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Intro:

I feel like there is one more thing that I really need to get a grasp of.

Let’s say I have a circuit with a 5V battery and a LED that has a 3V forward voltage. Then before the LED is a resistor with say 100 ohms, the resistor should drop all of the voltage, should it not?

Math:

I=V/R, I=5/100=0.05A

Then we can figure out the the voltage drop across the LED,

V(of the resistor)=0.05A*100(Resistance of the resistor)=5V

Explanation: So, my math shows that the voltage drops completely along the resistor. But if you build this circuit the LED still lights up.

Question: So How/Why Does This Happen?

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  • \$\begingroup\$ Because it doesn't have all the resistance in the circuit. \$\endgroup\$ – user207421 Apr 19 at 23:38
  • \$\begingroup\$ resistor should drop all of the voltage, should it not? .... so that there is no voltage across the LED? ..... the only way to do that is to short-circuit the LED, therefore taking it out of the circuit \$\endgroup\$ – jsotola Apr 20 at 0:36
  • \$\begingroup\$ @jsotola That was part of the question... \$\endgroup\$ – BeastCoder2 Apr 20 at 0:52
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    \$\begingroup\$ you asked three questions .... first two are making an assumption that the resistor should drop all of the voltage ..... better question would be one without assumptions, such as Does the resistor drop all of the voltage in a series circuit? \$\endgroup\$ – jsotola Apr 20 at 1:00
  • \$\begingroup\$ @jsotola That is the questions tilte rewritten? \$\endgroup\$ – BeastCoder2 Apr 20 at 1:18
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Your first equation assumes that the full voltage is across the resistor. Then your second equation finds out that the full voltage is across the resistor.

What's really going on is that within a pretty broad range of currents, the voltage drop across your LED is roughly 3V. So you assume that 3V drop. Then you find out that the voltage across the resistor is \$V_r = 5\mathrm{V} - 3\mathrm{V} = 2\mathrm{V}\$. Then you solve for current: \$I = \frac{V_r}{100\Omega} = 20\mathrm{mA}\$.

Edit: then you make sure that the resulting current is within that broad range where you can expect a 3V drop -- there's some (really) lower limit where more current will flow through parasitic parallel resistances than through the junction, and a high limit where the diode will burn up.

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  • \$\begingroup\$ That is valid but I am going to delete the question, because my assumption was that the circuit does work but it actually doesn't so it turns out I was misunderstanding anything. But you did make a slight error in you answer (totally fine), I said that the resistor was before the LED, and that circuit actually doesn't work, but the way you made it (resistor after LED) does work, sorry for the confusion! \$\endgroup\$ – BeastCoder2 Apr 19 at 23:17
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    \$\begingroup\$ It doesn't matter whether the resistor is "before" or "after" the LED. The circuit works the same either way. Search the site for the dozens of times that has been asked about before. \$\endgroup\$ – The Photon Apr 19 at 23:18
  • \$\begingroup\$ But I just tried it on the arduino uno, and it only works if the resistor is behind it? \$\endgroup\$ – BeastCoder2 Apr 19 at 23:20
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    \$\begingroup\$ See here. Probably you did something else wrong for the configuration that didn't work. \$\endgroup\$ – The Photon Apr 19 at 23:20
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    \$\begingroup\$ Also here \$\endgroup\$ – The Photon Apr 19 at 23:22
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If you put 0 V across an LED, then 0 amps will flow through it. If it's in series with the resistor, though, it must have the same current as the resistor. The current can't be both 50 mA and 0 at the same time. So this is obviously not the right answer.

Like any diode, for a wide range of currents, the voltage will be pretty close to the specified forward voltage. In your case that's 3 V. So there's 2 V left for the resistor. 2 V / 100 ohms = 20 mA, and this is pretty close to what you'll measure.

The difference will be because the LED forward voltage isn't always exactly 3 V. It will change due to operating temperature, and increase slightly as the forward current increases.

