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I am trying to make a simple circuit for a nametag for kids to solder at a workshop. The LED I was planning to use is 151033BS03000 it has a graph of forward current vs forward voltage. What is this? Don't I determine the current of the LED by my resistor value?

graph

circuit

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    \$\begingroup\$ Why do you think the LED doesnt impact the behaviour of the circuit? \$\endgroup\$
    – Jun Seo-He
    Commented May 5, 2022 at 18:54
  • \$\begingroup\$ Both things are right. Have you ever applied a resistor load line to a chart like that? \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 18:54
  • \$\begingroup\$ @jonk I am confused by what you mean here \$\endgroup\$
    – Feynman137
    Commented May 5, 2022 at 18:58
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    \$\begingroup\$ @Feynman137, google the term "load line" and most of the examples you find will be about setting the bias point of a diode (whether it's an LED or regular silicon diode doesn't change the method) \$\endgroup\$
    – The Photon
    Commented May 5, 2022 at 18:59
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    \$\begingroup\$ CR1220 batteries have 100-300 ohms internal resistance, so in addition to that 300 ohm resistor in your drawing you have another similar sized resistor built into the battery. This will make calculating the current through the LED even more complicated, especially if you don't know the internal resistance of the specific battery you're using. \$\endgroup\$ Commented May 5, 2022 at 19:11

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The resistor limits current and the LED curve is right

When I mentioned that both are right, I mean it this way:

enter image description here

Assuming, for example, that your voltage source is a low-impedance \$3.3\:\text{V}\$ (note the circle on the x-axis at that point), then the green line represents a \$330\:\Omega\$ resistor and the light blue line represents a \$180\:\Omega\$ resistor.

Let's look at where these lines intersect with the LED curve:

  • Green: approximately \$2.5\:\text{V}\$ and \$2.5\:\text{mA}\$ -- so the \$330\:\Omega\$ resistor drops near \$825\:\text{mV}\$, which is just enough to make up the difference to reach \$3.3\:\text{V}\$ given that the LED is running at \$2.5\:\text{V}\$.
  • Light Blue: approximately \$2.6\:\text{V}\$ and \$4\:\text{mA}\$ -- so the \$180\:\Omega\$ resistor drops near \$720\:\text{mV}\$, which is just enough to make up the difference to reach \$3.3\:\text{V}\$ given that the LED is running at \$2.6\:\text{V}\$.

That's what a load line does for you. Makes it easy to find the operating point.

And you can see that, yes, the resistor limits the current, and, yes, the diode curve is correct, as well.

Creating a load line

To draw the load line for any resistor, just start at the \$V_{_\text{CC}}=3.3\:\text{V}\$ point on the x-axis (or, whatever your \$V_{_\text{CC}}\$ value happens to be) and then compute \$\frac{V_{_\text{CC}}}{R}\$ to pick out the y-axis value and mark that position there. Then get out a ruler and draw a line between these two points.

CR1220 as a power source

Tony does bring up a very important point in comments below my answer, given that you've specified a CR1220 battery rather than a low-impedance power supply. These batteries do have significant internal resistance (one that varies over time, as well) just as Tony wrote. He's quite right to suggest that the above analysis would only apply with a low-impedance power supply.

The internal resistance of the CR1220 might be as high as around 1000 Ohms, skimming a datasheet or two and doing some guess work. Even assuming that the datasheets are overly conservative, it's still going to be in the hundreds of Ohms.

Let's assume that the CR1220 has an internal resistance of \$470\:\Omega\$, just to pick something, and that you don't use an external resistor. Then the new chart might look like this:

enter image description here

I've added the red load line for the assumed internal resistance of the CR1220, above. (And it's nominal voltage starting point, which is \$3.0\:\text{V}\$.) Here, you can see that it intercepts the LED curve at somewhere between \$1\:\text{mA}\$ and \$2\:\text{mA}\$ and an LED voltage perhaps a little less than \$2.4\:\text{V}\$.

It may very well be that an external resistor isn't needed. If you supply one, you could consider just shorting it out to see what happens. (Given the CR1220, I would not be worried about attempting that experiment.)

Summary

But the earlier point above remains. Regardless of what's incorporated to make them, load lines make it quite easy in the end to work out the operating point when facing non-linear curves such as those presented by LEDs.

P.S. Looking at the LED curve, the function that seems to approximate it well is a simplified Shockley diode model of about \$I=1.96\times 10^{-10}\left(e^{^\frac{V}{0.153}}-1\right)\$. (That sets the emission coefficient to \$\eta\approx 5.9\$ and limits the bulk impedance to about \$2\:\Omega\$, I think.)

