Need ideas for PCB repair to fix LDO thermal shutdown issue - Page 1
Forum section: electronics / beginners

author: rbm
author_rank: Regular Contributor
timestamp: November 12, 2016, 12:31:09 pm
content: Hi,I've encountered a problem with the current iteration of my design, whereby the 5V LDO I'm using to drive a 7-segment display is dissipating too much heat and shutting down. I need ideas how to quickly correct the problem hopefully only by a slight modification to the PCB and not a complete redesign if possible.I build a motorcycle gear display circuit that uses a Digistump Digispark Pro Arduino as a core component. It can drive a variety of LED-based gear displays including a single digit 7-segment display. In a previous iteration of the design, I used an outboard MCP1703-5002E LDO in a SOT223 package mounted on the PCB to provide the +5V rail used to drive the display. Voltage into the regulator was the 10V - 13.5V from the motorcycle electrics and the solution worked well. Any heat generated by the voltage drop across the LDO dissipated adequately through the PCB.In a fit to add an additional display option, I replaced the outboard LDO with a current mirror, with the intention of using the inboard LD1117-5V LDO in an SO89 package integrated on the Digispark to drive the 7-segment display. This output is advertised to provide 5V with up to 500mA current. The 7-segment draws a maximum of 158mA worse case so I believed I had enough headroom to safely replace my regulator. To be clear, the reason I didn't keep the regulator was because I am imposing a limitation on board size of 5cm X 5cm and thus PCB real estate is limited.Full Schematic of modified design for referenceTypical beginner move I ordered new PCBs without fully testing my new design. I'm finding that the LDO heats up and can thermally shutdown with the Digispark Vin at 12V. If I reduce the supply into the PCB to 9V, the LDO can handle the heat generated and the circuit works. I have been trying to come up with a way to reduce Vin to the Digispark or find another means to drive the 7-segment display without major surgery on the PCB. I've thought about a way to drop the voltage to the Digispark by using a resistor in series with Vin but feel that won't be the best solution. I've considered cutting the supply line to the 7-segment and soldering in an MCP1703 on the board's backside, feeding its output to the switch. Don't know yet how doable that is.Can anyone advise me as to whether they can see a better quick fix? That, or tell me I have to go back to the drawing board. Thanks.

author: LA7SJA
author_rank: Regular Contributor
timestamp: November 12, 2016, 12:53:29 pm
content: Try a 7808 type regulator on a small heatsink to reduce the 13.8V down to 8V.http://www.bravoballroomdancestudio.com/b3d773-7808-regulator-quick-shoppingJohan-Fredrik

author: dmills
author_rank: Super Contributor
timestamp: November 12, 2016, 12:58:34 pm
content: Replace D1 with a ~4V zenner diode turned around the other way (A 2W part would be indicated)?@10V input it will leave 1V across the LDO (Should be ok, but check the datasheet), and @ 13.5V will leave 9.5V at the LDO input.An alternative would be to make S1 in 7 seg mode connect the raw 12V output to ILED. Personally I would separate this selection of ILED source from the 7 segment mode as this would allow the use of individual bulbs without needing to bodge in an extra wire.Regards, Dan .

author: sleemanj
author_rank: Super Contributor
timestamp: November 12, 2016, 01:20:05 pm
content: Quote from: rbm on November 12, 2016, 12:31:09 pmAs you have learned the hard way, "linear regulator can provide 5v 500mA" is meaningless without knowing input voltage, with 13.5v that's 8.5v drop and even at 100mA that's nearly a watt of heat to dump, even a to-220 would struggle with that unassisted.Anyway, I would get a mini buck converter for a few dollars off your favourite online marketplace set it's output voltage to 5v and use that to power your entire system.

author: amitchell
author_rank: Regular Contributor
timestamp: November 12, 2016, 02:21:48 pm
content: These are slick, mini buck SIP drop in.http://ca.mouser.com/ProductDetail/Murata/78SR-5-2-C/?qs=35WhQNrE6p3H6nhgWoUFlQ%3D%3D&gclid=CLvVz_-0o9ACFYqBfgodBI0CxQ

author: David Hess
author_rank: Super Contributor
timestamp: November 13, 2016, 07:20:09 pm
content: Quote from: rbm on November 12, 2016, 12:31:09 pm"Whoops! There was a problem displaying this image."I had the same problem regulating car power down to 5 volts for LED drivers but I knew it would be an issue from the start so I used a massive off board 3 amp TO-3 LM309 5 volt regulator and heat sink which was probably larger than your entire device. And it still ran hot but I was driving a lot more LEDs than you are.I doubt you can save your existing printed circuit board so what I would try is using either a big external low voltage zener diode or a lot of rectifier diodes in series to drop the voltage before it reaches your regulator. It looks like you are dealing with 1.5 watts so a 1N5338 5 watt 5.1 volt zener in series would be about right but keep in mind that this does *not* lower the total power dissipation; it just redistributes it.

author: rbm
author_rank: Regular Contributor
timestamp: November 13, 2016, 08:51:52 pm
content: Thank you for the great suggestions. Some are not viable and I'll explain, but others are very good and I hadn't thought of them.It's not evident from the circuit, but the 12V input drives a start enable circuit and a neutral detect circuit. This circuit detects when the motorcycle transmission is in neutral and send an enabling signal to engage a relay in the motorcycle's ignition. This relay is 12 Volts and draws 537mA in addition to the current drawn for the circuitry plus the display. That's about 700mA. Therefore, I can't drop the input voltage at D1 because (1) it would affect the function of the neutral and start enable circuits and (2) the power dissipated across the zener would be 4*0.7 = 2.8W. That's a big diode. What I can do is introduce the zener in the Vin line to the Digispark. That would never exceed 167mA current draw. I'm going to also experiment with the 7808 regulator idea and look for small cheap buck converter to try that idea.In the long run though, I'm going to redo my design and chock this up to experience.

author: rbm
author_rank: Regular Contributor
timestamp: November 13, 2016, 08:57:17 pm
content: Quote from: dmills on November 12, 2016, 12:58:34 pmDan,Are you suggesting driving the 7-segment display from the +12V? I have 1/8W SMD resistors on the display card and I don't think they are large enough. Each would need to dissipate almost 200mW. I chose the 5V LDO to limit the voltage needed to be dropped by the resistors and therefore the power.

author: dmills
author_rank: Super Contributor
timestamp: November 14, 2016, 01:42:43 am
content: Those resistors being 125mW was not on the schematic.....A high efficiency display at 10mA per segment would not be bright enough (~100mW per resistor)?If you are going to respin, TI have some lovely (and very easy to use) buck converters in tiny packages which would be the efficient way to do this).Protip, pick an inductor that is in a package that you can source from several manufactuers (Looking daggers at TDK here).73 Dan.

author: rbm
author_rank: Regular Contributor
timestamp: November 14, 2016, 09:32:53 pm
content: Unfortunately, I can't do a thing about the Start relay; it's OEM. The NDT2955 P-ch MOSFET in my design acts as a switch to engage/disengage that relay as required.I do select automotive grade components where possible. I am using what I think is an efficient display but I'm not sure how you are defining efficient. I will be redesigning the part of the circuit that drives the 7-segment display for the next generation of the board and relaxing my dimensional limitation to open up more real-estate for components.It usuallly isn't a problem with the elements because this circuit is usually installed in the relay box. It is protected and has all signals and power readily available. The board will be subject to 40*C temperature maximum and normal humidity. It's possible that the temperature in that area could rise above 40 if left sitting but it quickly cools off due to air movement.