I am trying to use a MC34063 switching regulator to step-up 5v to 12v.

I have wired it up as shown below, I think its the same as the datasheet.

But I only get ~4 volts out.

enter image description here

I know the capacitors I am using are greater than what is in the datasheet examples, but its the closest I can get. I do have capacitors that are below the datasheet values too, but those don't seem to make a difference.

(I am eventually using this on a battery, but currently the 5v I am using is from a power supply (stable constant DC current))

What am I doing wrong? What can I try next?

  • 1
    \$\begingroup\$ Can you give a schematic? Also, many switch-mode power supplies are very dependent on layout, it is plausible that the chip cannot work on a breadboard. \$\endgroup\$
    – Zuofu
    Jan 20, 2014 at 8:06
  • 1
    \$\begingroup\$ My experimentation shows that MC34063 can in fact work on a breadboard, but actual arrangement of components matters a lot if clean output is required. Try to use as little jumper wires as possible and to pack components as tight as you can. Also I'd recommend using low ESR capacitors for the input and output filtering. \$\endgroup\$
    – AndrejaKo
    Jan 20, 2014 at 10:25

1 Answer 1


If \$C_t\$ is actually 100uF, as shown on your rendering, it will never work. You need to have something close to the 1nF range. It is not a bypass capacitor like the other two, rather it a timing capacitor that sets the switch "on" time \$t{on}\$.

Looks like you've also left out the current sense resistor? And the inductor may be on the small side, even for 1500pF \$C_t\$ . You may have damaged the chip in that configuration. I assume you're using the recommended 1N5819, slower diodes such as 1N400x series will not work properly.

enter image description here

Above is most of the information (left column) you need to do the calculations, directly from the datasheet. You can use something like 0.5V for \$V_f\$, maybe 4.5V for \$V_{in(min)}\$, \$V_{sat}\$ is about 0.5V typically. \$I_{pk(switch)}\$ should be kept well under 1.5A. Try running f higher if you want to try to use 100uH.

If that's too much trouble, here's a javascript calculator that does the heavy lifting:

For your 5->12 converter, and 200mA output current, I get

  • Ct=1180 pF
  • Ipk=1140 mA
  • Rsc=0.263 Ohm
  • Lmin=104 uH
  • Co=1062 uF
  • R=180 Ohm
  • R1=1.5k R2=13k (12.08V)
  • 2
    \$\begingroup\$ There is also an executable that can be downloaded to be used offline and an alternative online calculator located here \$\endgroup\$
    – alexan_e
    Jan 20, 2014 at 11:27

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