Looks like I need help understanding how to use the OKR-T/1.5-W12-C DC/DC voltage converter, b/c I just burned two of them. I am obviously doing something wrong, but I don't understand what.

Here's the part info: http://www.digikey.ca/product-search/en?x=-989&y=-73&lang=en&site=ca&KeyWords=811-2782-ND

And here's how I hooked it up: enter image description here

The regulator begins heating up rapidly and burns itself out. The first one lasted two days, the second one died in 5 minutes. Turning it on or off with the On/Off pin makes no difference to the heating process. The regulator does supply the expected +Vout until it dies.

P3 is a AA NiMH battery that I am trying to charge.

What am I missing?

EDIT 1: Since the problem is not turning out to be obvious, I think it is time to update this post with complete info. Here's the schematic (I hope you can read that): enter image description here

As you can see, I've added the capacitors and a constant current sink. The regulators continue to work for a minute or so and then break to the point where Vout just sits around 1V, not reacting meaningfully to Rtrim anymore. Vout does change when I change Rtrim, but these changes are tiny.

EDIT 2: I'll also add the physical design picture, just in case there are any clues there: enter image description here

EDIT 3: I have implemented many of the suggestions provided here with a different buck converter IC, but still no luck! Details here: Strange oscillation in buck converter (AP3431)

  • \$\begingroup\$ What is your load and how much current does it draw? Does the regulator still heat up with no load connected? \$\endgroup\$
    – kwc
    Apr 19, 2014 at 4:47
  • \$\begingroup\$ What kind of battery is P3? If +Vout=1.5V, minus the diode drop across D2, is higher than the battery voltage, the battery may be drawing a high charging current unless there is some current limiting in the voltage regulator circuit. Have you measured the current into/out of the battery (in addition to the load, as mentioned above). \$\endgroup\$
    – Matt B.
    Apr 19, 2014 at 4:49
  • \$\begingroup\$ Have you considered putting a replaceable fuse at the input and output? \$\endgroup\$
    – Matt B.
    Apr 19, 2014 at 4:58
  • 1
    \$\begingroup\$ Just in case you assumed pin 1 on the device is at the left when looking at the front of the device, it isn't it's on the right hand pin. \$\endgroup\$
    – Andy aka
    Apr 19, 2014 at 12:39
  • 1
    \$\begingroup\$ A USB-sourced +5V rail is not going to be able to support high current without appropriate resistors on the +D and -D signals. I would check for input sags as well as the previous suggestions of appropriate input and output capacitors. \$\endgroup\$ Apr 19, 2014 at 14:23

2 Answers 2


If I were you, this is what I would do first:

  • Read every page of the datasheet.

  • Is "pin 1" really what I think it is? The pin numbering on p. 12 of the datasheet runs in the opposite direction from what I expected.

  • Do I have every pin connected correctly, according to the input/output connection table on p. 12?

  • p. 14 recommends an input capacitor and mentions a "minimum external output capacitance required for proper operation", but neither is shown on the above schematic. Do I have at least the minimum recommended input and output capacitors?

  • Directly connect pin 1, the On/Off Control input, to GND or +Vin. The p. 3 of the datasheet mentions "a 49.2 KOhm pulldown to GND", and p. 4 mentions "The On/Off Control input should use either a switch or an open collector/open drain transistor references to -Input Common". The 10 MOhm on the above schematic is outside those guidelines. A 10 MOhm pulldown is weak enough to be easily overcome by exactly the sorts of high-current transient noise that I expect a switching voltage generator to produce.

  • Consider looking at each pin with an O'scope -- is it what I expected? (smooth, constant input and output voltages, without any significant transients or pulsing or ringing).

