Negative DC to Positive DC Converter

Discussion in 'General Electronics Chat' started by cackharot, Feb 2, 2011.

  1. cackharot

    cackharot Thread Starter New Member

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    Hi,

    I need a circuit for converting -48V DC to 24V DC capable of 1.5A output.
    I search the net for a while but nothing found. Can you guys help me out.

    Thanks,
  2. RmACK

    RmACK Active Member

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    Welcome to the forum. Have you considered a buck-boost converter? If you look around, you will find that they are usually drawn to show a positive input and a negative output voltage, e.g. http://en.wikipedia.org/wiki/Buck-boost_converter.

    To get a negative input converted to a positive output, there is actually no difference except where your common earth is, so simply swap the polarity of the input and the diode.

    How experienced are you in electronics?
  3. cackharot

    cackharot Thread Starter New Member

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    RmACK, Thanks for u reply.

    I looked into buck-boost converter. Its seems to solve my problem!

    But could you suggest any IC with my required Input and Output range?

    I just now finished my B.E. :)
  4. RmACK

    RmACK Active Member

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    Hi cackharot,

    The large voltages relative to the max allowed by many PWM chips is an issue I foresaw in implementing the control for a buck-boost for this task. Personally, I would look at deriving a power rail between your ground and -48V say at around -12V, using a zener or a 3 terminal linear regulator to power your PWM chip. I'd probably just go for a TL494 since they're so simple to use, very forgiving and have uncommitted transistor outputs which can simplify level shifting.

    Then there just 2 considerations: feedback and drive.
    -For the feedback, I would put a resistor divider between the +24V output and the -12V rail such that its output is within the common mode range of your PWM chip's error amplifiers. Then it's a case of going through the typical design steps for your feedback network to ensure it responds appropriately to changes in load but not to its own output ripple.

    -For the drive, I would configure the 494's transistors to pull down the base of a pnp bjt whose emitter is tied to ground. Its collector would pull up the gate of the MOSFET, thus performing your level shifting. You would need to of course limit the base current of the bjt (and provide it a pullup resistor), as well as limit the gate voltage on the MOSFET and provide the MOSFET with a pull-down resistor. You could even incorporate the pulldown into your voltage limiting.

    That would be my general approach, but it is only but one of many ways you could do it.
  5. cackharot

    cackharot Thread Starter New Member

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    Thanks RmACK!,

    Could you please this explain a bit more...
  6. t06afre

    t06afre AAC Fanatic!

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    -48 volt is the telephone line voltage. I hope you do NOT waste time believing that you can take 1.5 A from a telephone line. The only outcome will be that your telephone line will be disconnected and some really grumpy people from the telephone company will knock on your door.
  7. RmACK

    RmACK Active Member

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    I'll draw you a little schematic of what I mean after work. t06afre raises a very good point, what is your application?
  8. marshallf3

    marshallf3 Well-Known Member

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    Try registering with http://www.national.com/analog and using their online designer.

    Any reason the input has to be negative? Swap the wires.

    By all means let us know what the app is and any other info you can provide.

    And as mentioned above, you can not use a telephone line voltage for anything more than just a minor amount of current, then you've got the problem of the ringing voltage which can run upwards of 100V not to mention the fact that if you draw much of anything out of it it will send out a busy signal to anyone wanting to call you. There are ways around this using a tiny trickle to charge up a battery if you don't expect to draw much of anything over time.
    Last edited: Feb 2, 2011
  9. cackharot

    cackharot Thread Starter New Member

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    Its not a telephone line application.

    Actually I've a requirement of designing two power suppliers one will get input from 240V AC 50Hz and another is -48V DC. As far as -48V DC they didn't provide me more information. So I've no clue where they ll get that -48V DC.

    Here is my Power Design spec. So you can help me better

    1) I/P: -48V DC and 240V AC, 50Hz
    2) Output
    a) 24V DC at 1.5A
    b) 5V, 3.3V for my embedded system operation < 1A
    c) -48V DC if the input is 240V AC.
  10. marshallf3

    marshallf3 Well-Known Member

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    That is a mystery but certainly not unsolvable.

    Any circuit to share? I'd love to see this one.
  11. RmACK

    RmACK Active Member

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    I've attached a schematic for both the zener and 3-terminal regulator approaches I mentioned for creating a floating -12V rail for your PWM chip to run from. R1 has 36V across it and needs to have its resistance chosen such that more current flows through it than your 12V circuitry needs. You must then choose the power rating of R1 to be able to handle this current.

    Alternatively, the 3 terminal regulator is also shown. Either way you'd of course want to put plenty of capacitance across the 0V and -12V points. It is also wise to place a diode across the 3 terminal regulator to protect it, anode to the -48V, cathode to the -12V.

    Attached Files:

  12. n1ist

    n1ist Active Member

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    As mentioned, -48V is very common in the telecom world (why negative? So the outside plant doesn't electoplate itself away...) In a CO, it comes from a rectifier with a large battery bank floating across it. Elsewhere, it comes from small power supplies.

    Depending on quantities and cost targets, I'd likely start with an isolated 48V to 24V switching converter brick. Then follow it with a switcher to get the 5V and 3.3V (maybe something in National's "Simple Switcher" line?) For the 240V option, just stick a 240V to 48V supply in front.
    /mike
  13. cackharot

    cackharot Thread Starter New Member

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    Why -48V is very common in Telecom? Could u explain?
  14. thatoneguy

    thatoneguy AAC Fanatic!

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    Power = I * V

    For same power, 48V will drop less power (voltage) over a distance than 12V. Hearsay: 50V was considered "dangerous" at the time the phone system was designed, so they made it a bit less. 4 12V batteries (though you can still get a shock in the right conditions).

    Reason for negative relative to earth was stated above.
  15. cackharot

    cackharot Thread Starter New Member

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    This may be silly question but I'm not very clear

    How can we get -48V with 4x12V batteries? Batteries have 0V at cathode and 12V at anode. if we cascade we get 0V and 48V then how -48V?
  16. SgtWookie

    SgtWookie Expert

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    It's all what you reference the voltage from.

    If the top of the string's positive terminal is connected to ground, then the bottom negative terminal will measure ~-48v.
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