I would expect the input capacitor to help - especially when you are driving the circuit from a battery - but probably not as much as you need. I see that the test circuit on the data sheet has a 220uF input capacitor.
Looking at Fig 7 on the data sheet, for 12v@250mA output from 5v input, you should be using an inductor with code L220 or L330 i.e. 220uH or 330uH.
I suggest you add a load resistor to draw at least 100mA from the output at 12v, because the regulator may behave very differently when unloaded.
The resistor attached to the comp pin (and possibly also the capacitor) needs to be adjusted to suit the conditions. For 250mA load I calculate 1K maximum, for 100mA I calculate 430 ohms. See page 18 of the application note.
If the output drops with time, then something is probably heating up - probably the chip or the inductor, if you are not running from a battery.
I see - I didnt realise that the inductor depended on what output current you needed - I assumed that the 100uH would also accomodate for current less than 800mA.
I will try a 220uH with the new resistor and such.
I'm still surprised that you are not getting a higher output voltage. Unless the input capacitor is more important than I thought, it may be that the inductors you have to choose from just aren't good enough. I think a lower value inductor than the optimum should still work, however it will make the regulator less efficient and the input capacitor will be more important. It should be easier to get the circuit working with a higher value inductor.
Will I try with the 220uH inductor and a 1k resistor, or do i need to get someone to work out if I need a different output capacitor also?
John
I would work out the minimum and maximum current that you need to draw. Then work out what inductor you need based on the maximum current, and choose an inductor. Then work out the minimum value of the output capacitor, and choose a capacitor. Then work out the minimum value of the compensation capacitor based on the output capacitor you intend to use. It's all described on page 18 of the datasheet. You may find that your existing output and compensation capacitors are OK.
It's possible that your regulator is unstable and that's why you aren't getting 12v output.
If you are switching the door strike, the current will presumably drop to 0 when it is off?
Do you really need to run the whole thing from 5v? Why not run it from a 12v supply instead and derive the 5v for the Arduino from the 12v? My guess is that the voltage to the door strike is not very critical and doesn't need to be regulated.
How about 10 x 1.2v NiMh AA cells? I'm assuming the door strike is only operated for a small fraction of the time, otherwise the capacity of AA cells will be too small. Or a small SLA battery? The only disadvantage I can see is that you will need to connect more solar panels in series to get 12v to charge them.
If you have a large current load on the 5v supply then you might want to use a switching regulator to get 5v from 12v efficiently, otherwise you can use the linear regulator already on the Arduino.
If you stick with the LM2577, then I suggest you connect the door strike permanently to the regulator output, and switch the input to the regulator circuit using a PNP power transistor. That way, the regulator always has a load of 210mA (making the design easier), and draws no current when the door strike is off.
Yes the strike is only on for 2 seconds every time someone enters the door from the outside, so the 4 x AA's should have been enough. I suppose using 2 solar cells in series and a small SLA would just make it so much easier to do.
Perhaps i'll give up on the idea of using AA batteries.
That's a shame, I was looking forward to hearing that you'd got the step-up regulator working! But using a 12v battery is certainly a simpler solution, unless you also have a large current draw from ~5v.