I have a nominal 10A DC-DC booster PCB with both Voltage/Ampere adjustment. It is intended for charging a 5C LiPo and is adjusted to output 17.5V and 5A for charging.

It works very well when powered with a 12V/20A switch-mode supply. But when powered with a 12V Lead-Acid battery, the DC-DC booster breaks down (short circuit on the booster input after a short while).

I simply don't understand this behaviour and I have now destroyed two boosters on this stunt. There must be something different with the switch-mode versus the battery - although both are a 12V source for the DC-DC booster.

What is going on? Can anyone explain this?

The picture shows the 3 parts (switch-mode supply, DC-DC booster and battery)

switch-mode supply, DC-DC booster and battery


The damage is done after a few seconds (5-10) presumably after the LiPo is disconnected on the output (while testing). Maybe the battery should be disconnected first, although this is not needed with the switch-mode supply. However something points to this may be the cause, as there seems to be no problem while the charging is ongoing from the battery (although this is only tried for the mentioned 5-10 seconds).

It's true that there is a swift rise in the PCB temperature, which is not seen with the switch-mode supply. The manufacturer only suggests adding cooling above 6A (its 96% efficient). So it's not a real cooling problem.

As described, the booster input becomes shorted, which may be from an excessive current that shorts the PCB to the underlying aluminium plate. I lean towards thinking there is a resonance phenomenon, but I have never heard about/seen that. The battery manufacturer states an internal resistance of 37mOhm at 1kHz.

I'm not too happy to try anything before things are more clear.


With a fuse on the input of the DC-DC, I have discovered that it blows when the output from the DC-DC is disconnected from the LiPo-battery to be charged. This only happens when using a Lead-acid battery as power source (aka not with the sw-mode).

The difference is, that the Lead-acid battery can burst up to 40A and by all means the DC-DC apparantly destroys itself by eating all this.

Key to solve this problem is: Why does the DC-DC sukcs all that Amp. when its output is disconnected.

The trivia is gaussian: when the current from a coil is removed, the coil tries to keep the current by raising the voltage and a runaway situations is happening in the control circuit (IC). The sw-mode apprantly prevent this by simply its overvoltage protection etc. which a LA battery doesnt.

The non triva is: Why did they design this DC-DC to self-destruct, well knowing that someone possibly would use a powerfull battery (and not a sw-mode) as the power source.

The not-so-happy-solution is to use a finger-resattable fuse on the input, but a better is recommended.

  • \$\begingroup\$ What is the maximum input voltage rating on your booster? A lead-acid battery will put out more than 12V when fully charged. Could it be that extra volt or two is enough to destroy y our booster board? \$\endgroup\$
    – AlmostDone
    Commented Jun 15, 2018 at 1:17
  • \$\begingroup\$ nope - the input range is 8.5-48V with an output range of 10-50V, whis says that there is a 1.5V voltage-drop. It works so well so that the output stays at the adjustment even at a varying input range (ofc with respect for the voltage drop limit).. Neverthelees, the battery can according to specs, burst deliver 44A for 5 secs, aka kick ass for a short while... but the booster should limit that, the switchmode are rated 20A and this works well, so... \$\endgroup\$
    – Gearlos
    Commented Jun 15, 2018 at 1:21
  • \$\begingroup\$ I am using what appears to be the exact same DC-DC converters for my portable lighting rigs. Power source for my units are Dewalt 20V 6AH or 9AH battery packs. Loads are Honeywell LED Barn Lights: at (IIRC) 47 Volts 1.3 Amps. No problems after many hours (days, weeks) of use. \$\endgroup\$ Commented Jun 15, 2018 at 3:25
  • \$\begingroup\$ Yes Dwayne.R - But my setup is not for powering LEDs that are pure power ballasts, but a LiPo battery which may be a more complex setup. Anyway, good to know others experience... \$\endgroup\$
    – Gearlos
    Commented Jun 15, 2018 at 5:49
  • 1
    \$\begingroup\$ Inrush current destroying the diode causing the MOSFET to fail? \$\endgroup\$
    – winny
    Commented Jun 15, 2018 at 11:36

1 Answer 1


A complete analysis of resonance is hard without datasheets and a model for each part. But the primary fault is system design with lack attention to thermal losses and cooling. It requires a fan and heatsink.

