I have spent probably around 8 hours working on this today, and I cannot make any progress.

I'm trying to use the TI MC33063A in a boost converter circuit, just to make a proof of concept. Right now I'm using a 9-volt battery and I'd just like to see any voltage boost.

I have followed the sample circuit layout on page 6 as closely as I can, but the voltage coming out is at most 8.75 volts. The only noteworthy differences are that I'm using a 22mH inductor, since it's the closest I have, and that my voltage divider is a 10 to 1 ratio, which is different from theirs. I'm using a 100k ohm and 10k ohm voltage divider. I have spent hours looking for differences and I see none, at all. My circuit does not produce any voltage boost at all, and my calculations show that it should be generating at least 13 volts.

A voltmeter shows that the voltage divider circuit is -- expectedly -- about a tenth of the V_out at all times, which means the comparator should be outputting a constant value. The #2 pin seems to be draining about 85% of the power that is sourced from the 9 volt battery, based on mA measurements I took.

Earlier, I doubled up on 9 volt batteries and pumped about 18 volts through the system. If I left the circuit ungrounded and just connected it to the positive lead on the series batteries, I actually did see 21 to 24 volts come out the output, but grounding the circuit quickly negated any transient voltage spikes.

Does anyone have any suggestions of what I could be doing wrong? After a full day of trying and failing to cause a voltage boost, and I spent some time trying with a Maxim IC as well, it's starting to seem like voltage boost converters can't possibly work at all... very frustrating.

  • \$\begingroup\$ Are you sure you've got the orientation correct? Note that in the application schematic, pin 1 is shown at upper right, when physically it's at upper left. For whatever reason, when they show functionality of pins and schematics, they show a mirror image of the actual pinout. Pin 2, for instance, should not have a resistor connected to it, so I can't see how you can say it's "draining" power unless you've got it hooked up wrong. \$\endgroup\$ Dec 14, 2014 at 3:28
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    \$\begingroup\$ The inductor is a critical component -- you can't just substitute a choke. The LI^2 energy storage and the series resistance are more important than the actual inductance value. The capacitors, FETS and diodes are equally critical. Trying to put together a SMPS on a solderless breadboard is generally unworkable -- that's why we make evaluation kit boards (disclosure: I work at maximintegrated.com. And yes Linear and Analog Devices also make SMPS controllers and evaluation boards.) What's most cricital is starting from a known working PCB layout and parts list instead of starting from scratch. \$\endgroup\$
    – MarkU
    Dec 14, 2014 at 3:30
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    \$\begingroup\$ @coder543 Switchers (switch-mode converters) are challenging. Switchers must not be treated lightly. Substituting a 170uH with a 22mH inductor (two orders of magnitude) and expecting it to work is not a good proposition. Prototyping a switcher on a breadboard often does not work because of the parasitic capacitance of the breadboard. \$\endgroup\$ Dec 14, 2014 at 3:31
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    \$\begingroup\$ Just skimmed that TI data sheet. What an embarassment; very little detail. This may be relevant, at least as an overview of SMPS supply layout: electronics.stackexchange.com/questions/115292/… And the AD user guide is a good read: analog.com/static/imported-files/user_guides/UG-189.pdf Shows good layout tips as well as what waveforms you should look for when checking performance using an oscilloscope. \$\endgroup\$
    – MarkU
    Dec 14, 2014 at 3:39
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    \$\begingroup\$ Please post a schematic and/or photo of the circuit you built. It will make it much easier for people to help. Right off the bat I am questioning the 22mH inductor. That's over a 100x larger than the example in the datasheet and makes me wonder if the circuit is unable to oscillate. Secondly, are you using a Schottky diode? \$\endgroup\$ Dec 14, 2014 at 3:47

2 Answers 2


First off, you're using an inductor that's 100 times larger than the one in the example schematic. That sort of change will break a lot of circuits, and switching power supplies are finicky to begin with.

Secondly, are you loading the converter at all, aside from the feedback network? It's better to have continuous current in the inductor instead of letting it drop to zero every cycle.

Finally, are you sure you've got everything hooked up right? Pin 2 is supposed to be a low-inductance path to ground, but it sounds like maybe you're driving current into it instead.


As others have pointed out, it's the inductor. That much inductance, 22mH, means it probably has a non-gapped core. Non-gapped cores have very nonlinear BH curves and can't take much if any DC current without saturating. After they saturate, they will have almost no inductance. It's like having just a piece of wire between the battery and the switch in the controller. That's why the current in the controller gets so high during operation, it's shorting battery voltage to ground.

Probably during operation the battery is less than 9V since it is being heavily loaded. Fortunately, the MC33063 series has a current limit for the switch. I hope it's a good one.

You might be interested in this OnSemi (Motorola) datasheet of the MC34063a. Page 9 has layout info, including inductor winding instructions, with core type. You could use that to figure out worthwhile inductor characteristics. Also you might want to read the TI app note SLVA061


What would be a worthwhile inductor here? The inductor in the sample design is 38 turns of #22Awg magnet wire on a Magnetics Inc 55117 core. That works out to be ~166uH with ~50mOhms resistance. Choose a part near that. Here's a sample search at Mouser of parts that would work. Finally, app note AN920 has lots of details about the MC34063 family. Read that and you'll be an expert.


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