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Topic: Converting 5v to 12v using an LM2577 - all i get is 4.8v? (Read 7 times) previous topic - next topic

dc42

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.
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idiotjohn

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

idiotjohn

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

dc42


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.
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idiotjohn

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

idiotjohn


idiotjohn

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

dc42

#37
Sep 05, 2011, 10:26 am Last Edit: Sep 05, 2011, 10:31 am by dc42 Reason: 1
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.
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idiotjohn

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

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

dc42

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.
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idiotjohn

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

John

dc42


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.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

dc42

Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

idiotjohn

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

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