LT1613 - some questions about its functionality - Page 1
Forum section: electronics / beginners

author: mariush
author_rank: Super Contributor
timestamp: June 09, 2013, 09:04:34 pm
content: I'm thinking of using LT1613 for a project, here's the datasheet:http://www.farnell.com/datasheets/1605123.pdfThe project is basically a sort of a 9V battery replacement.Basically, I'm planning to use a couple of 25F 2.7v supercapacitors, either in parallel or in series. The idea is to charge the battery either from usb (5v) or from 2-3 AA batteries (minimum 2 x 1.25v = 2.5v or 3x1.25 = 3.75v) or maybe a Li battery (4.2v)I want to get the IC to output about 5.8-6.5v at up to 10-15mA, basically just high enough for a multimeter to not show "low battery" indicator and work properly when such a fake battery would be inserted.This chip advertises working as low as 1v so it would be great as I'd get as much use of the supercapacitors, but what I'm not sure of is..would the regulator actually be able to output at least 10mA with 1v input?The typical application on the first page (see attached picture) shows it doing about 5v @ 60mA with 1.5v input... has any of you guys experimented with this part or a similar one, could you estimate how much it would do at ~1v?Basically, any tips and suggestions would be useful, before committing to buying one and designing pcbs for it... it's a 1.4 mhz switcher, so it won't run on breadboards, have to make pcb for it (or do it bug style).Or do you have a suggestion for another IC for this purpose, reasonably cheap (this is about 4$ a piece) ? Remember, low voltage, ~6-7v out, probably under 10mA output needed 90% of the time..later edit:Also, what would be more suitable for this:- 2x25F 2.7v supercaps in series to have 5.4v therefore higher efficiency on switching regulator, but may not be able to charge caps to 5.4v with 2aa batteries or 5v so less capacitance... and on the upside I would only need a sort of current limiting resistor when charging from usb, instead of adding a voltage regulator to limit to about 2.65v.- 2x25F 2.7v supercaps in parallel to have more capacitance but less voltage... but it would take much more time to charge to the max (i estimate about 4-5 minutes to charge to max, i charged a 50v 2.3v capacitor to 2v in about 3 minutes with peak currents of about 2.2A)So is going more voltage and better efficiency worth the drop in capacitance, would the voltage drop faster than capacitance?even much later edit: I found it in ltspice iv so I'm going to make up a simulation later, but I'd still really appreciate any feedback, ideas, suggestions etc.

author: cyr
author_rank: Frequent Contributor
timestamp: June 11, 2013, 10:59:04 am
content: You don't get more energy stored just because you connected the caps in series (if you did you could charge the up in parallel, flip a switch and discharge to get free energy!).You forgot that you get half the capacitance (12.5F) in the series case.

author: codeboy2k
author_rank: Super Contributor
timestamp: June 11, 2013, 11:14:53 am
content: Quote from: cyr on June 11, 2013, 10:59:04 amoh damn... I'm rusty

author: mariush
author_rank: Super Contributor
timestamp: June 11, 2013, 02:54:09 pm
content: Quote from: cyr on June 11, 2013, 10:59:04 amThank you for feedback, but I think you misunderstood something.It's 2 25F 2.7v supercapacitors.In series, I would get 25F 5.4vIn parallel, I would get 50F 2.7vI realize that, I never said in series I would get the same 50F value, I'm just wondering if the increase in efficiency in the switching regulator (regulator is more efficient the closer the input voltage is to the output voltage) would make up for the loss in capacitance.As for charging in parallel then connecting them in series, yes, I thought of that, but it's a bit difficult as I'm thinking of using only 3 wires to the outside of this fictional battery (voltage, ground, shutdown) - the shutdown wire when tied to ground kills the boost regulator (useful for making a sort of auto power off for multimeters that don't have this feature or it's not convenient enough).So these three wires would come through a tiny hole drilled in the back of whatever the battery is used for and attached to a tiny pcb that charges the internal battery.

author: JackOfVA
author_rank: Unknown
timestamp: June 11, 2013, 03:42:08 pm
content: Quote from: mariush on June 11, 2013, 02:54:09 pmYou've got the math wrong - two 25F caps in series = 12.5F, not 25F.In any event, you have the same energy storage - CV2/2With two 25F caps in parallel (50F) at 2.5V, the energy stored = 156 watt-seconds (Joules) and you obtain the same value with the series configuration of 12.5F at 5V. So there's no increase in energy storage by series configuration.In terms of extracting the maximum amount of the stored energy, if the converter will operate down to 1V, you can extract more of the stored energy from the series configuration if charged to 5V than the parallel charged to 2.5V.

author: mariush
author_rank: Super Contributor
timestamp: June 11, 2013, 04:02:18 pm
content: Damn, you're right. I don't know why i kept saying 25F ...1/Ctotal = 1/C1 + 1/C2 ... so Ctotal = C1xC2 / (C1+C2) = 625/50 = 12.5 FYes, the LT1613 goes down to about 1v (datasheet says typical 0.9v, 1v max) , SHDN pin needs 1v to keep IC running... but I did some simulations in LTSpice and it's a bit worrying, if my simulation was right at 1v it can only do about 4.6v at around 9 mA ... with 1.5v it was doing up to 8-9v at 15-20mA just fine (the graph says 1.5v to 5v up to around 50mA)So I guess if I stay with this one, it's safer to just put them in series and expect the boost regulator to crap out when it goes down under 1.3-1.5v. Should still give about 30 minutes to 1 hour of life to a multimeter if my math is right.I guess for the initial "prototype" I can use a low drop fixed 5v regulator to charge the capacitors from USB (using regulator just for safety in case I'll use some phone charger that outputs 5.5-5.8v) with a resistor to limit current to about 1A. Maybe later make some sort of SEPIC regulator to charge the supercaps from as low as 1AA battery to a 7.5-9v classic power adapter.Thinking of using a PIC or Atmel running from a 3v battery to add voltage check (show voltage in supercaps or percentage/estimated time left on 2 digit 7-segment display ) and also add auto sleep (press on a button to select auto shutoff at off, 10m, 30m, 1h, 2h, 4h intervals) ... basically the pic can run at 32-512khz using uA's of power and when time expires pull the shdn to ground and kill the boost regulator.