How to limit charge rate of large output capacitor?

I am designing a board to provide 5 volt power to some custom Mega2560 boards from LiPo batteries. I am looking at a TI DC-DC step up converter (TPS61092) to do the job which has a 1 amp output which is plenty for the board. The trouble is that sometimes these Mega boards will have a GSM daughterboard attached which pulls about 2 amps at 3.8 volts (from an LDO being supplied by the 5v rail) transiently when it transmits.

To cater for the high load I wanted to put a 470-1000uF cap across the output of the TPS61092 to buffer power for the transients. However, the cap will be like shorting the TPS61092 on startup. To minimise the inrush current I could use a resistor between the TPS61092 and the cap however when the GSM transmits I would like it to have unchoked access to the full current from both the cap and the TPS61092.

I could go for a more powerful DC-DC chip however they require more components, more space, more cost and dont' run efficiently at the ~20ma draw of the Mega2560 board when its not transmitting (i.e. most of the time).

Has anyone else had this problem in the past and solved it?

http://www.digikey.com.au/Web%20Export/Supplier%20Content/Vishay_8026/PDF/VishayBeyschlag_CircuitProtect.pdf?redirected=1

Series PTC resistor.

Weedpharma

Use an inductor in place of the series resistor.

Use an inductor in place of the series resistor.

Thanks Mike, but I don't understand how the the inductor (which would have very low DC resistance?) would stop the capacitor from drawing the initial large current from the DC-DC converter? What am I missing?

@weedpharma

With the PTC; would it cut the TPS61092 off from the cap with the high initial load, the cap would then be drained by the GSM module before the PTC reset and the the TPS would start charging the cap again. The current would rise again, the PTC reset, etc going into a loop?

but I don't understand how the the inductor (which would have very low DC resistance?) would stop the capacitor from drawing the initial large current from the DC-DC converter? What am I missing?

You are missing the fact that the effect of an inductor is to slow down rapid changes. So when the current rapidly rises the effective resistance ( inductive reactance ) is high but when it slows down to DC it is low ( just the resistance of the windings ) which is exactly what you want.

Ah, I see. How many uH would the inductor need to be to limit the draw to 1 amp at 5 volts into a 1000uF cap?

There are formulae on the internet for calculating inductor size for buck/boost converters but I can't find any for this 'smoothing' type of application.

lemming: @weedpharma

With the PTC; would it cut the TPS61092 off from the cap with the high initial load, the cap would then be drained by the GSM module before the PTC reset and the the TPS would start charging the cap again. The current would rise again, the PTC reset, etc going into a loop?

If that is what happens, they would not be used. You need to select the correct one for your use.

The R increases with the current at switch on and limits the current. As the voltage on the C increases, the current reduces and so does the R.

Weedpharma

Whoops! It seems everyone has missed the mark.

Figure 9 and section 9.3.5 of the datasheet explain that there is no problem charging a large output capacitor.

Since this is a boost converter, there is a "pre-charge" phase while the output capacitor is charged to the input voltage before the switch-mode boost function can start. This is deliberately current limited - read section 9.3.5 of the datasheet.

And once the switch-mode converter starts up, the current is strictly limited also. In fact, such current limiting is an inherent part of a boost converter (or switch-mode supply) in any case.

So, just wire it up!

That's radical, read the data sheet before answering a question??

Weedpharma ;)

While the chip's data sheet might not mention a capacitor limit for start up every commecrcal DC / DC converter I have seen has mentioned one. So it depends what data sheet you are talking about.

AFAIK, common DC/DC boost converters use an inductor and diode in series between in and out. So even if the boost converter is not working, you have the DC-in (minus schottky diode drop) on the output. Leo..

Wawa: AFAIK, common DC/DC boost converters use an inductor and diode in series between in and out. So even if the boost converter is not working, you have the DC-in (minus schottky diode drop) on the output.

Ah, but you see, it is not a diode*!

Hey - tell you what!

Read the datasheet and find out! :roll_eyes:

(*Because a diode wastes too much power.)

Ahh, I see. This IC uses a mosfet rectifier switch for increased efficency. Leo..