LED's are current driven

Discussion in 'General Electronics Chat' started by cjdelphi, Jun 24, 2010.

  1. cjdelphi

    Thread Starter New Member

    Mar 26, 2009
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    Audioguru stated that LED's are current driven but is that really true? I don't see how that's completely true.

    If that was true, you'd be able to power an LED from just 1volt but you simply can not it's impossible, if you have 1v in a circuit and 1 ohm resistor, you'd have 1 amp going through the circuit, now if you place an LED in the circuit, you should get a pretty bright light and with 1 amp going through the LED it's going to go bang.. but yet, it does not, you need voltage to make it go pop not current.. so an LED is happy providing it gets the right voltage and you don't need a resistor at all if the correct voltage is given, although it's best to use a resistor to protect the LED.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well you it's true that it's much easier to driven the LED's form the current source then from a voltage source.
    LED's require current limiting (resistor) when driven from a voltage source (Voltage source drop his voltage when discharged so current also will drop) .
    And if we use the current source we can forget about luminous intensity variations or changes in forward voltage drops between LEDs and temperature influence on forward voltage drops.

    But there is no such thing as a current source without power supply (voltage source)
    So we must build the device that is act like a current source but is supplied with voltage source. And this voltage source must be larger then Vf
    http://en.wikipedia.org/wiki/Current_source
    or we can use a switching regulator.
    http://focus.ti.com/lit/ds/symlink/tps61042.pdf
     
  3. retched

    AAC Fanatic!

    Dec 5, 2009
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    [edited as a need not to confuse readers]

    If you dont have enough voltage to jump that gap, it wont operate to draw the current required.

    If the led was a voltage controlled device it would be brigher with 24v @ 20ma then with 9v @ 20ma.

    as long as the forward voltage is satisfied, it is a current controlled device. You change the current to control the brightness.
     
    Last edited: Jun 24, 2010
  4. beenthere

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  5. Wendy

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    Then there is this...

    LEDs, 555s, Flashers, and Light Chasers

    Chapters 1 and 2 focus on LEDs.

    There are many devices that require a breakover voltage before they start working, regular diodes being one of them.
     
  6. whale

    Active Member

    Dec 21, 2008
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    At first voltage criteria should be satisfied, then LED is controlled by current upto its absolute maximum current value to change the intensity of light emitted.
    you should also remember that when a LED is connected to higher voltage above 1.7v, even 30v with a very low current source( such that the current delivered by the source is very less than the maximum current limit of LED )nothing will harm the LED.
     
  7. Darren Holdstock

    Active Member

    Feb 10, 2009
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    Meaning no disrespect to m'learned peers, but this does need clarifying. Audioguru is correct, indeed LEDs are current-controlled devices, and if forward biased with a current there will *always* be a forward voltage present. I see where you're coming from cjdelphi - the issue you have is with the compliance of your current source; that is, you need more than 1 volt of available voltage to put 1 amp through your LED, given that the forward voltage drop of the average LED is 2V when it's biased at a few mA. Bear in mind that the forward voltage is a product of the forward current, and trying to control an LED with a voltage is only ever going to end in tears.

    Let's assume a perfect current source, able to deliver an infinite voltage if necessary. Bias the LED at a few pA - a uselessly small current admittedly - and the forward voltage drop will be a few mV, or thereabouts. This is because the LED is a semiconductor junction, not a spark gap or a switch, and there are equations that give the forward voltage drop as a function of forward current. The Shockley equation is a pretty good approximation; other more complex and less popular equations are available. Draw a graph of junction current versus junction voltage drop and it starts off at zero amps, zero volts. With both these axes on a linear scale it appears as though a moderate forward voltage suddenly snaps in at a moderate current. This is an illusion of scale, as the voltage versus current characteristic is non-linear. Draw this graph again with a logarithmic current scale, and we now have a (more-or-less) straight line, though the slope of this does tend to fall at higher currents as series resistance effects start to dominate.

    Have a look at this handy document; Advanced Electrical Design Models, from Agilent and Philips, who should know about this stuff. Look at the left-hand curve of Figure 3.2A: At 10 nA the forward voltage drop is about 1.2 V; increase the current to 10 mA and Vf is 1.8 V. If the graph were extended downwards the trend would hold, all the way down to 0,0. Corroborating but nasty equations are supplied for the mathematically curious.

    This holds for all semiconductor junctions, whether diodes, LEDs, semi lasers, transistor base-emitter junctions etc. The characteristics will vary according to the materials used (silicon, germanium, gallium-arsenide or whatever), but the principles are the same.
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    The debate of current controlled vs voltage controlled is semantics in my opinion.