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    \$\begingroup\$ You did not account for the battery resistance which is significant! And the bulk resistance tolerance for Vf on the LEDS can also be significant changing the slope of the curve. \$\endgroup\$ Commented May 5, 2022 at 19:32
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    \$\begingroup\$ @TonyStewartEE75 The CR1220? \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 19:34
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    \$\begingroup\$ @TonyStewartEE75 Thanks, Tony, for the catch. I've updated my answer to incorporate your comments here. \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 19:41
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    \$\begingroup\$ @Feynman137 Yes. Something like that would be true. The battery resistance isn't exactly fixed, though. And it isn't the same between any two of them, either. So, if as Tony suggests, the resistance of the CR1220 is high (and I think it is high and he was right in this case to take note) enough, then the variations between batteries will mean variations between "this unit" and "that unit." Perhaps enough to see, even. For toys, that's fine, I suppose. \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 22:59
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    \$\begingroup\$ @Feynman137 The internal resistance of the CR1220 might be as high as around 1000 Ohms, skimming a datasheet or two and doing some guess work. Even assuming that the datasheets are overly conservative, it's still going to be in the hundreds of Ohms. So if you do plan on that particular battery, Tony was right to call my attention to that fact. You may not need a resistor, at all. Worth trying out. If you do try it, let us know what you get in terms of measuring the resulting current. Please! \$\endgroup\$
    – jonk
    Commented May 6, 2022 at 1:56
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All batteries have ESR resistance and LEDs can be modelled with an ESR of 1/2Pmax minimum = 1/2(0.1W) = 5 Ohms Vf = 2.6+ I*5 ohms, when using rated current of 20 mA. But the tolerance on that can be much higher causing Vf max to be 3.6V or deltaV... 1V/20mA = 50 Ohms !! But these tend to be outliers. My experience with 65 mW LED is the diode Rs for blue /white tends to be around 12 to 15 ohms @ 20 mA and 10x higher ESR at 2 to 3 mA.

If your target current is 20 mA The CR1220 will not be able to sustain this for long due to its mAh is rated for a 20h drain down to 2V. This means you might get 0.5h bright then dimming rapidly and then drawing less current so it is dim for a day.

Choosing a brighter LED with 16000 mcd is wiser and drive it at 3 mA. With Vf=2.6V nominal. The variation in LED Vf is not the knee threshold but the variation in ESR or Rs or incremental resistance due to manufacturing tolerances. These can be binned or you can get lucky. I happen to have lots of binned LEDs > 16Cd.

It is not possible to calculate the ideal series R unless you have expectations on battery power vs operation time due to the large variables of battery ESR, LED ESR and Iv luminous intensity ratings of the LED along with your expectations as a backlight to a nametag.

White would be brighter with the phosphor over blue as the eyes are more sensitive. I have many bags of 200 white LEDs in the 20,000 mcd 30 deg range.

But if you want to use what you have, I would suggest 3 to 5 mA for longer duration with an avearge Vbat of 2.8 and range of LEDs from 2.6 @ 5mA to 2.5V @ 2.5 mA (Nominal) thus 2.8-2.5V = 300 mV / 3 mA = 100 Ohms . With a fresh battery 3V-2.6V= 400 mV/100 = 4 mA . This is almost equivalent to a 3V/4mA= 750 ohm R load which you can look up on battery capacity to 2V which is not useable so take say 3/4 of that capacity for a CR1020 of 15 to 20 mAh so it might be bright for 5h then dim for another 20h. https://ca-en.alliedelec.com/m/d/6ec5c9a8833414eb34f877f612fc890f.pdf

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    \$\begingroup\$ This was helpful, I was assuming battery life would be short. I was told by the teacher that 3hrs would be a good mark to shoot for... I think I am assuming this correctly but are you telling me that I should use a 100 ohm resistor to achieve 3-4mA LED operation? Because the dV is actually 300-400mV not 3V as I assumed.. \$\endgroup\$
    – Feynman137
    Commented May 5, 2022 at 20:05
  • \$\begingroup\$ As mentioned in the first paragraph and commented by @user1850479 the battery will most likely have 100+ ohm resistance. So in light of this is there even the need for a resistor? \$\endgroup\$
    – Feynman137
    Commented May 5, 2022 at 21:49
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    \$\begingroup\$ NO if you want it bright for a couple hrs then dim. YES if you want a brighter LED and longer life \$\endgroup\$ Commented May 5, 2022 at 22:39
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    \$\begingroup\$ But the LED across the battery edge is easy to do. It just kills the battery at half capacity in a couple hours then decays to weak and weaker down to <2.5V in a day . So you get half of the lost capacity back by limiting the current with 100R \$\endgroup\$ Commented May 5, 2022 at 22:44
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It is the IV curve of a LED.Due to the fact that the LED is a semiconducting device you cannot use Ohm's law to determine the current inside a LED by its voltage drop.

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    \$\begingroup\$ You can develop a closed solution using standard laws, such as Ohm's law. See here for a case that would apply here. It's just that the mathematics is non-linear, is all. The usual rules still apply. It's just that the math gets a little more complicated, is all. \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 18:59
  • \$\begingroup\$ I agree if the current is very small using the Taylor series e^x -1 = x.But it is still a approximation. \$\endgroup\$
    – Jun Seo-He
    Commented May 5, 2022 at 19:04
  • \$\begingroup\$ No, the Lambert W function develops an exact solution. Not an approximation. \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 19:06
  • \$\begingroup\$ what is a Lambert W function? \$\endgroup\$
    – Jun Seo-He
    Commented May 5, 2022 at 19:07
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    \$\begingroup\$ I provided a link to my answer in my first comment. Go there, where you can see the function applied to this very situation. I also reference an academic paper on the topic there. You can also look up the Lambert W function at the Wolfram site. \$\endgroup\$
    – jonk
    Commented May 5, 2022 at 19:09
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LEDs drop a voltage when you pass a current through them, and that's given by the curve you have reproduced. The curve is for the LED alone.

The voltage across your resistor will be the supply voltage from your CR1220 minus the LED's voltage. So the LED affects the current flowing through the circuit. But the current through the circuit also affects the voltage across the LED. Attempting to set the current exactly can be quite tricky, but most people don't bother too much as LEDs work over a very wide range of currents.

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Don't I determine the current of the LED by my resistor value?

I assume you're thinking about the simplified model where you assume that the voltage drop across the LED is some constant, which you can subtract from the supply voltage to get the voltage that you'll need to drop across the resistor: $$V_{\rm resistor} = V_{\rm supply} - V_{\rm LED}$$ and then use Ohm's law to calculate the resistance \$R\$ of the resistor you'll need to get the current \$I\$ that you want to flow in your circuit: $$R = \frac{V_{\rm resistor}}{I}.$$

As you've noticed, this simplified model isn't exactly true, because the voltage drop across the LED does in fact vary with current, and the graph you've found tells you exactly how much it varies. But as you can tell from the graph, it doesn't vary much, so in practice you usually can just approximate the voltage drop for your LED as, say, 2.5 V:

Forward Current vs. Forward Voltage graph for the 151033BS03000 LED, taken from the question above, with a vertical line at 2.5 volts superimposed

Here the red dotted vertical line is at a constant 2.5 V; comparing it with the blue line showing the actual voltage drop vs. current, you can see that it stays between 2.0 and 3.0 volts for the entire range of currents for which the LED is rated.

If you use the graphical method shown in jonk's answer to calculate the actual current through your circuit for a particular LED and resistor, you'll note that it's typically pretty close to the value calculated with the simplified assumption of a constant voltage drop. In fact you can estimate the error visually from the graph, just by applying the intersection method but using the vertical line at 2.5 V (or whatever voltage you prefer to use in your approximation) instead of the true current vs. voltage graph.

Typically the biggest error occurs for low resistor values, where the current vs. voltage line for the resistor is closest to vertical. But those are also the cases where other sources of error (such as the internal resistance of the battery and variations in the LED current vs. voltage characteristics) are most significant, so you're unlikely to get very accurate results in those cases anyway, at least not without actually measuring the characteristics of the components you're using.


In any case, as long as you make sure that the approximate constant voltage drop you're using in your calculation is less than the actual LED voltage drop at the current you're aiming for, you can be confident that the approximation will lead you to overestimate the resistor value, which is fine — all it'll do is make the actual current slightly lower, and thus the LED slightly less bright, than if you'd used the true current vs. voltage graph to calculate the resistor value. Unlike underestimating the resistance, this cannot damage either the LED, the resistor or your battery.

(And of course, as also noted in other answers and comments already, for this particular circuit the internal resistance of the battery is likely to be high enough that the LED will most likely be fine even if you leave the resistor out entirely!)

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It is amazing how it is possible to show so many practical considerations for the humble LED "flashlight". Obviously, in this world, a lot of complexities are hidden behind every simple thing…

I will contribute by explaining the idea of ​​this "grapho-analytical" representation, usually called "load line".

The LED … has a graph of forward current vs forward voltage. What is this?

IVcurve. The common name for this graph is "IV curve". It graphically shows the relationship between the voltage across and the current through a 2-terminal element (LED here). As you can see, it is nonlinear.

The rest of the circuit consists of resistor(s) and a voltage source in series. The IV curve of this network is a line that is shifted to the right with the value of the source voltage and inclined to the left with the value of the total resistance.

Only to note that its conventional name "load line'' is misleading. It is rather a "generator (source) line" since it is associated with the source. More precisely speaking, the LED IV curve is the "load curve" here since the LED is the useful load.

Don't I determine the current of the LED by my resistor value?

Graphical solution. To solve this circuit graphically, we superimpose the two IV curves on the same coordinate system (because the voltage across and the current through the two parts are the same). The intersection (operating) point is the solution.

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