your inductor size is correct, not too big? Its best to do on pcb and have some small ceramic right near the input pin (100nF). and also near the feedback pin with another ceramic there (100pF).
Beadboard is OK for linear regulators, but this is a switching regulator & need very careful PCB design. You can't expect to get output from a switching regulator with this lengthy wires & connections. The inductor position & schottky diode positions are most important to make this circuit to working. Even with PCB, proper design is required to get the perfect output. You must avoid to use breadboard for the testing & for worst case you can use a through hole regulator & components which you should directly plugged into the breadboard without the lengthy wires involved.
I recommend you to test this circuit with a small size of general purpose veroboard. Note you have to take care of the inductor & schottky diode arragement and you can refer for this from datasheet PCB design sample & internet search also. Good luck.. Don't use this type of connection to test a switching regulator.
Oh my friend!, what you show reminds me of my first ever smps.....something I like to forget.
Please repeat and do on Eagle free pcb layout and then here is pcb layout instruction for smps attached.
Sometimes you can get away with wireboard if you make the connectons really small...but smps really need pcb
That's the advantage of using a pcb software, you can check the schematic and see if its hooked up right. Sorry but the hookup of post#7 is not the way to go, it will cost you in the long run, its gambling with noise etc.
I'm with Treez on this one. That breadboard will only give you grief.
Some individuals in this and other forums claim that they have built SMPS on breadboards successfully. To which I respond: some people can walk on a tightrope, most of us can't.
Additionally, one is depending on so many unknown variables.... For instance the leakage inductance and the contact resistance can easily trigger the ICs shutdown mechanisms.
attached now is design document for buck with lm2596...what is you inductor value and peak curren trating, it should be about 33uh and 4a rated
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looks to me like you have got wrong inductor in DR125
..YOURS SAYS "0.47", THIS CANNOT POSSIBLY BE 47Uh....so you have got wrong inductor. (sorry about capitals)
are you sure that i will get 5.00 output when i replace the inductor with 33uh
so now i am feeding this circuit from 12dc 2A adapter, but we are going to use this circuit with solar panel to charge battery bank solar panel is giving output upto 22V and able to provide 1.5A current to my battery bank.
my battery bank will take 5.0 v and upto 2A for charging
so the calculations you gave with this configuration will work with that also??
your vout = 5v, and iout(max) = 2A?
yes that will work, with either 47uh or 33uh...as long as its RATED TO 4a
What is you vin?...I though it was 12v?
in photo, your inductor shows "0.47"...thatcannot possibly be 47uh.
And are you sure your diode is the right way round, and is of sufficient current rating?...and is schottky or ulta fast diode (if UF , then trr <50ns?)
perhaps you have the adj version which explains the 1.2 Vout if you connect the Vout direct to the f-back pin, you need to divide it down with resistors to get 5.0V out.. say, 120 ohms to gnd off the pin and 390 ohms from the pin to Vout...
the fact that when you connect directly to that pin you get 1.22 V tells me it is an adj version - put the resistors in - it can't hurt, and if you get near 5V you know the part has been wrongly printed (it does happen..!)