Easiest Way to Raise DC Source's Voltage?

  1. What would be the easiest way to raise the voltage of a DC source? (Mainly batteries)

    For example, say I have a 9V battery and would like to convert it to 120VDC. What would be the best way of doing this without just hooking 13 or 14 9V batteries is series?

    If this is impractical, please feel free to educate me.

    If you recommend a component to buy, please try to include a link.

    Last edited: Mar 17, 2014
  2. jcsd
  3. berkeman

    Staff: Mentor

    Is that 120Vac (like AC Mains), or 120Vdc?

    Either way, that is a dangerous voltage level. What background do you have in working with high voltage? What is the intended use?
  4. DC, sorry, I'll fix that.

    And I have a some experience with HV (up to about 500V), but its all been in a lab with HV power supplies. And that's just an example voltage, it wouldn't necessarily be that high. But I'm not sure I would call 120V "high voltage". The IEC defines HV for DC being in excess of 1500V. 120V is on the bottom of the Low Voltage range technically. Sorry if that sounded a bit rude; I appreciate your concern and will take proper precautions.

    As for the use, its pretty simple. I need to construct a Helmholtz Coil and want the magnetic field to be decently strong.
    Last edited: Mar 17, 2014
  5. Last edited: Mar 17, 2014
  6. berkeman

    Staff: Mentor

    For the purposes of the PF, IMO, anything above SELV is "High Voltage". We get way too many newbies on here who have no idea of the shock and fire hazard presented by voltages above SELV, and also no idea about the safety regulations for dealing with AC Mains voltages when building projects.

    It does sound like you have some experience, and Baluncore's advice is good. I've never built a boost converter up to that voltage level -- the ratio of Vout/Vin does seem a bit problematic. How many turns are you planning on for your Helmholtz coil? It's obviously the current that matters, and you can use bigger coil AWGs to get good currents without so much voltage boost...
  7. berkeman

    Staff: Mentor

    BTW, you know that at best, DC-DC converters are near 100% power efficient, right? So that means that the current you draw from your 12V source will be about 10x the current that you supply at 120V. Is there a reason that you need to start with a 12V source? Are you wanting to make this battery-powered, instead of using AC Mains as the power source?
  8. I don't see why such high voltages are needed at all. The magnetic field depends on the current and not the voltage. If the resistance of the coils is too high to get the maximum current that your supply can deliver ,use less loops of thicker wire.

    If you use half the number of turns,and wire with twice the cross section, you can use half the voltage, twice the current, and get the same field and use the same power and the same amount of copper.

    googling up some figures for a Helmholtz coil


    show that it has 2.1 ohm resistance and a maximum current of 5A, so you need only 10.5 V to get the maximum allowed current through the coil. (If your power supply can deliver 10 A for 2 coils)
  9. russ_watters

    Staff: Mentor

    Minor typo with a major impact there; you meant 1/10th the current.
  10. Baluncore

    Baluncore 3,305
    Science Advisor

    I think not.
    10A * 12V = 120 W
    1A * 120V = 120 W
    The English language is elegantly reversible.
  11. russ_watters

    Staff: Mentor

    Oy, you're right. I got confused by the order of the phrasing -- that's for posting first thing when I woke up.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted