Converting 5v to 12v using an LM2577 - all i get is 4.8v?

Okay! So I bought a new chip and soldered it to some stripboard the way the schematic suggests. I am using 4 x AA batteries (1.2v 2450mAh each) into the LM2577.

I am now getting approx 7.2v out! What have I done wrong now!?

EDIT: It seems as if the longer the batteries are connected to it, the more the voltage drops. Its now at 6.99v.

Thanks
John

Do you have a large capacitor say 470uF or 1000uF connected across the input to the regulator, in parallel with the 0.1uF one? If not, try adding one.

If it still doesn't work with the cap added, then the inductor is probably saturating or otherwise unsuitable. Is it or the chip getting warm? What load do you have on the output?

No I dont - I'll try doing that - I only have the 1000uF one between the output and ground.

I dont believe either were getting warm.
EDIT: The chip is getting a little warm, but not very much.

I was just testing with my multimeter - so not much load - would the fact that there was little load affect the output voltage?

John

Having no load shouldn't cause the output voltage to be low. My guess is that the inductor is unsuitable. If you can tell me the dimensions of the toroidal core and the approx number of turns, I can probably work out whether it is saturating. Also useful would be its dc resistance and the current you want to draw from the +12v.

Ps is the diode the same type as shown in the regulator datasheet? Is it or the output capacitor at all warm? If the problem is iron losses in the inductor, it will be heating from the core and it may take a while for the windings on the outside to get warm.

Okay so the inductor has about 32 turns on a
Outside dia: 23mm
Inside dia:12mm
Width: 7mm

Yep the diode is the same. No the capacitor isnt warm.

Interesting though - when powered from the 5v of my arduino at ~4.9v the output is about the same, but when powered from batteries at 4.1v, it gives 7.2(ish) volts out.

Just then when I was testing it, I managed to bridge +12 and gnd, and it gave a small spark if that helps too.

Thanks
John

Just looking at the reciept for the inductor, it says only 42uH! I swear I picked up a 100uH one.

Would that be my issue?

John

idiotjohn:
Interesting though - when powered from the 5v of my arduino at ~4.9v the output is about the same, but when powered from batteries at 4.1v, it gives 7.2(ish) volts out.

I suspect that the peak current that the regulator is drawing may be greater than the current that your Arduino can supply - that is why it works better from a battery, and why I suggested the input capacitor (have you tried it yet?).

Using 42uH instead of 100uH would definitely be a problem. What load current are you intending to draw? I suspect that you might be better using an inductor greater than 100uH.

Well I went to Jaycar (electronics store) today to get the 100uH inductor. connected it all up and im getting an even lower voltage - about 5.6v now.

Unfortunately I forgot to grab the capacitor - would it change the circuit to the point that it could work?

John

All I need is 250mA at 12v.

John

I just tried it with a 5.14v 2.5A input. I got 9v to begin with but then it started dropping as i mentioned before.

John

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 will try adding a resistor on the load too.

John

Gee thats so confusing!

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

Don't forget the input capacitor as well!

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.

Okay ill go and get the input capacitor, a new inductor and a new resistor tomorrow!

John

idiotjohn:
Gee thats so confusing!

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.

btw what is it that you need to drive from 12v?

Okay so the electric door strike im using is 215mA - so say min current is 200mA and max current is 230mA.

Ill try and work everything out now...

John