7824 supplied less then Vin minimum??

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

  1. wishIknew

    Thread Starter New Member

    Jan 7, 2014
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    Regarding your typical 7824, I understand it needs a minimum voltage differential (Vin-Vout) to regulate (typically 2V), and if Vin drops below this threshold it will `stop regulating`. I need to know what that means practically, i.e. will stop passing current completely, or will pass current and Vout will simply be equal to Vin, or funky crazy unpredictable electrical stuff will happen, or it will go into thermonuclear breakdown and blow up the world??
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    I think your question deserves to be on a thread of its own. Hence I have started a new thread.
     
  3. MrChips

    Moderator

    Oct 2, 2009
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    Welcome to AAC. We hope you enjoy a long extended stay with us.

    All linear voltage regulators will exhibit a dropout voltage. That is just simply the nature of the beast. This means that the output voltage will be lower than the input voltage.

    Once the input voltage falls below a minimum the chip will continue to conduct but the output voltage will be lower than the specified regulated voltage.

    Since voltage regulators are often used to remove the ripple in AC rectified power supply circuits what this means is that the power supply ripple will leak through the regulator IC resulting in a DC supply with a substantial amount of ripple.
     
  4. ScottWang

    Moderator

    Aug 23, 2012
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    Unless the drops voltage of datasheet is shows that less than 2V, otherwise you should count it >3V is better.

    The important thing is the current should be enough to avoid the voltage drawing down.

    The basic current request formular as :
    Iout_dc = Iin_ac * 0.707 + 20%

    The 20% is for backup, it depends on what the load needs, some loads are need more and more.
     
  5. wishIknew

    Thread Starter New Member

    Jan 7, 2014
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    Okay. Cool thanks for that. I am 90% there on understanding this.

    My application - DC environment. 24VDC power regulation to small water pumps. Supply 24VDC batteries. 1-2A load. The pumps don't like the high voltage fluctuation when the batteries are under charge. The voltage differential across the reg will be sufficient during charge. When I put a large (2000W) load on the batteries, the terminal voltage is going to drop to ~23V. This voltage is fine for the pumps (they just run a little slower, no issue), but what it does to the reg is my concern.

    So, from your responses, am I right to say, the reg will simple continue to 'try' to regulate, but Vout will be lower then the specified Vreg, AND naturally continue to be lower then Vin?

    If this is the case, and nothing else undesirable or significant is doing to happen, it should still be adequate. As all I am trying to do is 'cap off', or 'shelter' the pumps from the higher voltages experienced under battery charging. 1VDC or so lower than 24VDC is fine for the pumps to run on.
     
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Or get a modern LDO (Low Drop Out) regulator. Even better, an SMPS that puts out 24V while the input goes from <10V to >35V at an efficiency of ~85% (no heatsink)
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Here's an example of an LDO regulator that should work for you.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    If you want a discrete design here's a relatively simple one. Note that the input and output voltages are differ by only a little over 300mV below the 24V regulation voltage (as determined by the ON resistance of the MOSFET). The P-MOSFET can be just about any MOSFET with adequate current and voltage rating (derate at least 50% for good reliability).

    Note that the MOSFET will need a heatsink with the maximum dissipation occurring at the maximum battery voltage and maximum current. The power dissipated equals the input-output voltage differential times the current.

    Edit: Here's an improved version with simplified compensation and added current limiting.

    LDO.gif
     
    Last edited: Jan 18, 2014
  9. #12

    Expert

    Nov 30, 2010
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    Exactly. All you have with a starved regulator is over-voltage protection.
     
  10. ScottWang

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    Aug 23, 2012
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    Why we need to use a regulator, because the load need a stable voltage, and the regulator can do that, the pump was made of the coil, and it can be provided by a voltage range, no need to setup for a fixed voltage as 24V, if the power source less than 24V then it just working slower and no harm to the pump.

    Why the 24V voltage will drawing down, because the current of the input power source is not enough, if you felt that the voltage drawn down is ok, then you just don't care about the '24V' voltage level.

    I build a frequency counter, and used 15 IC and 6 digits 7-segs LED, I thought that the LEDs drawn too much current, so the input of 7805 can't keep the voltage that it should be, and also caused +5V generated a little ripple, but it won't affecting the function, so I just let it be, maybe someday when I want to modify the functions, during that time that I may changing a transformer which can be get more current, so this situation for personal use is ok, of course we can't do that for the commercial.
     
  11. wishIknew

    Thread Starter New Member

    Jan 7, 2014
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    @ MikeML - SMPS is a little overkill for what I want. A LDO reg is a great idea though and exactly where I have headed.

    @ Crutschow - You are a champ! Thanks for the link to the LDO, and also for your circuit! I would have built it except I had already ordered some LDO's. Dam!

    This is what I bought: KA278RA05C, its a 2A LDO Nominal fixed at 5V or adjustable between 1.27VDC - 32VDC.
    http://www.fairchildsemi.com/ds/KA/KA278RA05C.pdf

    My problem now is, I am not sure on how to work out the appropriate resistors to build the following circuit (found on page 13 of data sheet).

    https://www.dropbox.com/s/jiuty60t7p5k6l1/24VDC adjustable LDO reg.jpg

    Can someone please tell me what values of R3 & R4 I need if I want Vo = 24VDC?

    (Sorry about the Dropbox link, I couldn't work out how to use the BB code in the FAQ's on how to attached/embed a picture.)
     
  12. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Comparing the R with RA version, you can see the internal resistors R1 and R2.

    [​IMG]

    [​IMG]

    As the resistor to the bottom is 600 Ohms, you can only decrease the value of R1 by putting an external resistor parallel.
    For 24 volts this would be around 40 Ohms.

    As for attaching images, read this from the FAQ:
    Attachments and Images

    Bertus
     
  13. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    Hola crutschow,

    What actually is Vtst in the circuit?
     
  14. bertus

    Administrator

    Apr 5, 2008
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  15. ScottWang

    Moderator

    Aug 23, 2012
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    KA278RA05C is a medium power package, the Pd Wmax is only 15W, so if you using it less then 0.5A is better when the output voltage is more higher and with heat sink.

    You can calculating the Iout = PdWmax/Vout.
     
  16. wishIknew

    Thread Starter New Member

    Jan 7, 2014
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    I need 2A continuous from the regulator. I believe the KA278RA05C will do it with a adequate heat sink adjusted for 24V, unless I have missed something??

    Regarding the quoted message;

    I do not 100% follow you. Referring to the following Application Circuit below, I understand both R3 & R4 are external resistors that I need to add yeah? If so, what values for both R3 & R4 do I need to create 24V?

    24VDC adjustable LDO reg.jpg
     
    Last edited: Feb 3, 2014
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