positive and negative power supply voltages

Discussion in 'General Electronics Chat' started by new2circuits, May 4, 2009.

  1. new2circuits

    Thread Starter Member

    Apr 22, 2009
    21
    0
    Hi:
    I was wondering if someone can explain the concept behind positive and negative power supply voltages ? For example some devices can accept +5V and -5V. Is this strictly for polarity purposes ? The magnitude is 5V in either case, I understand.

    And, why do some devices accept 2 different types of inputs while others only accept one ? For example, one device accepts 5V while another accepts +- 5V ?

    Thanks for your assistance.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    That is so broad a question that a simple answer is quite difficult.

    In general, logic IC's will be unipolar in operation. They will switch between +5 and 0 volts. Other voltages are possible with logic, but single-supply operation is nearly universal.

    Analog IC's often have dual polarity supplies so their outputs can swing both positive and negative. While some may be limited to +/-5, +/-12 is probably most common still.

    You seem to be specific about some devices that utilize voltage only in the +/-5 volt range. Perhaps you could say more about the devices in question?
     
  3. new2circuits

    Thread Starter Member

    Apr 22, 2009
    21
    0
    Hi: Thanks for your response.

    I do not have anything specific in mind. Just happend to use 5V for the example.

    Appreciate your response.

    Thanks
     
  4. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    One example of the usage of negative voltage is to increase switching speeds in digital circuits.
    Most multiplexers have the +5V Vcc and ground (0 V) power pins but some multiplexers also have a negative -5 voltage pin for the sole purpose of enabling certain transistors within the circuit to switch faster on/off. You can look up emitter coupled logic for more details.
     
  5. new2circuits

    Thread Starter Member

    Apr 22, 2009
    21
    0
    Thanks for your response.
    I'm having a hard time trying to quantify "negative voltage" like -5V.

    How would you describe negative voltage (whether it is a power rail or an input signal) to a novice ?

    Any good reference material on this is welcome as well.

    Thanks again.
     
  6. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    Voltage is defined as the energy each charge aquires when moving from a to b. I assume you havent taken physics courses in electromagnetism so i'll try not to complicate things.

    The term voltage is a bit difficult to grasp at first because it's all about relative voltages. Let's describe voltage between 2 points as the tendency of charge to move between those 2 points. If we arbitrary choose location B to be at zero volts and location A at 10 volts. The charge will have a motive to move from A to B (from higher to lower voltage).
    We chose 0 and 10V for simplicity but we could aswell have chosen B to be at 30V and A at 40V. The voltage difference however is still 10V as before and the charge will have the same tendancy as before (0 and 10V) to move from A to B.

    So let's imagine a circuit with 3 points A, B and C which have voltages +5V, 0V and -5V in the same order.

    The voltage difference between A and B is 5V. As mentioned before you can look at this as the tendency for the charges to move between the locations. You might have realised by now that the voltage difference between A and C is 10 volts. (5V-(-5V)) = 10V. So the charges have an even greater motive to move from A to C then from A to B (actually double the tendency).

    Zero voltage is always decided at an arbitrary location (we can choose any location we want as long as relative voltages are maintained). Conventionally this is what we call ground. Usually when you look at a circuit you imagine the current flowing from a source with positive voltage to ground (0V).

    Hope this helped.
     
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  7. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    In order to have positive and negative voltages, you must first define what your reference point is. In many cases, earth ground is a very good 0V level, but that need not always be the case.

    In a battery operated system, you can define the positive side, negative side or middle of the battery voltage to be your 0V level. Depending on what you choose your 0V level to be, the other voltages become mathematically positive or negative with respect to it.

    In a battery operated system with 2 9V batteries, you can define the system multiple ways.

    0V -battery+ 9V -battery+ 18V

    -9V -battery+ 0V -battery+ 9V

    -18V -battery+ -9V -battery+ 0V

    Now you can also connect you 0V to earth ground and change the batteries to other power sources or storage elements.
     
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  8. Jack Bourne

    Active Member

    Apr 30, 2008
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    Can you create a +-9V supply from a 18V supply with poitive and ground as its terminals?
     
  9. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    Certainly. You can make a virtual ground at the midpoint of the supply. Just remember that the virtual ground and the ground on your supply, and possibly other test equipment, are not the same. The inputs and outputs will have to be adjusted accordingly.

    Or, you can use a DC/DC converter to give you an isolated +/-9V supply and then ground the center voltage.
     
  10. Jack Bourne

    Active Member

    Apr 30, 2008
    39
    0
    So use a potential divider both resistors same value and use the mid point for ground reference? That would be good.
     
  11. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    That is right. Use a resistor divider to create the new "ground" voltage and follow it with another op amp, which can be as simple as connecting divider voltage to (+) terminal and connecting output of the op amp to the (-) terminal, so that it becomes a point that can source and sink current without altering the divider voltage.
     
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