negative power supply

Discussion in 'The Projects Forum' started by rohitgeek, Sep 29, 2008.

  1. rohitgeek

    Thread Starter New Member

    Sep 29, 2008
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    Hi everyone,

    I wish to have a dual power supply i.e. +- 5 volt . I have a 12v dc adapter. I have generated these 2 voltages seperately without any difficulty using regulators. But when i try to embedd both the regulators in the same circuit, the universal problem of ground point(reference point) stands in the way. Kindly suggest any idea as i need it urgently .

    Thanks
     
  2. iONic

    AAC Fanatic!

    Nov 16, 2007
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    If you can and are willing to get inside the 12V Wall adapter, you can remove then obtain the ground, or reference point you are looking for, but in reality you already have that, what you are looking for is the negative side of the AC signal from the full wave rectification circuitry inside the adapter. If the adapter is center tapped then you would be in luck and the attached image would be a good start to getting the job done. I am assuming that you were using the 7805 and the 7905 to build you circuit. Unfortunately the majority of wall adapters are not center tapped.

    The solution would then require an IC that is capable of inverting the voltage from positive to negative. Not knowing your application need and just how exact you need the "+/-5V" and whether or not you need to be able to individually or simultaneously adjust the output voltage in any way makes it difficult to suggest a suitable part.
     
    Last edited: Sep 29, 2008
  3. DickCappels

    Moderator

    Aug 21, 2008
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    You can just connect the two power supplies together, with one "upside down", if there is no DC connection between the two power supplies (if they use two different transformers, of if they are powered from isolated windings).
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    You will need two wall warts. Unless the transformer secondary has a center tap, then there is no way to make both positive and negative supplies using one transformer winding.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    See the attached; it is a "split supply" using an LM317 adjustable regulator and a power op amp.

    The advantage of using a single regulator is the elimination of the dropout voltage of a negative regulator. An LM317 has a nominal 1.7 dropout voltage. In the attached schematic, the LM317 has been set up to output 10V; but it can be adjusted down to about 1.25v via R2. The output of the LM317 is measured between +V out and -V out.

    One half of an L272 power op amp is used to provided an artificial ground reference. R3 can be used to adjust where the ground reference occurs; anywhere from about -V+1.5v to +V-1.5v.

    The 2nd half of the L272 is unused.

    An LM675 can be used for the power op amp; it is capable of 3A vs the L272's 0.7A capability. However, the LM675 is considerably more expensive. The L272 is available for under $1 directly from Fairchild Semiconductor. The LM675 is around $6 to $9.
     
    Last edited: Sep 29, 2008
  6. iONic

    AAC Fanatic!

    Nov 16, 2007
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    I think, bottom line, that we need to know what the power requirements are and how close to +5V and -5V he needs.

    What will the supply be operating in the end???

    SgtWookie: I didn't see your post prior to posting my recent comments, but it looks as though you have a nice solution. I like it anyways!

    Is the + and - voltages vrey close to one another?
     
    Last edited: Sep 29, 2008
  7. rohitgeek

    Thread Starter New Member

    Sep 29, 2008
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    0
    @ionic

    what basically i need is separate simultaneous two voltage levels(+5 and -5). furthermore, i am able to generate these in ac domain( utilizing center tapped transformer and negative cycle). Seconldy, i need exact 5 v and -5 v and current rating around 500 mA. I am using 7805/905 IC but as i told , they are not working in unison.

    @sgtwookie

    this circuit looks the solutions to what i require...actually when i was using7805/905 , i know the problem of ground will arise but was not getting the virtual ground that i could use to measure the + and - voltage with and take this ground (artificial ) to our circuit. will be studying the circuit.....



    thanks a lot
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Do you mean as far as amplitude? If so, they can be equalized via R3.

    If the offset voltage of the L272 were 0, and one wanted +V to be the same amplitude as -V, then pot R3 could be replaced by two identical resistors.

    However, the L272's offset voltage is nominally 15mV, and may be as high as 60mV. Using a single multi-turn pot makes for an easy way to compensate.
     
  9. iONic

    AAC Fanatic!

    Nov 16, 2007
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    OK. Having a center tapped transformer is great for your cause. The question that arises now is, does the center tap position read in the neighborhood of 6V or 12V? If it measures about 6V then by the time you add rectification and regulation you will not reach the 5V you are looking for. But if you have about 12V then you have all you need to apply the method I posted originally.
    I'm the schematic I posted you would have to tweak the two trimmers R1 and R4 to match the voltage levels.

    As for the 500mA requirement... I think the newly posted circuit can do that...
     
