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Circuit questions (polarity and LEDs)

  1. Apr 14, 2010 #1
    I'm going to go under the assumption that whatever I say is right, even though I know it's probably wrong (it saves me phrasing the entire thing as dozens of questions that way), so... correct me where necessary:

    If I have a +X volt battery and I connect it up to a circuit with an LED in it (forward biased), and the LED itself has a switch-on voltage of +X volts, then it will light up. However if I have the same circuit, but reverse biased, it won't work.

    Now, if I say my battery is a -X volt battery (the LED still needs +X volts to switch on), is this not the same as saying that it's a +X volt battery, except with the current flowing in the opposite direction? If so, would a reverse biased LED with a switch on voltage of +X volts switch on if the voltage of the battery was given as -X volts?

    I don't think I'm wording it very well, but... hopefully someone gets what I'm asking. I asked my physics teacher and he said no, however I don't know if he's understanding what I'm trying to ask, and, well, he's not done this stuff in years. He could be wrong.
     
  2. jcsd
  3. Apr 14, 2010 #2
    There's no such thing as a -X volts battery. A battery will give its positive side X volts more potential than its negative side.
    A LED needs its positive side to be at X volts more positive than the negative side to conduct current and give off light. (and you better use a resistor to limit the current)
     
  4. Apr 14, 2010 #3
    Hence the polarity, right? If you SAY you have a -X volt battery aren't you essentially saying a +X volt battery that's simply fitted into the circuit 'backwards' (opposite direction that you'd normally fit the battery)?
     
  5. Apr 14, 2010 #4
    LED lamps must be driven with current, not voltage. To drive an LED with a voltage source, i.e.a battery, a series resistor must be employed. Forward biased p-n junctions are thermally unstable with constant voltage sources, but stable with current sources as well as a voltage source plus a sufficient resistor. An LED is a current-driven device.

    Claude
     
  6. Apr 14, 2010 #5
    And the voltage is the driving force of a current, isn't it? A positive voltage drives the current in one direction, and a 'negative' voltage is the driving force of a current in the opposite direction to the positive one, right?

    My question would probably be a lot easier to ask with diagrams. I might draw up some of my thoughts on my white board to try and get what I mean across.
     
  7. Apr 14, 2010 #6
    wouldn't a -x volt battery imply that it's taking in more energy than it gives out? not sure if I'm picturing this correctly.
     
  8. Apr 14, 2010 #7
    From what I understand when something is said to have a negative voltage it's simply a positive voltage in the opposite direction. For example, an AC current can have a range of -5 to +5 volts because the current is going 'backwards and forwards'.

    So, if that's the case, I figure a 'negative voltage' in a circuit with a reverse biased LED should work (because a negative voltage is essentially the same as flipping the battery around from my understanding). Essentially you'd be making the LED reverse biased in relation to the battery first (turning it around in the circuit), and then you'd be turning the battery itself around.

    DSCF9332.jpg

    Are these diagrams correct?

    Top left is forward biased with a positive voltage. Current (conventional current) flows in the direction of the arrow.

    Bottom left is reverse biased with a positive voltage. Current won't flow because it's reverse biased, however, IF it were to flow it would flow in the direction of the arrow (as it's from positive to negative).

    Bottom right is reversed biased with a negative voltage. Seems to me like the bottom right and top left SHOULD work (while the bottom left and top right won't).

    Wait, I'm tired. I ****ed up one of the diagrams. The two diagrams on the right hand side are essentially the same, just flipped. My bad. The top right diagram should have the LED flipped around.
     
    Last edited: Apr 14, 2010
  9. Apr 14, 2010 #8
    Voltage doesn't "drive" current, and vice-versa. Both are needed for the LED to illuminate. Optical power is outputted when electric power is inputted. The law of conservation of energy is immutable. Input power is voltage times current. Both must exist or no light is produced.

    "Current driven" does not mean that voltage is not needed. It means that the source providing power is constant current in nature, not constant voltage. If a sufficient resistor is added in series with a constant voltage source, then this approaches "current drive" behavior. I was simply addressing the fact that an LED should never be directly connected to a constant voltage source without a resistor in between. That is all that "current driven" means.

    Even when current driven, the voltage is still present and indispensable for LED operation just as is the current. Neither is more important than the other, for LED lamps, and everything else. Does this help?

    Claude
     
  10. Apr 14, 2010 #9
    Usually current-limiting resistors are needed to limit the current in an LED, even when using a battery. Even so, if the battery is hooked up backwards (wrong polarity), the LED won't light up. But there are two-color LEDs available which are green with one battery polarity, and red with the other (there are actually two back-to-back junctions inside), so the LED is never really reversed.

    Bob S
     
  11. Apr 14, 2010 #10
    You can have voltage without current, but, you cannot have current without voltage.
     
  12. Apr 15, 2010 #11
    I'm thankful for the explanations, however, while they may help me with understanding the reality of the situations that isn't really going to help me when it comes exam time, unfortunately. We've not really delved into this stuff past equations and circuit diagrams in school, and, unfortunately, that's all the exam is going to focus on: not the 'reality' of the situation.
     
  13. Apr 15, 2010 #12
    Sure you can. Superconductor. This issue has been beaten to death. Only under static (dc) conditions can one exist w/o the other. Under dynamic (ac) conditions, they are inclusive. You can't have one w/o the other, like that song "love and marriage".

    Although the LED current cannot exist w/o voltage, we still drive the LED by controlling its current, not its voltage. A constant voltage source, CVS. connected directly across the LED leads results in thermal runaway. A forward current is established, the temp goes up, the diode scaling current "Is" increases, resulting in temp increase, then Is increases, temp increases, etc, etc.

    With a resistor, as soon as temp increases and Is begins increasing, the forward current increases so that the drop across the resistor increases. But then the forward voltage across the LED decreases, so that equilibrium is achieved. If the resistor is large enough, and the current limitations of the LED are observed, thermal stability is achieved.

    This is why LED lamps are always "current driven" and never "voltage driven". With current drive, the voltage is always present, but never driven directly across the LED w/o a resistor. Any LED maker will concur that a resistor must be used.

    Claude
     
  14. Apr 15, 2010 #13
    All current must be driven by voltage.
     
  15. Apr 15, 2010 #14
    Not so. A ring of wire of zero ohms can have current go from zero to one amp without any voltage drop between any two points of the ring. All it takes is a change of magnetic flux though the center.
     
  16. Apr 16, 2010 #15
    Yet, voltage must be present in order for current to "flow"
     
  17. Apr 16, 2010 #16
    Not at all. Just curious, how much formal education do you have in circuits & fields? No offense, but I just want to know your basis. What is your reference, or source of info? Do you do EE for a living? Your teachings are at odds with known scientific findings.

    Do you presume to know more than EE practitioners & profs?

    FWIW, current must be present in order to establish a voltage. They are mutually inclusive. Do you know what it means to say that 2 entities are "mutually inclusive"? This issue has been known since the 19th century.

    Before telling the whole scientific community what's what, I suggest that you brush up on what's what. Again, nothing personal. I'm not attacking your intelligence, just telling you that electrical science takes years of intensive study to even begin to master. I'm 54, been practicing EE for 32 yrs., am a part time Ph.D. student, and I'm still learning all the time. I still have not "arrived". BR.

    Claude
     
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