## Server Bug Fix: Tantalum Capacitor Explodes When Engine Starts

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I designed a circuit using a SMPS voltage regulator based on the tps65261rhbr triple synchronous buck converter. The circuit is rated for up to 18V. It is connected to a 12V lawnmower battery, which also starts and powers a gasoline engine that returns charge to the battery with an alternator.

The circuit has worked perfectly over many hours of testing and switching power on and off before I connected and started the engine. Immediately upon starting the engine it failed: one of the 25V 47uF tantalum capacitors (TPSD476K025R0150) feeding the SMPS exploded.

My oscilloscope has a max of 10V, so I can’t see what the input voltage waveform looks like when the engine starts and runs. I tried anyway, and I see that when the engine is started the voltage briefly dips under 10V, but outside of that brief blip it is clipped to 10V so I can’t tell what’s happening. I assume the voltage must have exceeded 25V for the capacitor to explode.

I’m considering switching to higher voltage (50V?) aluminum polymer bulk capacitors and a high input voltage rated 12V LDO before the SMPS to protect against input voltage spikes.

Does this seem like a good approach? Should I get a better scope or build a voltage divider to see what’s really happening? Does anyone have any experience with powering circuits from an engine alternator and battery in parallel that can weigh in on this power supply design?

Tantalums are very sensitive to overvoltage so you have to derate them if you want to use them. They are already typically derated by 30-50% in normal use but you are connect them up directly to a gas engine. Gasoline engines are a very harsh source of power so you should be installing transient suppression and the like anyways such as TVS diodes or MOVs to suppress voltage spikes. Regardless, you probably shouldn’t have chosen tantalums in the first place as the input decoupling capacitors knowing they would be directly exposed to something so harsh as a generator.

No LDO, or any type of linear regulator that matter. Having a linear regulator defeats the purpose of having an the efficiency of an SMPS and they are too delicate for the protection task anyways. Furthermore, 18V to 12V with a linear regulator is too much heat for any remotely moderate levels of current.

Get a big TVS diode with a working voltage (not a breakdown voltage) as close to but greater than the battery voltage at full charge. It would help if you could scope to see what the startup transients, and the transients in general are like. There’s a chance the TVS diode won’t be able to handle the power in which case you need to go with a metal oxide varistor (MOV). But if a TVS diode can do it, then a TVS diode will be better. MOVs do not not clamp as well as TVS diodes and have an inherent wear out mechanism each time they conduct so you don’t want to accidentally undersize it if you expect it to be constantly experiencing strikes or else it will wear out early, but they can be made a lot bigger (like bricks!) so can found in much higher power levels.

And go ahead and toss in that 50V aluminum polymer. You probably don’t need quite so high as 50V though. Aluminum polymers don’t need very much derating.

Might as well toss in a fuse while you’re at it.

Tantalum capacitors have a very low effective internal series resistance. A quick, low impedance fed change in voltage from the alternator can cause the current through the tantalum capacitor to exceed its maximum allowed value and blow up the capacitor.
I’ve experienced this behaviour once in a 40 kW IGBT inverter, (repeatedly) blowing out the IGBTs, and the remedy was to insert small series resistances in series in order to limit the equalisation currents to acceptable values.
Could it be that you are using an old fashioned alternator with carbon brushes and commutator?

## Server Bug Fix: Calculate peak input current of switch mode power supply

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I have the following problem.

MY ATTEMPT

The switch-mode power supply looks like it’s a boost converter. I found that the DC-transfer functions for the boost converter to be:

$$frac{V_C}{V_{in}}=frac{1}{1-d} : :$$ and $$frac{I_L}{I_{out}}=frac{1}{1-d} : :$$ where I presume that $$d$$ is the duty cycle.

My idea was first to find $$d$$ with the first equation: $$d=-frac{V_{in}}{V_C}+1=frac{-2.7 text{V}}{5 text{V}}+1=0.46$$

From here, we can plug $$d$$ into the second equation and find $$I_L=frac{I_{out}}{1-d}=frac{2.3 text{A}}{1-0.46}=4.256 text{A}$$

And since the $$V_{in}$$ and the inductor is in series, we conclude that $$I_{in}=4.256 text{A}$$.

But is what I have found the $$underline{peak}$$ input current or something else? I am in doubt, because I don’t use the information about the switching frequency at all.

I hope someone can help me with this.

First thing to notice is that you have a a very small output capacitor, which is used to supply the load during the on phase (when the inductor is being charged and there is no power transfer between the primary and secondary sides).

The required capacitor to hold the output voltage long enough until the next cycle can be calculated through the following:

$$C=dfrac{P_ocdot D cdot 2}{f_Scdot (V_o^2-V_{o,min}^2)}$$

where:

$$P_o$$ is the output power ($$5Vcdot 2.3A=11.5W$$)

$$D$$ is the duty cycle $$1-dfrac{V_{i}}{V_o}=0.46$$

$$f_S$$ is the switching frequency

$$V_o$$ is the nominal output voltage

$$V_{o,min}$$ is the minimum output voltage

Plugging in your values, you are gonna find out that with your current capacitor, the output voltage drops to zero at every cycle because of the high load. This also means that at every off cycle, the output capacitor will have to be recharged from zero again, as it was completely discharged during the on phase.

Anyway, disregarding that for the moment (assuming that indeed there is a stable output voltage $$V_o=5V$$ and output current $$I_o=2.3A$$) you can use this application note from Texas Instrument to calculate what you need. Here is a summary using your values:

The duty cycle is given by:

$$D=1-dfrac{V_i}{V_o} = 0.46$$

The ripple current of the inductor can then be calculated:

$$Delta I_L=dfrac{V_i cdot D}{f_S cdot L}=936mA$$

The maximum peak current is then given by:

$$I_{SW,MAX}=dfrac{Delta I_L}{2} + dfrac{I_o}{1-D}=4.727A$$

EDIT #1

Here is a small simulation just to double check it. The output capacitor was increased in order to account for the aforementioned problem regarding the high load: