How to solve for current given resistor value for LED circuit?

Question

So I have the following LED circuit:

I know how to find the forward voltage and current from the datasheet of the LED and how to then solve for the appropriate value of \$R\$.

However, what I really want to do is fix the resistor value at \$R=100\$ (I'm just picking a resistor value that exists in a kit I have) and solve for \$I_F\$:

The only way I can currently think of to solve this is:

$$V_F+V_R=V_{CC}-V_{OL}$$ $$V_F+I_FR=V_{CC}-V_{OL}$$ $$V_F+100I_F=4.3$$

I found \$(V_F,I_F)=\{(2.3, 20), (2.8,15)\}\$ to be 2 arbitrary solutions to the equation. Then I graphed a line on the VI plot to determine all possible solutions to the equation. Finally I found the intersection of the line and the VI plot to determine what was the true solution to the equation is.

From this it looks like \$V_F=2.1\$ and \$I_F=22\$.

However, this was a lot of work for such a basic circuit. Is there an easier way to solve for or at least nicely estimate what \$I_F\$ would be given \$R\$? The value of the current matters a lot to me because I need to make sure I don't fry my microcontroller!

Answer 1

Is there an easier way to solve for or at least nicely estimate what \$I_F\$ ... because I need to make sure I don't fry my microcontroller!

Two easy solutions:

• Choose a target current that's well below the limits of the micro or LED, like 5 mA instead of 20 mA, then just assume \$V_F\$ is close to its nominal value (say, 2.2 V in this example). If the LED doesn't have to be visible in very bright light, 5 mA should be adequate (or you should be able to find a different LED that is adequately visible with 5 mA)

or

• Use a transistor buffer with much more than 20 mA capability to eliminate risk to the micro.

Before you try to calculate the diode current any more precisely, you should consider that the diode forward voltage is likely to vary by 100 mV or more as its temperature changes.

Answer 2

You're on the right track. You can simply overlay the load-line for your chosen resistor value onto the LED IV curve. Here's one I did:

Figure 1. Various resistor and LED load-lines for a 5 V supply. The intersection of a resistance and IV curve gives the operating point. Source: LEDnique.com.

I'm still working on the curves which are somewhat off. (The blue and white, for example, should be much closer.)

There are some worked examples on the linked page.

Answer 3

You are wasting your time, assuming this is a simple visual indicator, for several reasons.

The LED forward voltage vs. current curve is not guaranteed, only usually a fixed point, or two. It varies from unit-to-unit and with temperature. You might want to change manufacturers in the future and the chemistry or internal resistance might be a bit different.

The visual brightness does not change all that much with current due to logarithmic eye response- a good LED at 5mA can be almost too bright, and a old school one at 20mA can be meh.

GPIO minimum drop is usually not specified at all, at any currents, just the maximum.

You should be nowhere near the current level that would "fry" your GPIO. Otherwise you are likely to run into reliability issues.

While there is a geek appeal in fitting a curve to the LED and to the GPIO output IV curves and hitting some arbitrary current target within a fraction of a percent, it's not really that useful a trick. If it requires that much effort it will likely be too variable.

Just pick a value that will give enough brightness and has plenty of margin.