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According to Page 7, Table 8 in the datasheet for MC34063, the startup voltage is 2.1V typical for A series and 1.5V typical for E series.

It is also said in Note 4 that the start-up voltage is the minimum power supply voltage at which the internal oscillator begins to work.

I am trying to implement a 2-cell Ni-MH battery to 5V,100mA step up converter, where the input voltage varies from 1.8V to 3V. I know that there are more suitable ICs like this one, but I want to learn how to make MC34063 work in low voltages if it is possible.

How can I make MC34063 work as a step-up converter with 2.1V input?

MC34063 Electrical characteristics - Table 8 - Total device

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  • \$\begingroup\$ Is there a question in there somewhere? \$\endgroup\$ – Dave Tweed Nov 2 '12 at 13:13
  • \$\begingroup\$ @DaveTweed Yeah, it seems a little bit hidden :) I will edit it out. But the title should have at least some hint about it. \$\endgroup\$ – abdullah kahraman Nov 2 '12 at 14:20
  • \$\begingroup\$ Using the ...E part would be a fine start. \$\endgroup\$ – Russell McMahon Nov 3 '12 at 2:18
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Worst case, you cannot do it without 'cheating'.
Vin_min is 1.8V and Vstart_TYPICAL is 2.1V.
Using the ...E suffix part would be far better.

There are several ways of cheating.
One is to use the "Start" signal, whether push button or digital signal etc to boost a pre-charged capacitor to supply a higher initial voltage to allow starting. Once started the IC can then self power itself from a higher voltage. The switch element must of course still draw power from the 1.8V minimum battery, but that is not a problem.

A capacitor charged via a Schottky diode will charge to within less than 0.2V of the charging supply. A Schottky diode will drop say 0.4V when discharging into modest load.

So:

  • Charge capacitor to 1.8-0.2 = 1.6V.
  • Step bottom of cap from ground to 1.8V with start signal.
    • Vcap is now 1.8+1.6 = 3.4V.
    • Feed capout to IC VCC = 3.4 - 0.4 = 3V.
    • IC will now start and Vcc can be bootstrapped.
  • Cap has to be large enough to power circuit during startup phase.
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  • \$\begingroup\$ Seems like a great idea.. You mean that something like capacitor doubler should be build. I will try that out once I fully understand it :) Man, I would be very pleased if you had some psuedo schematics.. \$\endgroup\$ – abdullah kahraman Nov 3 '12 at 10:01
  • \$\begingroup\$ How do I bootstrap Vcc, can you explain in detail? How do I connect pin 8 which is DRC (collector of the transistor that drives the switching transistor)? \$\endgroup\$ – abdullah kahraman Nov 3 '12 at 12:35
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You're not going to have much luck with that, unless you build some sort of step-up converter to ensure that the supply exceeds the worst-case startup voltage. Also note that the worst-case startup voltage isn't specified - there's absolutely no guarantee that it will even be 'close' to the 2.1V typical voltage that's specified. To me, you should just do the design with a more appropriate controller.

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  • \$\begingroup\$ Yeah, designing with MC34063 is not something that someone sane would do :) Good point on "typical". \$\endgroup\$ – abdullah kahraman Nov 3 '12 at 9:58
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The data sheet says that your appliction will "typically" work but is not guaranteed to work. There's no way around that.

Also, the MC34063 (which I've used, too!) is a pretty old, BJT-based part. It has significant losses, and will never be very efficient. If you just need a ready boost converter, buy a ready-made one from Pololu, or buy a DC-DC converter module from Digi-key.

If you're really teaching yourself switching power supply design, then you should go look at the modern CMOS based switchers, which have much lower losses. Also, the best way to learn is to read the application notes from places like Linear Technologies. There's all kinds of things you need to worry about, such as PCB layout, and selection of suitable capacitor dielectric materials.

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