    They are strictly related and one-to-one. If you have 1 V at 1 mA you will have 1 mA at 1V.
    Diodes (and hence many semiconductor devices) are considered current controlled because it is very straightforward to pretend the voltage is constant and set current by the circuit and much more complicated to fix a voltage to get the current you want. It's simply how the non-linearity is set up, where tiny changes in voltage create huge changes in current necessitating very high accuracy if you want to control it with voltage.
     
  9. retched

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    well spark gap was a bad choice of words, but it is still a gap. The distance between the P and N is what gives the LED the color, prior to a lens.

    You have provided more information and math to show why, and I will loose more hair reading it again.


    The light that is emitted from an led is from the PN junction. The distance from P to N is what determines lens-less color.
     
  10. Ghar

    Active Member

    Mar 8, 2010
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    What do you mean by distance?
    The P and N must be physically bonded since they are a junction.

    I was under the impression it's from the junction potential (could say distance in terms of energy/potential I guess) which is a function of the doping concentrations and the material itself.
     
  11. Darren Holdstock

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    Feb 10, 2009
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    True, but it's very practical semantics. Good luck trying to control a semiconductor junction with a voltage ;). You quite correctly gave the reason why it generally isn't done this way.

    retched is quite right about the pn junction dimensions and emission wavelength, but there's still no insulating gap as such, more of a band-gap, in eV. The diode-as-switch assumption leads to many a circuit that won't behave as the designer intended. But I can't complain, I earn a living as a troubleshooter, and I can keep paying the mortgage as long as there are circuits that need fixing. The devil is in the detail in a good solid, reliable electronic design.
     
    Last edited: Jun 24, 2010
  12. retched

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    Agreed. I need to do more math. ;)
     
  13. Ghar

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    And I guess I need to go read those optical device chapters in my semiconductor physics text I never got through.
     
  14. cjdelphi

    Thread Starter New Member

    Mar 26, 2009
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    just for the record I was not disagreeing about an LED not being current driven I was just questioning what i pointed out about the fact you needed a certain voltage it's not strictly current because of the point i made...

    but then, without current very little could be powered, for example, if I limit the voltage to 9v with a dc to dc converter, then use a LM317 and then limit the current to say 10ma, if you then read the voltage, it will read '9v' but the interesting thing i found out was that you can then connect a 3v circuit providing it uses more than 10ma the voltage drops to 3v and supplies 10ma ... without damaging it. if the circuit used less than 10ma the circuit would probably get damaged as 9v would be pushed through it.

    So without current voltage means nothing anyway right?...
     
  15. Bosparra

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    Feb 17, 2010
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    Maybe, a better explanation for someone not familiar with the maths, is as follows:
    An LED brightness is current driven as long as the forward bias voltage is met. If an LED has a forward voltage of 2V then any increase in voltage will be useless, as the forward voltage will remain 2V. In the real world it does increase slightly, but lets ignore this for the moment.

    The increased voltage will be dissipated over the current limiting resistor, causing the resistor to get hot. Now, according to Ohm's law this WILL increase the current, but this is inefficient due to excess energy being converted to heat. The right way to increase the brightness is to decrease the resistance of current limiting resistor, thus increasing the brightness. Thus, an LED is current driven.

    The brightness of an LED is more or less proportional to amount of current supplied and LESS proportional to the voltage.

    Stated another way, current can't exist without voltage, but any increase in current will have the same increase in brightness, an increase in voltage will not have the same increase in brightness.
     
  16. Wendy

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    This isn't quite on track, chemistry, what the LED is made out of, is much more important as to color.
     
  17. retched

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    Well.... bill..... that is what determines the distance between the junctions.

    Using different doping agents and such can tint the light, but the basic wavelength is determined by that distance. generally.

    I dont dismiss what you say, you are most likely right.
     
  18. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    Per your setup of having 1V thru 1 ohm and getting 1A, if you add a diode (.7V) you now have 1V - .7V = .3V across the resisitor, which is now 300ma through the circuit. You've changed the circuit and its operating points. Note that an LED would have too high a voltage drop to be used in this example.

    No, that is not correct. Try taking a standard 20-30mA LED and a 5V @ 1A (minimum) supply. Do a little math: 5V - 1.7V = 3.3V, and therefore, 3.3V/1A = 3.3 ohm ballast resistor.

    Now you will be putting 1 amp through the LED -- see how long it lasts.
     
  19. eblc1388

    Senior Member

    Nov 28, 2008
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    Try this cjdelphi.

    I have given you a 10mA current source. Without load you measures the voltage across the current source and the voltmeter reads 998V DC.

    That's nearly a thousand volts....:eek::eek::eek:

    Now connect your LED(in forward polarity) to the current source, see how long it will last?
     
  20. Ron H

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    Apr 14, 2005
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    A 10mA current source with 998v compliance should not damage an LED. The current source, however, will be dissipating about 10 watts, so it might get a little warm.:rolleyes:
    Now, if the current source is already on when you attach the LED, and the output has significant capacitance, there could be damage...
     
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