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LED2000 shorts itself when attempting to use the DIM pin

dimin
Associate II

Today I received several of my LED2000DR based LED boards. Everything worked flawlessly when using the chip as a plain LED driver. However, after trying to feed a PWM signal into the DIM pin the dimming worked for about a second but then the board went into a dead short. The input measures 0.7 Ohms on my DMM now. The only thing on the board is the LED2000DR and the LEDs it's powering. I tried this on several boards afterwards with the same result, some fail open, some fail as a short. The PWM was a 1000Hz 3.3V square wave with a variable duty cycle generated with a microcontroller (I also tried a PWM with full input voltage swing with a signal gen, but same result). Connecting the DIM pin to the positive/negative supply without any PWM worked fine. The ICs were obtained from DigiKey and have the following markings on them: "LED2K0", "G9302". Bellow I am attaching the schematic of the circuit in question.

Please advise me how I can get dimming to work.
Thank you.

1 ACCEPTED SOLUTION

Accepted Solutions
RhSilicon
Lead

Look in the datasheet for the implications of selecting the components, in particular on the output capacitor and the decoupling capacitor.

Your schematic shows 2x 10uF capacitors (20uF), but the app says 2.2uF, that's about 9 times the difference (The ESR value of the capacitors is also important).

"In fact, when dimming enables the switching activity, a small capacitor value is fast charged with low inductor value. As a consequence, the LEDs current rising edge time is improved and the inductor current oscillation reduced. An oversized output capacitor value requires extra current for fast charge so generating certain inductor current oscillations"

Layout matters, check out the part that talks about it.

"The input capacitor connected to VINSW must be placed as close as possible to the device, to avoid spikes on VINSW due to the stray inductance and the pulsed input current."

"In order to prevent dynamic unbalance between VINSW and VINA, the trace connecting the VINA pin to the input must be derived from VINSW and design local ceramic bypass capacitor (1 μF) as close as possible to the VINA pin."

 

 

LED2000_AN1a.png

View solution in original post

6 REPLIES 6
RhSilicon
Lead

What is the input voltage?

How is the design of the board layout?

The voltage was 12V, The layout is attached below, the ground pour was removed for clarity (connections still shown).

RhSilicon
Lead

Look in the datasheet for the implications of selecting the components, in particular on the output capacitor and the decoupling capacitor.

Your schematic shows 2x 10uF capacitors (20uF), but the app says 2.2uF, that's about 9 times the difference (The ESR value of the capacitors is also important).

"In fact, when dimming enables the switching activity, a small capacitor value is fast charged with low inductor value. As a consequence, the LEDs current rising edge time is improved and the inductor current oscillation reduced. An oversized output capacitor value requires extra current for fast charge so generating certain inductor current oscillations"

Layout matters, check out the part that talks about it.

"The input capacitor connected to VINSW must be placed as close as possible to the device, to avoid spikes on VINSW due to the stray inductance and the pulsed input current."

"In order to prevent dynamic unbalance between VINSW and VINA, the trace connecting the VINA pin to the input must be derived from VINSW and design local ceramic bypass capacitor (1 μF) as close as possible to the VINA pin."

 

 

LED2000_AN1a.png

RhSilicon
Lead

Another thing is about the DIM pin, your schematic shows a pull-up resistor connected to the VCC, being 12V, so the DIM pin has 12V. It is not recommended to connect this DIM pin in a 3.3V circuit without taking care of it. That's why the application circuit has a pull-up resistor (no external PWM signal) or pull-down resistor (with external PWM signal) selection option.

LED2000_pup.png

To work around this, perhaps a zener diode (3.3V) can be used on the DIM pin, so that the pull-up voltage is not incompatible with the control voltage that the MCU supplies. Or maybe an open collector NPN transistor control output, to keep the DIM pin scheme.

Thank you for the information, I indeed read both of those point, however I am just a bit perplexed that the failure mode of having a slightly larger capacitance than expected is the complete destruction of the IC. Since as far as I understand the only thing affected by the output capacitance would be the raise and fall time, simply limiting the dimming depth (at least that's what the datasheet suggests).

Also regarding the 12V microcontroller thing, I was aware of that and in fact had a clamp setup but didn't mention it in the initial post.

I will try the above suggestions by bodging my boards and post an update regarding it.

I have adjusted the output capacitors value to 1uF and that seems to have done the trick. I still do not understand why it would destroy an IC with a simple increase of capacitance, considering that this is mentioned in the datasheet:
As soon as the dimming input goes low, the low-side is kept enabled to discharge COUT until the LED current drops to 60% of the nominal current. A negative current limitation (-1 A typical) protects the device during this operation (see Figure 12).