LED current limiting resistor and Ohm's Law

Question

Yesterday I was trying to figure out how many ohms of resistor should I be using for my LED, and found this post right here in the forum.

The formula states that one should subtract the voltage drop of the LED from Vcc, and then divide it by the target current (A), in the example, it was (5-3.4)/0.005 = 320 ohms resistor should be used.

My question is this: why the resistance is calculated not with whole 5V but only fraction of it? Suppose the resistance of entire circuit would be at the lowest level of 320 ohms if LED is 0 ohms, and if I used Ohm's Law on this circuit: 5/320 = 0.015625 (A), which is way off the target current I was hoping for.

It's like a mystery for me, and I hope someone could really explain it to me.

EDIT: Another way to put it: If one imagine a blackbox payload consumes 5V, in order to supply 5mA through the circuit, the blackbox needs to be 5/0.005 = 1000 ohms. And I know the blackbox is composed of 2 components: the LED and resistor, and the sum of their resistance must be 1000 ohms. Let R of LED is x, and R of resistor is y, they have the relationship x+y = 1000. That means whatever resistor I put in it, the R of LED automatically changes to a fitting value so their sum is 1000, is that possible?

Doesn't one need to know the resistance of LED, or the LED's resistance is really dynamic?

Answer 1

All LEDs can be modelled as zener diodes with a colour/substrate specific forward voltage Vf and a series resistive Rs, where they combine both to give the Vf at rated current. Rs tends to be small so you can neglect it for approximations of adding current limiting series resistors. (see below)

Therefore the current is non-linear and proportional to the voltage difference between the supply and the Vf drop at desired current.

Batteries with low voltage variation are ideal such as Lithium primary cells. Most White flashlights using 3V per LED use these without series resistors as the Li cell is also 3V. However they may be specify a sorted bin of LED's to achieve this.

My Rule of Thumb is to string arrays of LED's such that the voltage difference is ~1V for the current limiting Resistor for a fixed regulator. If the supply range has a wide range, e.g. 10 ~15V then a constant current sink circuit is best.

Additional Info

For more accuracy over a wider range of currents, you can determine the Rs value from the specsheets for a given temperature. The Vf forward voltage also is a function of temperature which affects the results slightly. THe Rs of LED's is much lower than the dynamic Rs of Zeners using silicon junctions.

• 20mA HB devices are <20 ohms.
• 300mA HB devices are < 2 ohms.
• 1Amp power modules are ~ 0.3 ohm.
• Rs for LED arrays , add in series, and divide in parallel.
• Old technology LEDs were much higher Rs values.
• Rs will reduce as the current increases but you can approximate it at the 10% of rated current value and extrapolate if the device stays at constant temp.
• Because of the Shockley effect with voltage variationmyou can actually calculate the junction temperature from the voltage drop of a calibrated LED.

Answer 2

Because not all of the 5V voltage is dropped across the resistor. Some is dropped across the LED ... if they used 3.4V in the example it must have been a blue or white LED ... leaving only 1.6V across the resistor.

Put a red LED in there (which drops 1.7V) and it would leave 3.3V across the resistor. So what resistor would you need to get 5ma now?

Answer 3

Kirkhoff's current law tells us that the current through devices in series is the same through all the devices. Whatever current is flowing across the resistor is the same current that is flowing across the diode. Therefore, we just need to determine the current across the resistor, which tells us what the current is across the diode also.

Since the voltage across the resistor is, in fact, approximately the supply voltage minus the voltage dropped by the diode, and we are applying Ohm's law to the resistor, we have to use that voltage.

We cannot use full supply voltage present across both the resistor and the diode to calculate the current across just the resistor.

Actually, maybe we can. There are situations in which you can obtain a useful approximation by assuming that the voltage drop across a diode junction is zero. Supose the supply voltage is something fairly high, like 30V. If you pretend that a diode does not drop any voltage when it conducts, i.e. it drops 0V instead of 1.2V or whatever, the error in the calculation is 4%. Now consider that some components have tolerances as much as 5%. In some circuits, it is good enough to use a model of a diode as a simple one-way valve. That is the first approximation. For instance, when we build a bridge full-wave rectifier in a power supply, we don't always give much thought to the voltage drop of the diodes. The second approximation of the diode is to model it as one-way valve with an offset voltage (like 0.7 for a silicon diode). That covers the majority of the remaining scenarios which occur. Then, the next model is something more exact, which models the diode's voltage-current curve, leakage currents and whatnot.

But there is a difference between making a mistake, and making a justified choice of model.