I'm using the NTE0506MC DC/DC converter which boosts 5V to 6V. The output voltage I'm measuring with no load is 9V not 6V. In page 3 of the datasheet they mention that it needs a minimum load of 10% of the rated load, which in my case is 16.7mA (0.1*167mA).

So I decided to put a dummy resistance of 232 Ohms (after trying several larger values) in parallel with the output decoupling capacitor. The measured output voltage in this case is 6.33V which is much better but still not 6V (my acceptable tolerance is +-300mV). The 232 Ohms should maintain an output current of ~27mA (6.33V/232) which is more than the required minimum current (16.7mA). Why is the output still not close to 6V? How can i solve this issue?

My circuit is exactly the same as the one in page 4 of the datasheet for the part NTE0506MC and my load needs a max of 20mA.

Update: 25/5/2013

I added two 200 Ohm 0.25W in parallel to the output of the regulator and the ouput voltage is now 6.18V which is within my specs.

  • \$\begingroup\$ What are you measuring the voltage with? \$\endgroup\$ – MandoMando May 22 '13 at 13:58
  • \$\begingroup\$ @MandoMando With "GreenLee DM-110" digital multimeter. \$\endgroup\$ – Abdella May 22 '13 at 14:06
  • \$\begingroup\$ Keep in mind that the DM-110 also has 0.5% tolerance and at 6V it's 30mV. so you have 6V + 300mV (regulator) + 30mV (meter) = 6.33V \$\endgroup\$ – MandoMando May 22 '13 at 14:19
  • 1
    \$\begingroup\$ As well as what Dave et al say - be SURE SMPS moise is not affecting your meter. A simple filter can help when noise affects the meter. eg add 1k in series to voltmeter and say 100 uF across meter leads. \$\endgroup\$ – Russell McMahon May 22 '13 at 15:05
  • \$\begingroup\$ Adding a regulator before the load would help? In this case a regulator shouldn't be wasting much efficiency whether a low or high load appears upon it's output. Plus it can give you much more accurate voltage, with almost no ripple \$\endgroup\$ – Dor May 25 '13 at 21:38

The converter is operating within its specifications. See the "Tolerance Envelopes" at the top of page 3 of the datasheet. The output is allowed to be as much as 10% high at 10% load, which could be as high as 6.6 V in your case.

Try other loads, up to 100%, and see if it generally follows the load line that they show.

  • \$\begingroup\$ Didn't expect the load regulation will be that bad, ok, I'll try other loads and see what happens. Thanks. \$\endgroup\$ – Abdella May 22 '13 at 14:09
  • \$\begingroup\$ Vout = 6.26V (with 150 Ohm load) and 6.16V (with 97 Ohm). Yes it follows their curves. I got one more of this part, will check if it can give better results because a higher wattage resistance will be hard to fit in my board. \$\endgroup\$ – Abdella May 22 '13 at 14:31
  • 2
    \$\begingroup\$ If you really need a precise voltage at low loads, you could add an adjustable LDO regulator at the output of the boost converter. \$\endgroup\$ – Dave Tweed May 22 '13 at 15:10
  • \$\begingroup\$ I like LED's as minimum loads, (If only for the light show.) Might as well do something with all that energy that's being burnt. Could be a really bright one too!!! But on a serious side could you go for a smaller converter? \$\endgroup\$ – Spoon May 22 '13 at 16:26
  • \$\begingroup\$ @DaveTweed Unfortunately, I got the PCB with the mentioned design, so I'm trying to find a simple workaround that also won't look like a mod on the PCB. I think I will get a 100 Ohm 0.4W 0805 and solder it side by side with the output 0805 decoupling cap. \$\endgroup\$ – Abdella May 22 '13 at 16:56

Another thing you need to watch on this device is that the output voltage, over the range of specified input voltages is just about proportional to input voltage - this means that if your input voltage is 5.5V (i.e. 10% high) then your 6V output will be 6.6V.

I've been tripped-up by this on a similar product from Traco.

It's all in the table called "output characteristic" and "line regulation" is the point I'm making - it specifies it has a typical line regulation of 1% per 1% and this, to my mind means if the input goes up 10% then so may the output.

Did you also see that if the output is unloaded (below 5% load), it may double the rated output i.e. rise to 12V.

  • \$\begingroup\$ The input is 5.01V (only 0.2% higher). Yes, I tried the other part I have and at no load it was worse than the first one, the output was about 11V! \$\endgroup\$ – Abdella May 22 '13 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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