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  • \$\begingroup\$ Check out the comment I left for the other answer, sorry for the confusion! \$\endgroup\$ – BeastCoder2 Apr 19 at 23:18
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Here is a graphical explanation

schematic

simulate this circuit – Schematic created using CircuitLab

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First off, on the practical issue of how to set up an LED, let's not forget that this is usually a straightforward matter. The original question is why doesn't all the voltage drop over the resistor. And that is a deeper question than what resistance should be put in series.

So, as to the practical issue. Current running through an LED will induce a voltage across the device. LEDs have the behavior that the voltage developed across the device stays about the same regardless of current, roughly about 2 Volts. So they don't as like a resistor, where voltage is directly proportional to current. The rules of circuit analysis is that voltage drops across the circuit equal to the supply voltage. Another rule of thumb is run your LEDs at 10 mA. Thus, a resistor and LED in series with a 5V supply will have 2V across the LED and the Voltage across the resistor will have to be 3V. (3+2=5). Since there will be 3V across the resistor and you want 10 mA running through the circuit, E=IR; 3=0.010R; R=300. So with 5 volt circuits you see the experimenter with 330 1/8 Watt resistors, since 330 is a standard resistor value. If you go with more modern 3-3.2 V logic, you will want 100 Ohm resistors. The 10 mA drive current is selected because it gives LEDs are usually bright enough at that drive and 10 mA is way below the absolute maximum of 20mA, which is how a lot of LEDs are rated. This will work any of the colored LEDs you want.

For the deeper question of why doesn't all the voltage over the resistor, you may have heard someone tell you that the LED acts like it has no resistance, just develops a voltage. Then, knowing that zero resistance has no voltage drop I could see that you'd think all the voltage should drop over the resistor.

An example of a device that has a voltage without resistance is a capacitor. When there is a charge on the capacitor there will be a voltage. To limit the speed of the current into the capacitor a resistor can be placed in series. Of course, as the current flow there is a voltage across the resistor, but there is a voltage but no resistance acorss the capacitor. As the current charges the capacitor the voltage drop across the capacitor increases and the current drops until the capacitor reaches the voltage of the current supply. This is the basic RC circuit that you will learn about some time.

The nature of the LED is to maintain a roughly constant voltage regardless of current. This is true of any diode which is forward biased and with current flowing. This forward voltage drop is a relative constant regardless of current. The LED is special because it glows! But is just a forward biased diode in usual use. If you wanted you could build a rectifier out of one, but they aren't too useful as a rectifier due to the high voltage drop and limited current flow. That doesn't mean there is no resistance. After all, if an LED has 2V over it and it is passing 10mA, it is acting as if it has a resistance of 200 Ohms. t is interesting that if you drop the current to 5 mA the voltage is still about 2V, so the LED is acting like a 400 Ohm resistor. That's right, the LED acts like a variable resistor serving to maintain voltage over the device at about 2V.

There have been some simplifications in the above discussion. In truth, the forward voltage increases slightly as current increases. At a current of 0.1mA forward voltage is 1.7V, 1mA/1.8V, 10mA/1.9V in one datasheet I pulled up. Also, the forward voltage can vary quite a bit, from 2V to up to 3V in some LEDs even with the same part number. The rule of thumb of driving the LED at 1/2 to 3/4 maximum current generally gives satisfactory results and long life.

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  • \$\begingroup\$ It wont, right? I'll delete this later so no one else can see but doesn't the LED also have tolerance, so it wont? \$\endgroup\$ – BeastCoder2 Apr 19 at 23:38
  • \$\begingroup\$ There is a tolerance for both devices. If the resistor is 90 ohms would the resulting current burn out an LED that might have a current limit a little lower than average? You figure out the worst case scenario and the check the math. The LED will have a maximum power dissipation of something +/- somevariance. \$\endgroup\$ – kd4ttc Apr 20 at 4:15
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    \$\begingroup\$ I am a beginner so I fear that I dont know the math... \$\endgroup\$ – BeastCoder2 Apr 20 at 11:52
  • \$\begingroup\$ I edited the response to include practical and deeper explanations. I was looking at some datasheets, and the forward voltage min, typical and max is usually listed, but Current is generally just a maximum permitted. If you are driving the LED at half maximum current the variability of the resistor will not be important. \$\endgroup\$ – kd4ttc Apr 20 at 20:57

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