LED circuit without a resistor.

Discussion in 'General Electronics Chat' started by Robert Smith_1437948150, Sep 13, 2015.

  1. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Hello,

    A few weeks ago I created a thread discussing the current in a simple circuit containing a battery, LED and a resistor. I am now wondering what the current would be in a circuit that contains just an LED and the battery.

    For example a 5V battery and an LED with a forwarding voltage of around 2V.

    The reason I ask is KVL states that 5V must be dropped across the circuit, so the LED has to drop around 5V, but looking at a voltage-current curve for an LED to drop that amount the current would have to be ridiculously high.

    What if the battery can't provide that amount?

    How would you calculate the current in the above example, I believe the internal resistance of the battery is used?

    I understand you would always you a resistor and the LED would pop in the above example, just curios what would happen in theory.

    Regards.
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
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    An LED has a small dynamic impedance which is delta V divided by delta I. Calculate what that will be from the highest current graphed for the LED and calculate battery current with that resistance (ignoring internal resistance of the voltage source which is ideally zero and LED forward voltage). It should be higher than the maximum LED current specification and the LED will burn out if the battery can provide sufficient current.
     
    Last edited: Sep 13, 2015
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  3. ScottWang

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    Aug 23, 2012
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  4. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    3V Lithium button batteries (CR3032 for example) has enough internal resistance to power a 2V red LED without noticable damage. The battery lasts about a month from what I remember. "Dead" AAA and AA battery pairs are also interesting because they last forever when lighting an LED (red, yellow or green). Blue and white LEDs are more difficult because they are 3 to 3.3V in general and will not give any substantial light from a little 3V lithium or pairs of 'dead' alkaline battery pairs. Adding a third battery gives too much potential and discolored the LED and quickly failed (within minutes to hours for red LED).

    If the LEDs are not expensive, do some experimentation. It's fun.
     
  5. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Thanks everyone, I now realise it's more complex than I thought, but can someone simply answer the following please -

    1.) Am I right in thinking the LED has to drop around 5V if the source is 5V, in line with KVL?

    If the above is true, then looking at the following image -

    [​IMG]

    2.) For the LED to be dropping 5V, the current in the circuit would have to be incredibly high, what if the battery can't provide that amount? How does the LED still drop 5V?

    3.) If the LED does have to drop 5V, you obviously can't calculate the current in the circuit just by looking at the above graph and seeing how many amps 5V equates to. I'm sure you will have to use the method dl324 explained. It seems as if the 'methods' clash with each other if you see what I mean, can someone explain when the current is calculated via dl324's method, how this then relates back to the above graph?

    Thanks.
     
  6. #12

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    Nov 30, 2010
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    If you don't read the information provided you will not know the answer. Read the article in post #3
     
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    The LED either does drop 5V or it doesn´t. Every non-imaginary power supply or battery has internal resistance. A stack of AAAs could have a few hundred milliohms, a PC power supply will have much much less. To be correct, when you are calculating the LEDs current you should include that internal resistance as if being in series with your standard current limiting reisitor. Since the current limiting resistor is zero, the only thing left to limit the current is the internal resistance.

    So, when you connect a LED right at the terminals of a power supply, the current might be a few tens of amps and things will start going south quickly. If you connect it to a stack of old AAAs, the voltage measured at the LED will be a lot less than the open circuit voltage of the batteries because of the drop across the internal resistance, so the current will not be that drastically high.

    Since all you have here is the graph, you can find out the resulting voltage graphically. All you need is to know the internal resistance, and the open circuit voltage.
    For instance, say your voltage is 5V and the internal resistance is 100 ohms. That gives you short-circuit current of 100mA, and open circuit voltage 5V. Mark these two on the appropriate axes, and where the line crosses the graph will be the operating point, so the yellow LED will run at roughly 25mA and 3V.
     
  8. jaydnul

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    Apr 2, 2015
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    In an ideal case, if your diode drops .7V, then (I=0 for V<.7) and (I=∞ for V≥.7).

    But really the remaining 4.3V (ignoring the resistance of the diode) will be dropped over the tiny ~1Ω resistance of the wire (still a huge amount of current).

    Edit: Plus the battery internal resistance which was mentioned previously. The point is, the resistance of the wire plus the resistance of the battery is still very small.
     
