I am developing a circuit to convert 230VAC to 12VDC. I ran into the problem that my smoothing capacitor for the full wave rectifier is pretty huge. The load is estimated to be around 500mA and therefore the capacitor needs a capacitance of several mF. My circuit consists of: transformer 230 - > 12 and a full wave rectifier. Is there a better circuit for such an application? Desired voltage ripple is no more than 0.2V.

  • \$\begingroup\$ Did you really mean "milliFarads"? Or µF? Just from looking at similar supplies I'd guess you'd probably need at most 100 µF. And keep in mind that if your transfomer secondary puts out 12 VAC then it will rectify to about 16 VDC. \$\endgroup\$
    – Rob Lewis
    Commented Feb 9, 2020 at 17:27
  • 1
    \$\begingroup\$ i did mean milli. Rectifying 12v yields an output of around 11v dc. And simulating it in ltspice and calculating the capacitor by hand shows the same result needing a huge cap. \$\endgroup\$
    – schgab
    Commented Feb 9, 2020 at 17:32
  • 2
    \$\begingroup\$ @schgab Note that transformers are rated in RMS voltage, so rectifying 12VAC will give you \$\sqrt{2} \cdot 12V -2V_F\$ \$\endgroup\$ Commented Feb 9, 2020 at 17:59

1 Answer 1


You can rectify and filter the voltage to a somewhat higher voltage, with a smaller capacitor and correspondingly higher ripple, then use a regulator to reduce the ripple. That is probably the most common approach.

If you don't need the voltage regulated, another possible circuit is a capacitance multiplier. Again, you throw away some power in an active device to reduce the ripple, but in this case you're not stabilizing the output voltage against changes in input voltage or load current.

But a few thousand uF electrolytic capacitor is not very large (for example, 18mm diameter x 20mm tall for 4700uF/16V) compared to the power transformer you would need, so I don't see why there would be an issue.

Even with a regulator or capacitance multiplier you will need a capacitor of fairly large size because it has to store enough energy to supply 0.5A during the relatively long period between AC peaks.

If you have a 12VAC transformer you would get about 16VDC out assuming a full-wave rectifier, and the transformer would need to be rated at about 13VA or 800mA. A 4700uF capacitor would yield 0.01s * 0.5A/4.7mF = 1Vp-p ripple, assuming a 50Hz mains.

If you added a linear regulator you'd get 12V with a couple watts dissipation in the regulator at 0.5A (so a heat sink). That might not be enough to guarantee it wouldn't drop out and start passing the ripple under worst-case conditions, depending on the dropout voltage of the regulator and your specification for mains tolerance, you'd have to do a careful design to be sure. A switching regulator could run much cooler, but that means more complexity and another much smaller filter capacitor.

Finally, what most modern electronics does is rectify the mains, use a capacitor of similar physical volume to a 2200uF or so capacitor, but much less capacitance and rated for perhaps 400VDC. The mains is rectified and filtered, and chopped at tens of kHz, passed through a much smaller and lighter transformer, and regulated with a closed-loop feedback scheme. The amount of material required is much less, the mass and volume is much smaller, efficiency is high, and regulation can be pretty good (and you can add a small linear regulator after it to further reduce it). The noise can be a problem in a few applications, but in most it's almost ideal.

  • \$\begingroup\$ I think my best approach is to go with the rectify the mains. Thank you for this great answer! \$\endgroup\$
    – schgab
    Commented Feb 9, 2020 at 19:04
  • 1
    \$\begingroup\$ Okay, but beware that designing an offline switcher is at least an order of magnitude more challenging and potentially hazardous, even if you copy a circuit out of Power Integrations or some other supplier of SMPS chips. Just buy a 12V wall-wart from a reputable manufacturer and be done with it. \$\endgroup\$ Commented Feb 9, 2020 at 19:07
  • \$\begingroup\$ Strongly recommend not messing around with mains voltages unless you really understand what's going on, at the very least from a safety perspective. Unless it isn't physically possible for humans to touch your output line, you should have some form of galvanic isolation in the loop. Your converter could let through pulses of line voltage if it enters a fault state(and you have to have a lot of specific education and put a certain amount of effort in to reasonably think it won't) or fails. Use fuse(s) and try to build something that fails(burns out) safely. \$\endgroup\$
    – K H
    Commented Feb 9, 2020 at 22:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.