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I am facing annoying troubles converting 24V AC to 12 V DC. The goal is to supply 1.2A current to 12V DC load. I am both limited in input power (I can only use 24V AC output of a transformer that can supply up to 15W), size and possibilities of heat dissipation (the circuit will be placed within a sealed box with almost no air circulation). I've got some help in building the converter (See picture), but the problem is that it is way too demanding on transformer power input. I've added Proteus simulation screenshot, you can clearly see the power that is taken from the transformer (V1 on the screenshot). Any suggestions on how to make it more efficient or is there any other way for me to reach my goal? Thank you. enter image description here

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  • \$\begingroup\$ You're trying to build a power supply which is 96% efficient (12V @ 1.2A = 14.4W and you have 15W available). A linear regulator is definitely not going to get you there in this case. Just the bridge rectifier at the beginning is more than likely to make this impossible. You probably need to rethink your requirements and/or constraints here. \$\endgroup\$ – brhans Jul 12 '16 at 18:44
  • \$\begingroup\$ Is the transformer 15 W or 15 VA? Power factor will come to mind too. \$\endgroup\$ – winny Jul 12 '16 at 19:25
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This just screams for a switching converter, you can buy these on ebay. I suggest an LM2596 based module.

enter image description here

You can remove everything in your circuit that is right of C2 and replace it with this module. Then turn the pot on the module so that the output supplies 12 V and you're done. Notice how this module does not need a heatsink !

BUT like brhans comments, you're really pushing it with 15 W in and 14.4 W out. You should really be using a slightly more powerfull transformer, like 20 W or so. When using a switching converter you can also just make the input voltage slightly higher (up to what the LM2596 can handle) to increase the power.

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    \$\begingroup\$ Yes - this will almost get you there, but without about 20% more transformer or less load you'll still be stuck. \$\endgroup\$ – brhans Jul 12 '16 at 18:50
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    \$\begingroup\$ Also note that these particular modules use a fake LM2596 (it might not matter much, but it runs at 52kHz instead of 125kHz, more like an LM2576). Probably more importantly, the input capacitor and inductor are undersized for the advertised current output. So one of these might take some tweaking if those get hot. Also, it comes uncompensated, so you may need to size and add an 0805 MCC feed-forward compensation capacitor on the provided but unpopulated pads, in the event it's unstable at your operating point. Also note the ripple might be quite large, maybe 400-1000mV (1V). \$\endgroup\$ – scanny Jul 12 '16 at 19:08
  • \$\begingroup\$ @scanny Then I guess you've been unlucky with these modules because the ones I use work fine and at the expected frequency and with a lot less ripple. So I'd rather say: Some modules might use fake/3rd party ICs. But certainly not all of them ! \$\endgroup\$ – Bimpelrekkie Jul 12 '16 at 19:16
  • \$\begingroup\$ Have you run any at 3A? I can't get mine much above 1A out before the input cap and inductor get pretty toasty. Mine has a 47uH inductor, so ripple is closer to 300mV, but pictured unit has 33uH so would expect higher. Also, mine oscillates at certain operating points. But has been very useful for learning :) But you're right, maybe I was just unlucky :) \$\endgroup\$ – scanny Jul 12 '16 at 19:47
  • \$\begingroup\$ @scanny No I would not use such a module at 3 A continously. You have to take the specs. of these cheap modules with a grain of salt. I'd use them up to 1 to 1.5 A perhaps. For 3 A continous I'd choose a 5A or 6A converter. These usually have a much larger inductor and some heatsinks also. \$\endgroup\$ – Bimpelrekkie Jul 12 '16 at 20:23

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