I'm making a power supply using a transformer and bridge rectifier, and I'm wondering whether it is a bad idea to put multiple (5) capacitors in parallel for smoothing.

The output should be capable of drawing 1.1A at 10.7V. I'm allowing for a ~2V peak to peak ripple, as this is fed into a boost converter.

$$ C = \frac{I }{2 f V_{p-p}} $$

$$ = \frac{1.1 }{100 *2} $$

$$ = 5.5mF $$

Further, what if I was to use different valued capacitors (e.g. a 4.7mF and 1mF)? enter image description here

  • 5
    \$\begingroup\$ Multiple capacitors of different values are often used for filtering out different frequencies. Multiple together like that act almost like a single capacitor, so no problem. Adding more smaller capacitors (100nF, 1nF, etc) can clean up other higher frequency noise. \$\endgroup\$
    – Majenko
    Commented Dec 11, 2014 at 0:12
  • \$\begingroup\$ @Majenko I have seen this all the time on schematics, but I always thought it was to reduce ESR. Why wouldn't the larger cap get rid of HF noise? \$\endgroup\$
    – tgun926
    Commented Dec 11, 2014 at 0:30
  • 2
    \$\begingroup\$ Reducing ESR is one effect, yes. As for frequencies, google "capacitor self resonance". \$\endgroup\$
    – Majenko
    Commented Dec 11, 2014 at 1:21
  • 2
    \$\begingroup\$ Take a gander at electronics.stackexchange.com/questions/3879/… for some more insights related to frequency response of real capacitors \$\endgroup\$
    – vicatcu
    Commented Dec 11, 2014 at 1:48
  • 7
    \$\begingroup\$ Also, check the polarity of your diodes. \$\endgroup\$ Commented Dec 11, 2014 at 2:14

3 Answers 3


It is common practice to parallel a bunch of capacitors of small values for noise suppression; each value suppresses a different frequency band. The values are typically provisioned in decades, e.g. 0.001µF (1nF), 0.01 µF (10nF), 0.1 µF (100nF) etc. For high-frequency devices like cell radios, you will see caps in the pF range.

However it is also common to parallel capacitors of higher values, either electrolytic or tantalum. This is done for several reasons. First, the value you want may not be available, but a smaller value that can be paralleled is. Or maybe the larger value isn't available in the tolerance you want, and again you can get that in a smaller value.

Then there are price considerations. Depending on a how common the value is, a larger value cap may actually cost more than twice what two smaller caps cost.

And finally, there are size considerations, particularly on a board with SMT devices. The manufacturer recommended adding 2000µF to a 3.6V rail going into a cell module. First of all, a single 2000µF tantalum cap wasn't available, just 2200µF. But it was actually bigger than two 1000µF caps, and cost more than the two smaller ones together. So I used two 1000µF capacitors.

  • 1
    \$\begingroup\$ And finally (2) you can place each capacitor close to different loads, to reduce impedance due to track length. Of course this depends on physical layout. This is also done with small 100nF'ish decoupling caps near (logic) IC's. \$\endgroup\$
    – jippie
    Commented Dec 12, 2014 at 16:44
  • 1
    \$\begingroup\$ An additional consideration from the "size" perspective is that many products are constrained in one or two dimensions but not all three. A product that needs to be long and slender might have room for a hundred 100uF caps but not have room for even one 220uF cap. \$\endgroup\$
    – supercat
    Commented Jan 20, 2015 at 18:03

No, it's not necessarily bad. A couple good reasons would be to get a form factor that fits (at some expense in total volume) or to get a higher ripple current rating. A bad reason would be to get the exact value you calculated- the tolerance is usually -20%+80%. In this case you can consider 6.8 or 10mF.


It is not "bad" to connect several smaller capacitors in parallel to make a larger capacitor (it is actually good). If size, volume, price, etc. need not be considered, then you are free to meet the required capacitance with any number of smaller capacitors (with appropriate voltage rating) that add up to the required capacitance.


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.