DC-DC booster fails with lead-acid battery source

Question

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)

Update:

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.

Update-2:

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.

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.