LED circuit: Need to understand how to calculate

Question

I'm facing to following problem: I need to turn on 5 LEDs (2 V, 30 mA) in series from a 12 VDC power supply using a MOSFET (2N7000G in this case). To calculate the resistor this formula was used:

12 V (power supply) - 5 x 2 V (series LEDS) = 2 V for resistor and MOSFET

From the datasheet: R_DS(on) is 6 Ω: 6 Ω x 0.03 mA = 0.18 V voltage drop across the MOSFET.

2 V (from resistor and MOSFET) - 0.18 V (across MOSFET) = 1.82 V for resistor

U=IxR -> R=U/I -> R = 1.82/ 0.03 -> R = 60 Ω

If the circuit is designed with a 60 Ω resistor, the LEDs are energized only with 25 mA (even if 60 Ω should provide 30 mA from my calculations). Also, the voltage across the LEDs is increased to 2.07 V each (in the LED simulation the properties used are 2 V and 30 mA; I double checked all of them).

My opinion is that 0.07 V makes my calculation wrong, but why does this happen? What did I do wrong?

Answer 1

The assumption you make about the LED forward voltage being constant is not right. The simulation is modeling the LED, not what you're assuming its forward drop to be.

The LED datasheet forward voltage is usually stated at the nominal forward current (e.g., 20mA). However, the IR drop across the LED, when forward biased, will increase with current, as the diode itself has some internal resistance. The simulation is modeling this, based on some form of the Shockley Diode Equation.

This answer discusses the LED I-V relationship in further detail: Does a resistor in a simple LED circuit actually make it more efficient or does the resistor just burn off the energy?

Even then, that LED forward drop given in the datasheet is nominal. It can vary from LED to LED with process, and also vary with operating temperature.

You can adjust for this by modifying the series resistor. But, your resistor overhead voltage as a percentage of the total IR drop is kind of small, so repeatability will not be so great.

Note however that the perceived brightness difference between 25mA and 30mA will not be very much for a typical LED. If this is your only string to illuminate, it may very well be 'good enough' to just reduce the value of the resistor a bit to achieve your target 30mA current, assuming your LEDs are rated for that higher current.

But if having an exact, repeatable LED current (and therefore, brightness) matters to you, you need a current regulator to compensate for the non-ideal forward voltage behavior.

As it stands, you have enough overhead voltage (about 2V) that you could use a 2-transistor current limiter. With a small modification the current limiter can also serve as your low-side switch.

Example (simulate it here):

Answer 2

lousy regulation with a resistor in this case

Resistors make terrible current regulators for long chains of LEDs. Escpecially so when you leave only a small left-over voltage to work with. The LED current variation you will expect is:

$$\%I_{_\text{LED}}=\%V_{_\text{LED}}\cdot\frac1{\frac{V_{_\text{CC}}}{V_{_\text{LED}}}-1}$$

So, in your case this would be \$\%I_{_\text{LED}}=-5\cdot \%V_{_\text{LED}}\$.

LED voltages commonly vary by 10%. So this means in your case that could mean 50% variation in the current. (In the opposite direction, so if the LED string voltage increased by 10%, from \$10\:\text{V}\$ to \$11\:\text{V}\$, then the current would decrease by 50%.

This is really very bad for regulation.

On the plus side, using 5 LEDs tends to make the statistical variations a little better as variations of one LED have a chance of being compensated by variations of another one. But it is still quite risky with only 5 LEDs in the string.

use a current source/sink

With that much overhead, though, you can follow the advice given by hacktastical and use that technique for providing a decent current sink/source for the LED chain.

how to design a current sink

By the way, if you are looking to actually learn how to design one of those rather than just being given one -- one that will work in a predictable way over operating temperature and device variations -- then see this link to an EESE answer on this topic.