The simple theory is the series resonance gain \$Q = \dfrac{X(f)}{R}\$ which is the reactive current gain also increases series losses \$I^2R=Pd\$ in its path such as RdsOn of the input MOSFET. Although the battery C gives a very long time constant \$T=C*ESR\$

The MOSFET can have a PTC tempco. and gets hotter with rising temperature leading to thermal runaway when spurious oscillations occur in the primary side switch. Powering LED's, no problem, powering a low impedance high Q load "big problem" for a switched voltage source.

This is the nature of high frequency charging batteries is that AC currents can also be high increasing ripple currents in caps and conduction losses in MOSFETs.

Unless there is a suitable system design of impedances and peak currents and resonances, many faults may occur. Reducing the resonant currents comes understanding how every part's impedance contributes to the system control loop.

This will not be a thorough analysis but just a low level discussion of some prinicples.

  • output impedance is "transformed" to source to a lower "average" impedance by virtue of boost regulator in voltage and \$\dfrac{V_{IN}}{I_{IN}}< \dfrac{V_{OUT}}{I_{OUT}}\$ for a boost regulator just as in an induction voltage transformer. Which means the low battery impedance looks even lower to the primary switch.

The DC-DC board is on an aluminum alloy substrate of about 2 sq. in. which in free air without heatsinking will normally get hot with about 2 Watts of loss should be mounted to a large heat sink for max power transfer with thermal grease. ( details held back)

Considering the application supports up to 85W output , what are your expectations ?? for worst case efficiency into an "almost" short circuit load. ( i.e. a battery) How hot did it get? how quikly did you react to shut it off?

THese are user errors not component faults.

  • the SLA battery has a ESR of 12V/44A or 0.25 Ohm minimum that increases as battery charge reduces or ages. This can stretch the primary current duration when the voltage dips during cycles. Boost regulators transform voltage up at the expense of more primary current and resonant impedances can increase energy losses. This may leading to oscillations with a battery that has a memory effect and high ESR resulting in unstable operation.

    • suggested solution: use appropriate low ESR caps that are rated for primary DC ripple current. e.g. 5Arms at the switching current frequency. Several caps are needed to support the one on board over a range of values low-esr tantalum and plastic to span spectrum of input current.

The battery also has higher ESR than step down regulators with ultra-low ESR output caps and even some with ferrite LC post regulation output filters.

This can cause primary boost current to resonate more. Also using the boost regulator to charge a battery with very low ESR now means an even lower impedance applied to input RdsOn switch meaning more load current.

Solution reduce current limit, add heatsink and microfan to board and add several large caps across regulator input with <=10mOhm ESR.

LED loads are low ESR with eg. 47V @ 1.3A = 61W implies ESR <<1 Ohm unless regulated with current sink.

But Battery loads equivalent to 100k Farads have an ESR according to voltage changes from step currents ΔV/ΔI=ESR which is closely related to short circuit current to Voc/Isc and will be xx mΩ range.

  • \$\begingroup\$ Tony.S. - I made an addition to the problem description after Your comment. \$\endgroup\$
    – Gearlos
    Commented Jun 15, 2018 at 5:51
  • 1
    \$\begingroup\$ +2 This is my assessment based on the information at hand. Simple test would be to put a scope on a low R current sense shunt on the input to observe waveforms and a recording voltmeter to see when the behaviour occurs. At a guess I would think that the ESR and discharge of the small SLA battery drops to close to 8.5V and the behaviour of the converter is no longer well defined and it fails in some design specific way. The suggested caps may assist but only to a point as the SLA discharges further and the input voltage rating is crossed again. \$\endgroup\$
    – KalleMP
    Commented Jun 15, 2018 at 16:51
  • \$\begingroup\$ Note to all: I have written an update-2 in the original question... \$\endgroup\$
    – Gearlos
    Commented Jul 3, 2018 at 21:24

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