Buck(-boost) with constant voltage?

Discussion in 'The Projects Forum' started by JMD, Apr 13, 2010.

  1. JMD

    Thread Starter Member

    Dec 9, 2009
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    I got a problem, that i hope any of you can assist me with.

    I got a supply of around 24V (varies a bit) and about 60mA (also varies a bit). I need a constant output (when the output is enabled) of 12V and 800mA (the load is a resistor @ 15 Ohm).

    The load needs to be connected for 10ms, with as high duty cycle as possible.


    So.. i figured i would use a buck-boost converter. But.. is that the best choice? Im open to ideas!

    Thanks in advance!
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    Think wattage. You have a supply of 24V an 60ma, that is 1.44W. Taking it down with perfect efficiency you would get 12V at 120ma (also 1.44W). Since nothing is perfect, 80% would be excellent, and you could get 12V at 96ma.

    TANSTAFL.

    You might be able to charge a cap, and get very short bursts of current. Is this what you're after?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    I'm wondering where you are going to get all of the extra power from?

    Let's say that you could build an 80% efficient DC-DC converter, which is reasonable.

    Your load is 12vx.8a = 9.6 Watts

    The theoretical 80% efficiency of a converter means you'll require 9.6/.8=12 Watts

    You say your supply is 24v at 60mA, which is 1.44 Watts.

    This means that you will need 1.44W-12W = 10.56 Watts over what your available power supply is.
    [eta]
    Gosh, when you stop to take out the trash, it really slows down your posting. :rolleyes:
     
  4. JMD

    Thread Starter Member

    Dec 9, 2009
    96
    0
    Bill --> What i need, is to dump 12V across a 15 Ohm resistor (9,6W), for 10 ms - and repeat that as fast as possible. Say, if i got a supply of 0,96W - that would mean i would have to "charge" for 100 ms, and then drive the load for 10 ms.

    What i need, is to be able to repeat those 10ms with the least possible delay (the delay is the time needed to store enough energy to drive the load at 9,6W for 10ms).


    SgtWookie --> The key-word here is "energy storage". The best case, would be where i had a supply of 9,6W - that would mean i could run the 10ms cycle 100 times pr. second.

    The case here, is that i got around 1,44W - meaning that in an ideal world, i would have to store that in a total of 66,67ms, in order to be able to drive the load for 10ms. Meaning that i would be able to repeat the cycle 13 times pr. second.

    Does it make more sense now? If not, let me know ;)
     
  5. JMD

    Thread Starter Member

    Dec 9, 2009
    96
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    Someone? :)
     
  6. ifixit

    Distinguished Member

    Nov 20, 2008
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    Does the 12V need to be regulated to maintain 12V during the 800mA discharge? If so, how accurate?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Why not just use a lead-acid battery for the energy storage, with some rather large caps on the output?
     
  8. ifixit

    Distinguished Member

    Nov 20, 2008
    638
    108
    To get 9.6W from 1.44W...
    1. Charge time with a constant current is: T = (C x ΔV)/I.
    2. Charge a 1500uF cap from the 24V with a constant current at 60mA.
    3. Allow 2V for constant current device headroom.
    4. Initial charge takes 1500u x 22V / 60mA = 0.55 Sec.
    5. Power a 1A, 3-terminal regulator from this cap and adjust for a 12V output.
    6. Allow 4V for the regulator headroom.
    7. With a 800mA load on the regulator output, the cap will discharge at 533V/sec. or from 22V to 16V in 11.3mS.
    8. Discharging for 10mS would take the cap V from 22V to 16.667V.
    9. Re-charging from 16.667V to 22V at 60mA takes 134mS.
    10. You could also just charge the cap with a resistor to 24V, but will take a little longer.
    Play around with this example and come up with a circuit that works for you.

    Have Fun,
    Ifixit
     
  9. Bernard

    AAC Fanatic!

    Aug 7, 2008
    4,172
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    A little more information on power source might be helpfull, ie, solar cells? I would use a buck converter to drop V down to about 15 V, & charge Sgt Wookie's 10 Ahr Pb-H2SO4 battery; would skip big cap on output. Might use sample and hold during on time to monitor battery V & adjust duty cycle as necessary to keep V above 12.
     
    Last edited: Apr 18, 2010
  10. JMD

    Thread Starter Member

    Dec 9, 2009
    96
    0
    That is preferred, yes.


    SLA's or any other batteries is not an option. We are to imitate the power grid - power source is a generator with 3-phases. (very tiny one).


    True, a LM7812 could be used - i just have to compare the power loss with a buck-converter :) Ill give it a shot.


    The power source is a 3-phase generator (actually a hobby motor). Batteries are not an option.
     
  11. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    Charge a large capacitor direct from the generator.

    Produce the 12V with a National 'Simple Switcher' IC to get decent efficiency and operation as the input voltage droops.

    I think? that some of the simple switcher ICs have a logic enable control. You could use that with a couple of comparators monitoring the input capacitor charge to switch the regulator on & off.
     
  12. JMD

    Thread Starter Member

    Dec 9, 2009
    96
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    Ive thought about the caps too - actually dual layer super caps. Didnt think of the ICs tho.. got any model numbers?
     
  13. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    LM2575T- adjustable or 12V, pin 5 ON-OFF.
     
  14. JMD

    Thread Starter Member

    Dec 9, 2009
    96
    0
    Thanks! Ill give it a look !
     
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