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So, I purchased a buck converter which was advertised to be able to range down to 0.8V but, as you can probably guess, it only goes down to 1.25V which I understand is the reference voltage for the chip it uses and quite common.

I'd like to know if there's an efficient and easy way that I can step down the output range further so that I can get down to that 0.8V or even 0V, without modifying the module too much.

I'll be using 24VDC as the input and the actual chip used in the converter, the XL4016, has a range of 1.25V to 36V.

The solution in EEVblog #221 is to replicate and replace the entire chip. That's just too much for me as I'm fairly new to electronics. Another solution might be to find another buck converter that has an even lower reference voltage but then it may not match up with my current requirements.

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    \$\begingroup\$ There are controller and regulator chips out there with lower reference voltages. 0.8 and 0.6 V are fairly common. Just return the one you got and buy one that uses one of those chips. \$\endgroup\$
    – The Photon
    Apr 28, 2016 at 16:57
  • \$\begingroup\$ @ThePhoton I've already initiated a return but I'd like to know how since many of these buck converters do have non-ideal minimums. \$\endgroup\$
    – Anthony
    Apr 28, 2016 at 18:42

1 Answer 1

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You can use a fast op-amp to amplify by 2 and feed it with the output voltage (say 1.25V). The output from the op-amp feeds the reference input and if you do this correctly, the switcher will force the pulse width down until there is 1.25V on its reference pin. This means the output MUST be at 0.625 volts.

In other words you have conned the switcher by using a gain stage. There can be stability issues so use a fast (possibly 10MHz) op-amp like the AD8605 and be prepared to add some filtering capacitors.

You can also do this by adding a DC offset into ref signal. So, take the output voltage and pull it up with say a resistor and diode. This raises it by ~0.7V and will force the switcher into producing an output that is 0.7V lower on the output. No gain involved here so stability probably won't be an issue.

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  • \$\begingroup\$ I understand the first option but wouldn't quite know how to wire it. It's discussed in the EEVBlog post and the stability issue is brought up too. I guess it's worse over the whole range of voltages but it can work. Why not go with the second option right off the bat? Less efficient? And why 0.7V? Ideally, I'd want 0.8V out at the bottom, so raising it by 0.45V would seem to make more sense, no? \$\endgroup\$
    – Anthony
    Apr 28, 2016 at 18:44
  • \$\begingroup\$ A little bit of imagination.... you have a spare 5V rail from maybe a 7805 - make a pot divider with the switcher's output to precisely set the feedback plus an offset that you can control with the ratio. You should be able to get all the way down to 0V. Watch stability although it should be stabler than the version using an op-amp. You also need to be aware that if the 5V gets disconnected (and becomes 0V) your switcher will try and make the output something a lot higher than 1.25V. \$\endgroup\$
    – Andy aka
    Apr 28, 2016 at 20:21

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