LEDs acting as photo cell, preventing SMPS startup - Page 1
Forum section: electronics / beginners

author: avector
author_rank: Newbie
timestamp: May 05, 2019, 09:35:14 pm
content: I'm building an outdoor display using LEDs. Indoors, the display starts up fine, but outdoors it does not. I'm using a TI TPS63071 SMPS chip to drive WS2813b neopixels, and have implemented the TPS63071 example circuit (9.3 Typical Application for Fixed Voltage Version) in the datasheet without modification.In my investigating, the LEDs are dropping Vout to about -0.3 or -0.4 V when exposed to the sun, before applying power. After powering up, I can touch the ground to Vout quickly and the display starts up. The datasheet for the TPS63071 shows the absolute minimum voltage for the Vout pin to be -0.3V, so I believe this to be the culprit.My thought to solving this would be to add a transistor between the TPS63071 Vout and the LEDs that would be switched on when the TPS63071 PowerGood pin was high. I'm not certain the transistor would keep the voltage greater than or equal to ground though, and testing this solution will require me to spin a new PCB. Any thoughts about how to keep Vout above or equal to ground, or any problems I may have missed would be appreciated.Datasheethttp://www.ti.com/lit/ds/symlink/tps63070.pdf

author: wraper
author_rank: Unknown
timestamp: May 05, 2019, 09:43:50 pm
content: You could simply add resistor between Vout and GND. Current from LEDs should be tiny. BTW what is resistance of feedback divider? I would expect it being low enough to sink current from LEDs.EDIT: I checked datasheet and suggested resistor values are quite high. You could simply decrease them to ensure enough of current sinking.

author: thinkfat
author_rank: Unknown
timestamp: May 06, 2019, 09:04:16 am
content: TI usually specifies a voltage divider with very high resistor values for devices with ultra-low quiescent current. If you specify a typical IQof 2 µA into Vout, that makes sense. But if you don't care for that parameter you can use a lower resistor voltage divider which at the same time makes the feedback input less susceptible to noise from the inductor.

author: avector
author_rank: Newbie
timestamp: May 06, 2019, 11:10:58 am
content: Oops, it seems I slightly goofed the part number, but I have the right datasheet linked. I should be TPS630701 instead of TPS63071. TPS630701 has Vout tied directly to the feedback pin, and the voltage divider is internal to the IC.Per Wrapper's idea, a 100 Ohm resistor between Vout and GND did allow the display to start up, but a 470 Ohm did not. I measured 1.25 mA coming from the LEDs in sunlight. If I treat them as a constant current source, I would need at most 200 Ohms, preferably a bit less to ensure stability, to keep the voltage difference below 0.3V. Since Vout is a 5V when running, this will use about 0.125 to 0.25 Watts depending on the resistance chosen. I was hoping for a more power efficient solution since the display is running on batteries.

author: wraper
author_rank: Unknown
timestamp: May 06, 2019, 11:29:31 am
content: I would guess then this is somehow related to negative voltage going into feedback pin. I would try either using TPS63070 which has voltage divider or TPS630702 which also has output discharge.

author: avector
author_rank: Newbie
timestamp: May 07, 2019, 01:55:22 am
content: I'm going to try using the pulse when the batteries are connected to the circuit to temporarily short Vout and ground together (see schematic attached). A high pass filter on Vin (battery voltage) is output to the gate of a transistor, whose drain and source are connected to Vout and ground, respectively. This adds only 3 components to the circuit: a capacitor, resistor, and transistor. With a correctly tuned filter, the transistor should only be on just before starting the TPS630701. This strategy works because the display is switched on and off by disconnecting the batteries via a switch.I will have to wait for the weekend to test the display in the sun with my idea. So far, it has worked indoors without shorting. Perhaps surprisingly, the LEDs do output a voltage under indoor lights, but at much lower current. I can see the voltage go -0.3V to 0 from the transistor turning on (I had disabled the TPS630701).

author: richard.cs
author_rank: Super Contributor
timestamp: May 07, 2019, 06:04:44 pm
content: Maybe I've misunderstood what you mean by negative, but when LEDs act in photovoltaic mode the voltage is in the same direction as in normal operation. I don't see how they could be the cause of the negative voltage unless your schematic is very different from what I am imagining.

author: avector
author_rank: Newbie
timestamp: May 07, 2019, 11:54:12 pm
content: Well, that would certainly make more sense, but now I'm bewildered as to what I'm observing. If I shine a light on the LEDs, I see my Vout go below my ground, but with only a little light, the voltage becomes more positive, but still less than my ground. I've tried shading other parts of my circuit to see if that had any effect, but only the LEDs affected the voltage. I've checked my polarity on my multimeter a couple times just to make sure.Since WS2813B LEDs have an IC inside the 5050 package, perhaps some of the transistors in the IC are acting as a photocell? The IC doesn't look particularly well shielded. Mine are gold, not black, unlike the website image.

author: richard.cs
author_rank: Super Contributor
timestamp: May 08, 2019, 07:23:35 am
content: It has to be the IC then. That's a bit weird, I would expect it to have black paint or other light blocking else I can't see how it would function correctly with the Led on.

author: Giaime
author_rank: Regular Contributor
timestamp: May 09, 2019, 07:22:30 am
content: Quote from: richard.cs on May 07, 2019, 06:04:44 pmIt could be that the ESD diode, in anti-parallel with the actual LED, is going into photovoltaic mode

author: mikerj
author_rank: Super Contributor
timestamp: May 09, 2019, 07:34:04 am
content: You could try a schottky diode from the regulator output to ground. This will clamp the negative voltage and won't consume additional power when the circuit is powered up (apart from a tiny bit of leakage).