Hi,
I've done tested this boost regulator IC according to the datasheet.
http://www.linear.com/pc/productDeta...31,C1060,P1897


but I just modified the output voltage to be almost 5 V by changing the resistor value, so that I can use it for my LCD display. With no load, the output is almost 5 V. When the battery is fully charged, it can power up the LCD and a PIC for hours (I'm not sure how long, just leave it there, maybe just 1 hour). When the battery drops to 1 V, it lasts for only a few second
Maybe the design is for 3.3 V output only.
I've measure the current drawn by the LCD and the PIC, it is not more than 2.5 mA. Just wonder is there any way to improve the running time?

BTW, I'm using a sony rechargeable battery, 2500 mAh.

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Hi Banana,
It is not working properly. Maybe you forgot to connect pin 2 to the positive input voltage. Then with a 2.5mA load it is 87% efficient so the 1.2V input will have a current of only 11.8mA. Then a fully charged 2500mAh cell should last about 212 hours.

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Uncle $crooge

 

As Audio says, it should run for way longer. How much current does it draw from the battery? Did you build it exactly like the app note shows or did you sub any components besides R1/R2 ? What diode did you use for D1? C1 and C2 cannot be plain old electrolytic either.
When a NIMH battery is at 1V it is dead and ready to be recharged. The voltage drops off very quickly at the 1v point so don't expect much run time after this point in its discharge cycle.

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I am at the top of a tall ladder in the dark with my flashlight powered by Ni-MH cells. My flashlight suddenly dims when its battery cell voltage drops to 1V and lower. Then I can't see anything.

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Uncle $crooge

 

Most likely you not putting PIC to sleep are you running the pic constantly or do you shut it down (not sure if PIC have sleep but going by AVR/atmel) active currents are quiet high remember also set unattached inputs to a level either with pull ups or pull down (another source of heavly power consumption floating pins)

 

Of course PIC's have sleep, and always have had, but consumption is low even without sleep, and is further reduced by slow clock frequencies - there's also no concern about floating inputs on a PIC - that's just on CMOS gates.

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Hi,
Thanks for the response
OMG.. I've seen the schematic wrongly. audioguru got it, I've connected the pin 2 to ground, but I've simulated with LTSPICE, both also working. And the datasheet recommends this pin to be high after the part has started up.
I've made 4 PCBs for this and burned one IC
For the diode, I'm using Schottky diode 1N5817 while for the capacitors C1 and C2, I'm using only electrolytic type.
Is there anything to do with the max current sourced by the battery? I know that there is internal resistance in the battery, but it is not measurable.

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Of course it's measurable, you just have to know (or work out) how to do it, it's not difficult to work out!.

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Measurable? There is no reading on the digital multimeter in resistance mode when I measured it. I don't think I can use an ammeter and short across for the max current.

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Does your meter still work?

Think about it! - it's very simple to do, if you work it out yourself then you will understand it much better.

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The datasheet says to use very low ESR capacitors like tantalum or ceramic. Your electrolytic caps are horrible at the high frequency of this voltage converter due to their high inductance (they are wound like a coil).
Try adding a 0.1uF ceramic disc capacitor with very short leads at the input and output in parallel with your electrolytic capacitors.

Maybe your inductor is too small and is saturating?

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Uncle $crooge

 

Use a resistor as the load of the battery, and measure the voltage across the resistor? Or measure the current of the circuit and find out the internal resistance? Am I right?
BTW.. the multimeter is working

Sigh.. I don't know whether I can get ceramic capacitor which is more than 1 uF over here, I have to find it out. I'm going to add 2 ceramic caps to my current PCB.
I think I'm using the ceramic or carbon type inductor (resistor looked inductor).

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Your inductor might be an RF choke. Then it will have a resistance too high and a saturation current too low.

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Uncle $crooge

 

Almost, place a voltmeter directly across the battery, write the voltage down, then place a low value resistor directly across the battery - the voltage will drop, take the voltage reading again. From the two readings you can calculate the internal resistance - bear in mind it will probably vary according to the load, and certainly will according to the state of charge of the battery.

If you're using an inductor that looks like a resistor it's not surprising it's not working, that's almost certainly just an RF choke, and unsuitable for this use.

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Which one is correct? I just googled ferrite coil and found these:
and
I think the green color is the RF choke.

Parts are really difficult to be got over here, as long as they got the value, the rest they don't care

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