I designed and created this circuit to get 1.5 V out of 18650 lithium ion battery cell (with protection board).


Circuit is built to use in a toy, so not a serious use.

It shows open circuit voltage as 1.575 V as expected. But rises to 1.7 when 6 Ω load is applied. Any idea why happening so?


Updated: Added layout and actual circuit on board

I built this circuit on a piece of perfboard. Here is its layout:


Hers is actual photo of the circuit:


I poured hot glue after my above test so it would be not affecting the results.

All components except inductor are on top of the board. Inductor is of package CD43 and is positioned in exact location per above layout.

  • 2
    \$\begingroup\$ It's either an artifact of your measurement plus the sampling rate changing, or there's serious ripple on the output (and measurement issues), or it's layout. Please edit your post to show the layout around the part, including L1, C3, and U1. \$\endgroup\$
    – TimWescott
    Sep 11, 2019 at 1:36
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    \$\begingroup\$ Have you looked at the output with an oscilloscope? \$\endgroup\$
    – TimWescott
    Sep 11, 2019 at 1:36
  • \$\begingroup\$ @TimWescott Not sure exactly about sampling rate changing. I can see it increases voltage as I connect load and it stays at 1.700 in DMM display. I hae added layout and photo of the circuit. Did not yet looked it in oscilloscope. I will try that later today. \$\endgroup\$
    – Junaid
    Sep 11, 2019 at 9:17
  • \$\begingroup\$ @TimWescott I looked at output with oscilloscope. You are right! There happening serious ripple of 720 mV. As per calculation I got required output capacitance as 17.9 uF. I used much more, 33 uF. \$\endgroup\$
    – Junaid
    Sep 11, 2019 at 14:10
  • 1
    \$\begingroup\$ Check your calculations of minimum inductor size- 4.7uH sounds awfully low for a 50kHz old-fashioned MC34063. \$\endgroup\$ Sep 11, 2019 at 16:29

1 Answer 1


Your inductor is probably too wimpy, and is saturating. Do a simulation of your circuit, or scrutinize the MC34063 data sheet for the peak and RMS current expected in the inductor, then select an inductor that exceeds those values. Or better yet, simulate the circuit.

And your cap looks like an ordinary ol' electrolytic, so it probably has a fairly high ESR for the task. If you know the ripple current then you can calculate the ripple voltage just from the cap's ESR -- a good starting place is probably to assume that the current is 500mA peak-peak (i.e., that the supply is just at the border between continuous and discontinuous). I absolutely cannot remember what the typical MC34063 circuit does, so you want to consult the data sheet there, too.

Get through those, come back with your updated circuit, and we'll start talking layout.


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