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I'm trying to make a 9v to 5v step-down voltage regulator using MC34063
I used an online calculator for the values and assembled this schematic for stepdown config:
enter image description here

The calculator gave me these values (for Vin=9, Vout=5, Iout=750, Vripple=1, Fmin=100):

Ct=257 pF
Ipk=1500 mA
Rsc=0.2 Ohm
Lmin=13 uH
Co=1875 uF
R1=1k R2=3k (5V)

I used a 1800uF capacitor and 10mH inductor but on the output I'm measuring 0.8V instead of 5v with voltmeter
What I'm doing wrong?

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  • \$\begingroup\$ I think the problem is your duty cycle, in the calculator they use 50% of duty cycle (based on the datasheet: ti.com/lit/ds/symlink/mc33063a.pdf page 16) \$\endgroup\$ – R Djorane Sep 14 '15 at 12:09
  • \$\begingroup\$ Possible issues I see: Can the the inductors you used handle the current ? Note that the 13 uH inductor needs to be one with thick wire and a core. Same for the extra filter inductor. Is the C0 capacitor a low ESR type ? \$\endgroup\$ – Bimpelrekkie Sep 14 '15 at 13:11
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I assume by 10mH (10,000uH) you actually mean 10uH.

10uH is way too low an inductance for the MC34063. You should be getting numbers more like 100-200uH.

Try an 'official' XLS spreadsheet from Onsemi.

I get more like a 170uH inductance, and maximum output current more than 500mA if you can stand 0.56W dissipation from the chip alone. Efficiency about 52%.

You need an inductor that will not saturate with 0.55A current.

enter image description here

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