Advice with 3,7VDC to 35VAC inverter

Discussion in 'General Electronics Chat' started by David Alba Ramírez, Aug 6, 2015.

  1. David Alba Ramírez

    Thread Starter New Member

    Aug 1, 2015
    2
    1
    Hello everyone.

    I'm working on something that needs to be powered at 35VAC (1.5A max) using a Lithium 4400mAh battery (3,7VDC). I've been searching and testing some DC/AC inverter circuits (using 555 timers, couple of mosfets, some reused transformers...) with little success, and i was wondering if what i'm trying to achieve is even possible. If it is, what would be the best approach? or, how would you solve this?

    Is there some kind of IC that can do the job, similar to a MIC4826 EL driver (but with a bit more of current output, of course)? I don't think so, but well, i have to ask.

    Thanks a lot, any help would be much appreciated, if you need more information just ask.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Going from 3.7v to 35v with a boost converter is not a simple task. There are various issues that come up. One is the inductor current rating. It has to be high enough to handle 10 times the output current at least. Going from 3.5 to 35 volts is a 10 fold increase, so current on the input will be higher than 10 times the output current. That means at least 15 amps on the input.
    For that kind of current you should really have an Li-ion cell that is rated for high current, like 30 amps. The run of the mill cells may have a much lower rating and thus would not work as well. Also remember that the cell voltage runs down and you still want the same output. That means 3.2v to 35v for example which is more than 10 times.
    For this kind of input output ratio, a transformer step up is really recommended, not an inductor based step up. This means the transformer will supply the necessary step up and then that gets rectified with high speed diodes and filtered with inductor and capacitors.
    The transformer primary has to be able to handle the input current too which would be more than 15 amps. At higher frequency that means several strands of lighter gauge wire to make up the full diameter of the wire required for 15 amps, which could be #12 AWG wire for the primary.
    Usually for this kind of current you use a controller IC chip and external switching transistors. The controller chip is generic, but the transistor ratings are chosen to match the application requirements.
    You should probably look up some reference designs to start with, or just look for something made commercially.
    Another thing to think about is run time. With 15 amps from the cell, the run time with 15 ampere hour cell would be ideally 1 hour, so with 1/4 of that cell rating the run time might be 15 minutes if you are lucky.
    What is it that you have to power?
     
  3. pwdixon

    Member

    Oct 11, 2012
    488
    56
    Sounds like you might be describing a dc-dc booster which is relatively easy at low currents but there's also the addition of the AC output requirement which makes everything even worse. Best to ask again what the output is needed for to try to reduce the requirements somehow.
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Yes he needs an AC output. We also then need to know the frequency and how clean the waveform has to be. For example, can the load tolerate what they call a modified sine wave (actually just pulses above and below zero).
    But he could build a DC to DC then later add an AC output stage.
    We might have to come up with a trick, like a DC to DC to get to 12v, then a commercial inverter that outputs AC, then a transformer that steps down to 35Vac.
     
  5. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    There is also the idea that the proposed battery will last about 10 minutes, if it works at all.
     
  6. David Alba Ramírez

    Thread Starter New Member

    Aug 1, 2015
    2
    1
    Thank you all for your help!

    The battery life is something i definitely overlooked... True, to get those 35VAC and 1A as output (35W), an input of 3.5VDC and 10A aprox. would be needed. And that would deplete the battery that i'm using really fast, or as stated by #12 maybe it wouldn't even work.

    I find MrAI's trick interesting, i think that i already have all the devices needed... I even made a DC/DC boost converter circuit with a MC34063 IC and some extra components (almost all of them salvaged from old power supplies and such... mosfet, capacitor, inductor,...). But i guess my battery wouldn´t still be enough.

    What i need to power is an ultrasonic transducer circuit, the ones used in humidifier, water atomizers... The one that i'm trying to use needs those 35VAC - 850mA. I forgot that i have another one that needs a bit less power... 24VDC - 1 A (24W aprox.), maybe it's a much better option (won't need an inverter and would get a bit more battery life).

    I think that i tried that last one, and didn`t get it to work, when i plugged it in, the voltage output from my boost converter circuit just plummeted, i'm not 100% sure why that happened, i need to dig a little more.

    I will post an schematic of the mentioned boost converter as soon as i can.

    Thank you again, anything else you need to know just ask.
     
    #12 likes this.
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