MC34063A Boost Help

Discussion in 'General Electronics Chat' started by blah2222, Jan 7, 2014.

  1. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    33
    Hi all,

    I have a set up a basic boost circuit on a breadboard using TI's MC34063A IC. I designed the converter to operate using an input 9V alkaline battery that can droop down to 6V and have an output of 26V (100 mV ripple) with a max output current of 100 mA.

    I chose components based on what I had on hand so the cap/inductor/resistor values are not exact. I had to make my own Rsc resistor by breaking apart wire-wound concrete resistor because I didn't have any sub an ohm. It's resistance is around 0.6ohm which is higher than I need but the best I could do for the moment.

    Unloaded this operates with better than expected ripple and is bang on 26V. The moment that I load this with a load resistor of 1kohm or less, the battery's voltage begins to drop relatively quickly, either the diode or inductor get hot (hard to tell because its really cold in my apt and I have no feeling in my fingers) with a noticeable smell, and after a while a little bit of a high-pitched noise starts up. Doesn't seem to be the IC though.

    The diode is rated at 1A/40V and the inductor is made of ferrite and rated at 180mA. I have a feeling this has something to do with the diode because I was reading -18V (anode to cathode) but the datasheet recommended using a toroidal inductor (which I don't have).

    So I figure this is either an inductor current rating issue, something to do with the diode (Motorola 1N5819), my Rsc that I made is too large, or a wiring issue.

    Wondering if anything sticks out to anyone. I have attached images of both the circuit and the wired breadboard (Rsc looks like a black jumper between pins 6 and 7).

    Thank you,
    JP
     
  2. #12

    Expert

    Nov 30, 2010
    16,337
    6,821
    Yes. Trying to suck more than 1/4 amp out of a 9V battery is not going to work very long. That's like 90 minutes to dead, but the internal resistance will limit you long before that happens.
     
  3. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
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    Even if you had a battery that could handle the current, I suspect the inductor listed would saturate because boost converters have peak currents much higher than the output current.

    You can use free design software on the TI website if you choose one of their simple switchers. It will select the external components including current ratings.
     
  4. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
    554
    33
    I actually only need the output to be on for 0.3% duty cycle (Ton = 100us out of 33.3ms period).

    If the voltage source were a power supply capable of that current, is there anything functionally wrong with this circuit?
     
  5. tindel

    Active Member

    Sep 16, 2012
    568
    193
    A couple more thoughts... disclaimer I don't know this chip specifically.

    I calculate >400mA with a 9V battery with >70% efficiency... it's important to know your battery impedance to know how this will effect you and how much MORE current that you'll have to draw to provide the required power.

    My other thought is that many SMPS's require very low ESR on the caps <25mohm. It looks like you have cheep electrolytics. Try some Conductive Polymer Electrolytics or Tantalums with very short leads. I've had weird heating things happen when using regular electrolytics.
     
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  6. KJ6EAD

    Senior Member

    Apr 30, 2011
    1,425
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    I think you need a much larger inductor (higher value and lower resistance).
     
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  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
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    I assume this is the problem:

    That's why I suggested running the simple switcher software to select the external components. They give a parts list with all values.

    Could be:

    1) Not enough inductance to store sufficient energy

    2) Inductor is saturating

    3) Caps to small or ESR too high

    4) Battery's impedance too high

    5) ALL OF THE ABOVE
     
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  8. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Absolutely! 33uH is totally unsuitable. An inductor of 220uH to 330uH would work well.

    Also, that inductor type and size is unsuitable for that current and power. You need an inductor that looks more like this type;

    [​IMG]
     
  9. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
    554
    33
    Darn, I figured I would need a more robust inductor. What is that inductor type called? Are these pretty easy to make or is it better to just purchase?

    Could I still get away with my current inductor based on my load requirements (30-50mA @ 30 Hz 0.3% duty)?
     
  10. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    784
    For the sort of current draw the OP is talking about - an old Ni-Cd 9V battery might actually give longer operating time than a Ni-Mh with 3 - 4x the Ah rating.

    Multilayer ceramic chip capacitors are probably about the lowest ESR you can get - values upto 47uF are not completely rare, I've seen brochures offering as high as 180uF.
     
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  11. KJ6EAD

    Senior Member

    Apr 30, 2011
    1,425
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    30 Hz?! Try 30 kHz.

    There's a design spreadsheet for this IC on the On Semiconductor website.

    http://www.onsemi.com/pub/Collateral/MC34063 DWS.XLS
     
    Last edited: Jan 10, 2014
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  12. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Sorry, I should have been more clear. My output load requires a 30 Hz pulse train of current, which is independent of the switching frequency of the boost converter.
     
  13. KJ6EAD

    Senior Member

    Apr 30, 2011
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    I don't know. Why don't you run the numbers in that spreadsheet I linked you to and find out? If your current and duty cycle figures are correct, which I doubt due to the constantly changing specification, you have an average current of only 150μA.
     
  14. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    The inductance of your inductor is the problem. The 34063 is a slow chip and its timing cycle is determined by the timing cap (which you seem to have a typical value). If you check the 34063 datasheet there are circuits presented for boost converters, you will find the inductor value is likely to be 220uH or 330uH.

    You can get away with a small RF style inductor like yours if you only need very low output current, BUT it's inductance will still need to be ten times higher at about 330uH.
     
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