2 questions regarding a simple circuit with an LED

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

  1. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Hello All,

    If I had a simple circuit with a 12V battery and an LED that drops 2V and needs 20mA, therefore I'd need a resistor to drop 10V and set the current at 20mA, so I'd do R = 10/0.020 = 500 Ohms.

    1. Is it possible to 'use up' the voltage before it reaches the LED. For example if you placed the resistor before the LED, and wanted it to drop the whole 12V, then if I used a 600 Ohm resistor, would it drop 12V and then leave nothing for the LED?

    2. If I wanted to 40mA to flow through the LED, then I could use R = 10/0.040 = 250 Ohm resistor. Looking at the voltage-current curve for an LED, to increase the current through it I'd also have to increase the voltage (albeit slightly), but I'd still only have 2V after the resistor. How can the LED have a current of 40mA, when it only has 2V available after the resistor, it goes against the voltage-current curve?

    If anyone could help clear things up a bit it would be appreciated.

    Thanks.
     
  2. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    LEDs are current driven devices.
    The resistor "sets" the current the LED will receive.. and it doesn't matter at all if the resistor is placed before or after the LED..
    Its setting the current for that whole circuit..
    Don't think of the resistor like its "dropping voltage" its setting the current in the circuit..
    An LED is eating the voltage (its forward voltage drop) but a resistor is just setting the current in the whole circuit.

    If you put a bunch of people in a circle and have them go around and around they can only move as fast as the slowest walker.
    Doesn't matter where that slow walker is but it effects the speed of the entire circle. The resistor is the slow walker..

    The "voltage-current" curve is an approximation of the forward voltage drop an LED will cause in a system at that current.
     
  3. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Hi mcgyvr, thanks for the response.

    I understand that the current is equal throughout the circuit, but obviously the voltage isn't, and I was thinking if you did place a resistor before the LED, could you have so that it dropped all 12V and left nothing for the LED?

    Also, If the LED drops 2V at 20mA, and I use a 250 Ohm resistor then current is 40mA, which means that the LED will drop more than 2V, but if the resistor is dropping 10V then the total dropped is more than the source, breaking KVL.

    Can't quite get it to add up.

    Regards.
     
  4. dl324

    Distinguished Member

    Mar 30, 2015
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    Yes, but not with resistors. For your example, if you put 10 LEDs in series, there wouldn't be sufficient voltage to light any of them. With a 600 ohm resistor, you would simply reduce the current in the LED. Using your stated forward voltage of 2V for the LED, that would give a current of I=V/R=10/600=16.7mA. So increasing the resistance by 20% just decreased the current in the LED by the same amount. The actual amount would be different because the forward voltage of the LED would be reduced at the lower current.
    The voltage across the LED can only vary between 0V and the max forward voltage. You look at the IV curve for the diode to determine what it's forward voltage will be at 40mA. Then you subtract that from the power supply voltage to find the voltage across the resistor. Using your 2V, that would leave 10V across the resistor. R=V/I=10/0.04=250 ohms.
     
  5. dl324

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    Kirchhoff's Voltage Law says the sum of voltages around a loop will be equal to 0. In your example, you have 12V from the power supply, minus 2V from the LED; that leaves 10V across the resistor. The resistor will determine the current in the loop.
    KVL will not be violated. The forward voltage of the LED will increase at 40mA, but it will still be close to 2V.

    IV characteristics of LEDs:
    ledIV.jpg
    As current is increased, forward voltage increases. When the LED is operated at 20mA or more, it's past the "knee" of the curve and the slope of the curve shows that further increases in current will result in smaller incremental increases in forward voltage.

    I don't have a datasheet handy that gives LED IV characteristics this way; they're all using a log scale.
     
  6. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Thanks dl324,

    This is going to sound really stupid, but when calculating the current you did I=10/600, and you used 10V because we can observe the circuit and see the LED has voltage drop of 2V, but if the resistor is before the LED, wouldn't the voltage for the resistor be 12? how does it only drop 10 volts so that it leaves 2 for the LED.

    This is why I asked if it mattered if the resistor comes before or after the LED, not because of the current which I know is equal throughout the circuit.

    Regards.
     
  7. dl324

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    Mar 30, 2015
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    FWIW, an IV graph with log scale for current:
    ledIVlog.jpg
    This shows that for super-red LEDs, the forward voltage at 20mA is about 2.3V. At 40mA it would be about 2.7V.
     
    Last edited: Aug 13, 2015
  8. dl324

    Distinguished Member

    Mar 30, 2015
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    You answered your own question. The current in a loop is the same for all components and their voltage drops will be determined by the current. The placement of the LED and resistor is more a matter of preference.

    In some cases, one order may be safer than another. For instance, if the LED in this circuit was off board (e.g. on a front panel), it would be better to have the resistor before the LED so that shorting the anode to ground would be current limited. If the resistor was after the LED, shorting the anode to ground could cause damage.
     