    Last edited: Sep 13, 2015
  9. Robert Smith_1437948150

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    Jul 26, 2015
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    I have, but there doesn't seem to be a part discussing the voltage drop of an LED in a circuit with no resistor.
     
  10. #12

    Expert

    Nov 30, 2010
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    That is because you are not supposed to connect an LED with no resistor!!!
    Only the strange cases using the internal resistance of a small battery can power an LED without a resistor.
    All this explanation and you can't figure out you're supposed to use a resistor?
     
  11. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    As I said in my original post, I know you're always supposed to use a resistor. It was just a theoretical question.
     
  12. dannyf

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    Sep 13, 2015
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    On an ideal battery (constant voltage, no internal resistance) and an ideal led (vertical I/V curve above its threshold forward voltage), the current in the circuit is undefined.

    On a real battery (constant voltage source + finite internal resistance) and a real led (sloppy I/V curve above its threshold forward voltage), the current is defined by the internal resistance + the slope of the I/V curve + the battery voltage, assuming that at the equilibrium point the led isn't burned out.

    It is not uncommon to drive an led without a resistor -> it is actually the preferred approach for power LEDs, to be driven by a voltage source whose output is current regulated: that's about 99% of the LED drivers out there.
     
  13. ScottWang

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    Aug 23, 2012
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    ThePowerSupplyAndTheEquivalentCircuitOfLED_ScottWang.gif
     
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  14. jaydnul

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    Apr 2, 2015
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    Technically it is infinite, not undefined. You could say the conductance is 'undefined' since the I/V curve is vertical and resistance would be 0.
    But even then it would be a step function, not a vertical line. Doesn't really matter though; we care about the real world.
     
  15. WBahn

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    Mar 31, 2012
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    What diode are you talking about. The TS is talking about and LED, not a silicon diode.
     
  16. WBahn

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    It's not that theoretical -- quite a few consumer products power an LED directly from a battery with no current limiting resistor. These are usually powered by coin cells (for instance, key fobs) where the internal resistance of the battery is sufficient to limit the LED current to an acceptable value.

    If the 5V source has low enough internal resistance at the currents needed to establish 5V across the LED, then the LED current will be limited by the I-V curve of the LED itself and will likely result in so much current that the LED fails very quickly (perhaps within milliseconds). If, on the other hand, the 5V source cannot supply that much current then it will appear as though there is a voltage being dropped across the internal resistance of the supply resulting in something less than 5V appearing across the LED. The LED may or may not be able to tolerate the resulting current.
     
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  17. jaydnul

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    Apr 2, 2015
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    It wasn't a diode specific answer. The point is, KVL does not lead to the diode dropping the full battery voltage. Sorry, should've quoted the question I was addressing:
     
  18. dl324

    Distinguished Member

    Mar 30, 2015
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    It might be helpful for your learning process if you connected an LED as you're describing. Then you'll understand why you're being told not to do that. The URL Scott gave in #3 had a video of the expected outcome when an LED is operated without a current limiting resistor from a supply that can supply more current than the LED can handle.

    If you want theoretical answers, you should do as much research as you can and then ask specific questions. If you pose a question with insufficient details (like the LED specifications - e.g. max forward current, IV characteristics (e.g. color), power source current capability), it's difficult for anyone to answer your questions. It would also be more acceptable if you did as much research as you can before you start playing 20 questions with numerous "what if" scenarios.
     
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  19. ScottWang

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    Aug 23, 2012
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    Draw the circuits with current limit resistor and without resistor, Do the practical experiments, measure the V/I of LED and the voltage of power or battery, measure the value of current limit resistor.

    Show the values of R,V,I,LED/V ,LED/I, what else you do?

    No practical parts to talk about that it just like breathing the air instead of foods.
     
  20. dannyf

    Well-Known Member

    Sep 13, 2015
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    "Can someone explain when the current is calculated via dl324's method, how this then relates back to the above graph"

    Fairly simple: take your graph, draw a line of Vf = 5v - R * I, where R is the resistance in the circuit. The point where that line intersects with the diode's I/V curve is the current through and voltage across the diode.

    Now, reduce R and see how the intersection moves.
     
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