Nov 16 '17 at 17:48
  • \$\begingroup\$ Yes, you're right \$\endgroup\$ – Julien Nov 16 '17 at 17:56

If you look at the curves on the PAM you will see the thing does not regulate the current till closer to 1.8V.

For the LED you are talking about you need a load resistor closer to 4.7R, not 2R.

Also, the diode is a Schottky diode, not a Zener diode.

enter image description here

If you plan on driving two of those LEDs in series with one PAM you are out of luck. The PAM can only go up to 6V, two of those LEDs at 700mA would require 6.6V.

  • \$\begingroup\$ Hum, I used 2R because I had no resistor on my bench that could handle 4.7Rx0.73Ax0.73A = 2.5W but with the graph you provided, 2R is non-sens. You're right. I don't understand why I can't drive 2 leds with this device. The sheet, in description, states that "The PAM2803 is targeted to be used for driving loads up to 1A from a two-cell alkaline battery", 2 cells is 3V. Doesn't it provide voltage booster? \$\endgroup\$ – Julien Nov 16 '17 at 17:56
  • \$\begingroup\$ @Julien it does, but only to a max of 6V. Two 3.3V LEDs and the internal over-voltage protection will kick in. See Absolute Maximums table on page 2. SW pin max... \$\endgroup\$ – Trevor_G Nov 16 '17 at 17:59
  • \$\begingroup\$ For my understanding, what's the purpose of the schottky? Avoiding reverse current to the led? \$\endgroup\$ – Julien Nov 17 '17 at 12:52
  • \$\begingroup\$ @Julien the device is a step up boost regulator. It turns on and off a switch to ground at the right end of the inductor. Every time the switch opens, the current in the inductor turns on that diode and pumps up the voltage on C4. When the switch closes, you don't want to short out that cap back to ground. BTW: That diode is why Vin can not be greater than the LED voltage + 0.2V. If it were, the diode would always be on even when the device itself is shutdown. \$\endgroup\$ – Trevor_G Nov 17 '17 at 13:06
  • \$\begingroup\$ @Julien actually, as far as I can see, shutdown is not really a Light-ON/Light-OFF control. It's more of a Bright/"Whatever the battery on it's own will do" control. \$\endgroup\$ – Trevor_G Nov 17 '17 at 13:13

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