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I am designing a field charger for charging lipo batteries while I am out flying and on the go and I already have a huge 16000mah 4S lipo available that I could use to charge my 1300mah 4S and 2200mah 2S lipos while I am out and about. I want a cheap, but more importantly compact solution so although there are multiple battery chargers available, none seem to meet my needs.

Not wishing to re-invent the wheel, I have come across this charger that a friend uses that is actually very accurate despite its cheapness (around £9GBP delivered - https://hobbyking.com/en_us/hobbykingr-dc-4s-balance-charger-cell-checker-30w-2s-4s.html)

I am thinking of using 4 of the boards from these so I can charge 4 batteries added at different times while I am flying.

My issue is that the operating voltage is 9-16V, fully charged 4S lipo is 16.8V.

I understand there is usually some allowance made in this figure so I opened up my friend's unit to inspect. I cannot seem to find any components that are obvious to me that cannot handle 20V.

Would anyone be able to assist me in trying to assess before I either buy 4 of these, or otherwise try it out and blow up my friend's charger? (always a risk I know).

All of the caps are rated 25V, of the smd fets I can identify, those are rated at 20V. My question is if looking at these two photos I should pay close attention to any particular components.

I COULD use a buck converter to convert down to 12V first, but that will add bulk and reduce efficiency, so I wish to avoid that if possible. I will obviously check that all chargers are charging to accurate voltages before deployment. (I am experienced with lipos even down to a scientific level where I have designed some, so I am aware of the dangers, this is just for my hobby, nothing else.) I also wish to learn more about how to assess circuits that are not my own / well documented.

Many thanks in advance.

Front of PCB

Rear of PCB

EDIT: The inductor for the buck/boost stage is pulsed using what appears to be an 'SI4606' - http://www.datasheetspdf.com/datasheet/search.php?sWord=Si4606

The control MCU is a Holtec HT46R066B - http://biakom.com/pdf/ht46r06xbv110.pdf But VSS is max 6V so like other ICs seems to be supplied by a reg... Other IC chips can be identified but they do not seem to see, from what I can tell, V_Input directly.

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  • \$\begingroup\$ The best would be to have a schematic of this board but I could not find one in a quick search. Since you can charge 4 cells with this device and it accepts a voltage below 16 V it must have a DCDC step-up converter. It is possible that this is upconverter cannot prevent charging current to flow into a stack of 4 cells when Vin > 16 V while the cells at the output are fully charged. You should list the type numbers on the ICs and add links to their datasheets. That will give a better clue how this is designed. \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 14:25
  • \$\begingroup\$ Hi, thank you... Yes that was my first thought, btw... but, the unit will charge 3S and 2S lipos, so it must also do buck as well as boost. Moreover, voltage for a discharged 4S may be as low as 12.8V, safely... so it must be able to switch between modes during charge if supply was 16V, for example. I could not find schematic either or I would have posted. Will try to list what ICs I can find... Are there any that would seem critical from the layout? \$\endgroup\$ – Rendeverance Sep 28 '17 at 14:29
  • \$\begingroup\$ I see the Holtek ht46r0.. microController, not so interesting. The 8 pin and 14 pin next to it, have the LED display on the other side, likely just for handling the display. The other 8-pin chips appear to be MOSFETs, I see that because some pins are used in pairs. I'm more interested in the architecture (overall design), my guess is that it is a current-feedback step up converter. So it does not make a fixed voltage but a fixed current. When charging 3 cells, one output is simply shorted (the one with no cell). \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 14:39
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    \$\begingroup\$ Instead of maintaining the voltage it simply maintains the current. Look up how boost a converter works, it charges an inductor and that inductor discharges (with a current) into a capacitor. The voltage can go very high of you do not load the output or stop switching. Most step-up converters must have an output voltage that is higher than the input voltage so that might give the 16 V limitation. I doubt that this design is a step up/down converter, it is not needed and would require more components. \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 14:50
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    \$\begingroup\$ No the energy is not wasted, the components would get HOT. The trick of a CC step-up converter is that it would automatically lower the voltage to maintain the same current. It does this by adjusting the PWM of the switch in the step-up converter. If the voltage was kept constant then yes, you would be right the power would be "burned". But because of CC the voltage will be lowered and almost no power is lost. Neat huh? \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 14:57

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