  9. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Thanks dl324 for all that, it's finally clicked.

    As a final example using arbitrary values, if I wanted the LED to be brighter and the current to be 60mA, and it turned out that at this current the voltage drop for the LED was 3V, then I'd do 12V - 3V = 9, 9/0.06 = 150. So I'd use an 150 Ohms resister.

    Going from a 500 Ohms resister has increased the voltage drop of the LED from 2V to 3V, but decreased the voltage drop of the resister so that it always equals 12V.

    Thanks again dl324 and mcgyvr.
     
  10. AnalogKid

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    Aug 1, 2013
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    The D in LED stands for Diode, and like all diodes, an LED is not a resistor. That is, its relationship between voltage and current is not linear. Ideally, a diode has a forward voltage (Vf) that is independent of current. For a Shottkey diode it is around 0.2 v to 0,3 V, for a silicon signal diode it is around o.6 to 0.7 V, and for a simple green LED it is around 2.0 V. For the purpose of this discussion, that voltage does not change no matter what the current is through the diode (until the diode burns up). So the series resistor can be 50, 500, or 5000 ohms and the voltage across the LED still will be 2 V and the voltage across the resistor still will be 10 V. This is why you can select the resistor to set the current through the LED.

    Your question about the resistor "using up" the voltage when placed ahead of the LED does not account for this question: In order for there to be 12 V developed across a resistor, there must be current through it. Where does that current go?

    ak
     
  11. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Thanks AnalogKid,

    That's the bit I'm still unclear with tbh. If you remove the LED from the circuit and just have a 600 Ohms resistor, the current will be 2mA and the resistor will drop 12V.

    When you add the LED, how does the resistor 'know' to drop the current so that the there is enough voltage remaining for the LED. The resistor was dropping 12V on it's own, but now because there is something further down the line that needs 2V, the current is reduced.

    Regards.
     
  12. dl324

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    Mar 30, 2015
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    Your calculations are correct, but don't plan on operating most LEDs at 60mA. Many specs are for 20mA. When you operate at higher currents, forward voltage will increase, color will change, and the LED may burn out.
    This is as you would expect. The voltage drop of the LED varies with current. For a circuit consisting of a 12V supply, a resistor, and an LED; the voltage across the resistor has to be 12V minus the forward voltage of the LED.
     
  13. dl324

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    Mar 30, 2015
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    The voltage across the resistor will be 12V - 2V = 10V. If you add another LED (in series), the voltage across the resistor will now be 8V.

    If you keep adding LEDs, eventually you'll get to the point where none of them will conduct so there will be no voltage drop across the resistor.
     
  14. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    The resistor doesn't know how to drop the current...

    Remember:
    1) There is only so much voltage to go around;
    2) Voltage does not drive an LED.
    3) LED's have a fairly constant voltage drop or Vf.

    You have an LED with Vf = 2V. With a 12V supply, there remains 10V across the resistor.
    If a 600 Ohm resistor has 10 V across it, the current is (10V / 600 Ohm) A of current or 17mA of current.

    No, no, no... The current isn't reduced because there is something further down the line; the voltage is reduced because there is something further down the line. Using Ohm's law, you can calculate the current through the resistor and hence the circuit. Don't mix up your currents and voltages.
     
  15. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    True, I was just using random numbers to make it easier.

    Thanks.
     
  16. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Hi djsfantasi, I didn't mean the current is reduced in the circuit, I mean't that from the previous circuit with just a resistor, the new circuit that now has an LED uses a lower current.

    Thanks to everyone, I understand now how it's calculated and where my initial mistakes were in my original questions. My question now is more what's actually going on physically in the circuit.

    I just can't picture how it works when the resistor is first. When the LED is first it goes from 12V at the battery, a 2V drop across the LED and then 10V across the resistor, which make sense. When it's the other way, there is 12V at the 600 Ohms resistor, and rather than setting the current at 0.02mA and dropping the whole 12V, it sets the current at 0.016mA so that the voltage drop is just 10V, and there is 2V available at the LED.

    What is physically preventing the resistor from setting the current at 0.02mA?

    I'm struggling to word it to be honest so god knows how I expect you to be able to answer it... :confused:

    Regards.
     
    Last edited: Aug 13, 2015
  17. AnalogKid

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    Aug 1, 2013
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    A diode, theoretically, is a constant-voltage device just like a battery. In your case, a 2 V device. So whenever it is in any circuit for any reason, and is forward biased, it has 2 V across it no matter what the current is. Get a grip on that, or this won't make sense. This is the reason a diode is consicered an "active" component rather than a "passive" component like resistors, capacitors, and inductors. Passives are never have a constant voltage or constant current as a part of their physics. they react to both voltage and current. A (perfect) diode does not. In this context, it does not react to current; for any current through it, it allows only 2 V to develop across it. This actually makes it a form of a negative-resistance device, but that's for another day.