And if all that fails:

  • Ask for help, perhaps on Electronics Stackexchange.
  • \$\begingroup\$ Thank you for the detailed suggestions. I am pretty sure I've got the pinout right (otherwise I would not be getting the voltages I expect), but I will follow all of your other suggestions about caps and resistor and post here after I try it with a new unit. \$\endgroup\$
    – Val Blant
    Apr 19, 2014 at 20:33
  • \$\begingroup\$ The caps made no difference. Read and re-read the datasheet, but still no closer to a solution. Question updated with latest info. \$\endgroup\$
    – Val Blant
    Apr 26, 2014 at 3:50

At the very least, you need a resistor between D2 and pin 1 of battery P3, to limit the charging current into the battery. As it stands, the battery can draw as much current as the regulator is capable of putting out. Besides being probably way more than you want, it isn't even a known quantity. It will be limited somewhere between the regulator's overcurrent protection and the maximum the battery can sink. It's similar to putting nearly a dead short on the output of the regulator, in parallel with whatever the intended load is, where the wire goes off the top of the schematic near the battery.

If the module does in fact need input & output capacitors as mentioned on page 14 of the data sheet (and comments above), as it stands you are to some extent using the battery as an output capacitor. You're using the output capacitor of the 5V regulator in the USB device as your input capacitor, but then probably some long wire leading to this circuit, which adds resistance and inductance. So it's possible the regulator could produce the required output but would not be stable under variation in the load. It's not clear whether the module has these capacitors on it or not, as ceramic capacitors wouldn't be all that big and might be the larger brown ones between the inductor and the pins. It wouldn't hurt to add more, up to some level, so maybe add some at the low end of the recommended range, 10uF input and 50uF output, in case they'll be in parallel with any that are on the board already.

You have no battery charge termination, meaning no circuit to turn off charging when the battery is full. If you intend a continuous trickle charge and/or you'll disconnect the battery manually after a reasonable time, or don't care if the battery gets badly overcharged and ruined, maybe the current-limiting resistor would be enough. If you're trying to charge the battery at all quickly (a few hours or less), you need a better charger circuit.

I asked about putting a scope on +5V and the On/Off pins in comments because it's possible they are oscillating for some reason, triggering the regulator to have to ramp up & down over and over, where it might look like the output is being regulated but the circuit isn't working as expected. +5V could sag or oscillate if it's being overloaded (as Madmanguruman mentions), On/Off pin could oscillate because R6 is too high a value (as davidcary mentions).

To start with, you could take D2 out of the circuit so the regulator has neither the battery nor load on its output, and scope every pin you can touch to see if things look as you expect, and stable. Then put say a 10W power resistor (calculate the value so it draws similar current as your actual load) between the regulator output and ground and scope things to see if they still look good. This tests the regulator unloaded and fully loaded. You could add tests for half load, pulsed load, etc. if you don't trust the regulator or want to understand its behavior better.

Once you believe the regulator can do its job, then try connecting the actual load and not the battery. Maybe first add a fuse between the regulator and load just in case. If the regulator can run the load properly with the battery out of the circuit, then try adding back the battery, with current-limit resistor.

  • \$\begingroup\$ Thanks for the detailed answer, Matt. Since I broke all 3 of my regulators now, I have to wait until I get more to try your incremental troubleshooting suggestions. In the mean time, I'll update the original post with all of the current information about the circuit. \$\endgroup\$
    – Val Blant
    Apr 26, 2014 at 0:05
  • \$\begingroup\$ OK, hope it helps. In the added schematic, can you explain what IC2 is and does? It's hard to read. Also, in the photo, what is taped onto the battery, is it a thermistor for charging? Is that on the schematic? \$\endgroup\$
    – Matt B.
    Apr 26, 2014 at 6:41
  • \$\begingroup\$ In my answer above, I assumed P3 connected to battery + on pin 1 and battery - on pin 2, which does not seem to be the case. I can't read the new schematic well enough to find P3 but maybe the other side of P3 also includes IC2 and not just the battery cell? \$\endgroup\$
    – Matt B.
    Apr 26, 2014 at 6:45
  • \$\begingroup\$ IC2 is an Arduino. A 1-Wire temperature sensor is taped to the battery. It is marked in the schematic as K3. The new schematic is actually a very high res image - I don't know why stackexchange is not letting you click on it. If you get the URL to it and open in another tab, you'll be able to see a lot of detail. \$\endgroup\$
    – Val Blant
    Apr 26, 2014 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.