    Last edited: Sep 30, 2008
  10. Tahmid

    Active Member

    Jul 2, 2008
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    Hi rohitgeek,
    Assuming your current requirement is <500mA, the attached circuit is supposed to be very easy (only 3 parts!) and reliable.
    Hope it helps.
    Thanks.
    Tahmid.


    [​IMG]
     
    Last edited: Sep 30, 2008
  11. scubasteve_911

    Senior Member

    Dec 27, 2007
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    Is that supposed to be a battery drainer circuit or something?

    Steve
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    Yep, simple and woefully inefficient.
     
  13. Tahmid

    Active Member

    Jul 2, 2008
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    Hi,
    The circuit is linear and has 80% efficiency at 500mA full load.
    1W is lost as heat. 1/5=20% therefore only 20% is lost at heat.
    Even if you put a linear regulator, 2v WILL be dissipated as heat. 2v at 500mA equals 1W loss. Same as the circuit I showed. So, Steve, could you explain how it is a battery drainer?
    Sgt Wookie, please explain how it is woefully inefficient in comparison with your circuit.
    Thanks.
     
  14. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Tahmid,

    You will get along with commercial product marketers well.

    Not even considering the diodes, the power loss across the 3.9 ohm resistor is 2V * R = 6W (i meant, I, so 1W). So, at ~1mA load, you're circuit is about 0% efficient. Linear regulators can have voltage drops in the 400mV range, so I don't know where your 2V is coming from. This isn't even considering the fact that zener diodes have intrinsic bulk resistance, which causes the voltage to dip under load. You rectify this by putting more current through the circuit, which this one does significantly.

    Steve
     
    Last edited: Oct 3, 2008
  15. Tahmid

    Active Member

    Jul 2, 2008
    344
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    Hi Steve,

    I think you have made a calculation mistake.

    According to formula,
    P(power) = VI = I*I*R = V*V/R , not 2v*R as you have mentioned.

    So according to formula, power loss across resistor at 500mA load,
    P = VI = 2v*0.5A = 1W
    or,
    P=I*I*R = 0.5A*0.5A*3.9ohm = appr. 1W
    or,
    P=V*V/R = 2v*2v/3.9ohm = appr. 1W

    or, at 10mA load, P = VI = 2v*0.01A = 0.02W

    So, in all cases efficiency is 80%.
    For linear circuit, I think it is an efficient one.

    Please look into the matter.

    Thanks.
     
  16. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Yeah, I made a mistake in my original calculation.

    I do not agree that in all cases it is 80% efficient, or you should go work for national semiconductor :p

    Under a very small load. Your current through the circuit is set by your resistor and the sum of the zener drops.

    12V - (2 * 5V) = 2V across resistor
    2V/3.9 = 513mA through circuit

    total power loss = power lost by zeners + resistive power loss
    resistor loss = V*I = 2V * 513mA = 1.02W
    zener loss = (5V X 2) * 513mA = 5.13W

    for a total loss of 6.15W under zero load

    The efficiency increases as the zener current equals the load current.

    Steve


    ps- dont discount experience by others, it is as important or more than intelligence
     
  17. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Oh, and I found this on wikipedia, which sums it up for you.

    This regulator is used for very simple low power applications where the currents involved are very small and the load is permanently connected across the zener diode (such as voltage reference or voltage source circuits). Once R1 has been calculated, removing R2 will cause the full load current (plus the zener current) to flow through the diode and may exceed the diode's maximum current rating thereby damaging it. The regulation of this circuit is also not very good because the zener current (and hence the zener voltage) will vary depending on VS and inversely depending on the load current.
     
  18. Tahmid

    Active Member

    Jul 2, 2008
    344
    25
    Hi Steve,
    My circuit is not for STMicroelectronics or Texas Instruments. It is for rohitgeek who wanted a circuit with 500mA load. Why bother about no load condition all the time? We all know that at no load condition, current drop is more through zener. Here nobody will test with no load condition. rohit's requirement is 500mA current at which power loss is 1W. So, it is a workable circuit, which I have tested myself.
    If you want a circuit for a small amount of load, it can be provided to you.

    One more thing, I like the companion of technically knowledgable personals, not the commercial personals of any company, as you have mentioned.
    Thanks.
     
  19. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Oh, just admit when you're wrong.. and it isn't about a no-load condition, it's about when there is little load..
    under 100mA load, there will be 1W delivered to the load, while about 5W consumed by the zener regulator.

    Steve
     
    Last edited: Oct 3, 2008
  20. Tahmid

    Active Member

    Jul 2, 2008
    344
    25
    In that case, always apply a load to the circuit (perhaps a dummy load) if you have to work at low currents. At higher current no need for a dummy load.
    Thanks for taking additional pain for providing me useful lesson from wikipedia.

    I said my circuit is easy and workable but not the most efficient one.

    With thanks.
     
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