    So the resistor has no choice but to react to the various combinations of voltage and current sources and drops in the circuit. In your case it is a simple, single loop. Battery, resistor, and LED are connected in a circle. In the classic water analogy, the battery is a water pump, pushing water around the circle of plumbing. It should be fairly obvious that in a closed loop, 100% of the water goes through each device, because it has nowhere else to go. The wires are fat pipes, the resistor is a section of narrow pipe, and the LED is a pressure valve. If the pump pressure is high enough, the valve opens and lets water circulate around the loop. In this loop, I hope it is clear that the order of the three components does not matter. 100% of the water goes through 100% of the devices.

    The LED circuit is the same. If the battery is less than the 2 V Vf of the diode, there is no electric current. For any battery voltage greater than 2 V, there is 2 V across the LED and whatever remaining voltage is across the resistor because there is nowhere else for a voltage to develop. Kirchhoff's Voltage Law covers this. The sum of all of the voltages around a loop must be zero. So if the battery has a fixed 12 V across it and there are no other batteries in the loop, then the sum of the voltage drops across each other component must add up to 12 V. Since the diode has a fixed 2 V drop, that leaves 10 V for everything else (the resistor). If the diode is not therre, the entire 12 V appears across the resistor. If the 500 ohm resistor is actually five 100 ohm resistors in series, then there is 2 V across each resistor when the LED is in the circuit. AND - it does not matter what order any of the components are in. The LED can be in the middle of the resistor string, and nobody cares.

    ak
     
  18. dl324

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    Mar 30, 2015
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    Solving circuits is often a matter of making assumptions to reduce the number of unknowns and iterating if it gives results that don't make sense.

    In your circuit, you only have 3 elements, so it's easy to solve. The first step is to determine what the forward voltage of the LED is. Since you don't know the current, you have to make an assumption. In this discussion, we've assumed the forward voltage was 2V. Without making that assumption, you can't proceed.

    It doesn't matter whether the LED comes first or last. The circuit is a single loop so all elements see the same current regardless of their position.
    circ.jpg
    If you have the components, you can wire it to prove it to yourself.
     
  19. Robert Smith_1437948150

    Thread Starter Member

    Jul 26, 2015
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    Thanks a lot AnalogKid, appreciate you taking the time to write that.

    Regarding the water analogy, you say "100% of the water goes through each device", with regards to the circuit, is it not that 100% of the voltage is used throughout the circuit, as KVL states, rather than 100% of the voltage goes through each device?

    I do understand the rules and laws, for example it doesn't matter whether the resister is first or last in the circuit, if you have two 250 Ohm resisters they will both drop 5V (in my original example) and the sum of all the voltage drops across the circuit must equal the same as the power source.

    That's how it works and I'm not trying to dispute it, I just wonder what's physically going on in the circuit. Going back to our circuit with a 12V battery and 600 Ohms resistor, as the first electrons flow through the circuit and hit the resistor, it's as if the resistor 'knows' whether there is an LED further along the circuit, if there isn't then it drops 12V and if there is then drop only 10V.

    Your example with five 100 ohm resistors in series and an LED, it's as if each device can see an overview of the circuit to know how much voltage to drop, if you see what I mean? We say the law stating 'The voltage drop across all devices in the circuit must equal the power source', but the devices don't know that, they can't get together and say "Well, we have a 12V battery, an LED that drops 2V and there are five of us, so if we all agree to drop 2V then we won't break Mr Kirchoff's law" haha. What's physically happening in the circuit that makes the law true?

    Again, I understand that's the way it works and I should just accept it, I just find it interesting. :)

    Regards.
     
    Last edited: Aug 13, 2015
  20. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    In order...

    No, voltage does not go through anything. Voltage pushes current through conductors. Voltage is pressure, which is why it's analogy is a pump. Water is the analog of current. In a DC circuit, electrons physically move through each component in series just like water through pipes.

    The resistor "knows" only one thing, the voltage across its terminals. It allows a current to flow through it that is proportional to that voltage, and the constant of proportionality is called "resistance". It doesn't know there is an LED in the circuit. For all it knows, there is a 1000 V battery and 495 LEDs in series with it. It sees 10 V, and acts accordingly. And since the resistor is in series with one or two constant voltage devices depending on the circuit, from the resistor's point of view the circuit with the LED looks exactly like a 10 V battery, nothing more.

    It is the current that is "happening". Each individual device, be it resistor or LED, responds to the current through it. If anything can "see an overview of the circuit", it is the battery. Reworking your statement, the battery sees the total impedance and "knows" how much current to push. More correctly, the battery presents an infinite source of electrons at a fixed pressure, and the impedances in the circuit determines the rate at which they flow. Going back to the five 100 ohm resistors, the circuit behaves exactly the same if the five resistors are not all the same value. If the resistors are 50, 75, 100, 125, and 150 ohms, the total resistance is 500 ohms and the loop current is 20 mA. All of the current goes through all of the components all of the time. That sentence allows tremendous flexibility in circuit design.

    ak
     
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