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I know that if an LEDs are to be connected in a parallel circuit, a resistor must be added to limit the current rate flowing through them. But let's assume that I have blue LEDs with rated forward current of 40 mA and these are connected parallel to 3V power supply. IV curvature shows that for 40 mA it is required forward voltage of 3V, which means that in such a case there is no need for additional resistors, right?

Typical IV curves for various colours of LEDs.

Source: LEDnique - IV curves

Also, if LED have nearly zero resistance, then in case if we add resistor in series with an LED, there will be no voltage drop across LED (all voltage will drop across resistor). As result you cannot reach voltage drop across LED that is greater than forward voltage and current won't flow through it, isn't it so?

For low-voltage applications a single resistor per LED prevents current hogging by the LED with the lowest Vf

Source: LEDnique - Parallel LEDs – the problem

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  • \$\begingroup\$ In theory nobody can stop you from omitting the current limiting resistors if you have a voltage source that is precise enough to be set at the output current of 40mA that you measure from the graph. However bear in mind that optical output of a LED varies quite allot with current, and from diode to diode there is variation. You could end up with LEDs from different manufactured batches with large differences in the optical and V-I characteristics, enough so that you could visibly see one LED brighter than another... \$\endgroup\$
    – citizen
    Sep 18, 2023 at 11:46
  • \$\begingroup\$ Additionally adding a larger resistance (say 330 Ohms), than that of the internal LED resistance (in its active state maybe a few Ohms) removes some of the natural variations in internal LED resistance, from one diode to another diode (again due to manufacturing tolerances), and this provides a more constant current. Don't forget, it's the bias current that sets the optical output power of a photo-diode and it's this you want to keep at a known value across all LEDs ... much harder to achieve when LEDs are in parallel, and worse if you don't provide a bias resistance ... \$\endgroup\$
    – citizen
    Sep 18, 2023 at 11:52
  • \$\begingroup\$ If you're buying LEDs in commercial-scale quantities (many 1000's), then you can ensure that they all come from the same batch, and are likely to have substantially similar characteristics. If you then also mount them on a PCB which spreads heat well (like an aluminum-core PCB) then you know they'll all be at (roughly) the same temperature. In such cases you can connect multiple LEDs in parallel directly and have them share whatever current-limiting mechanism you want to use (simple resistor, constant-current supply, ...). \$\endgroup\$
    – brhans
    Sep 18, 2023 at 19:58
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    \$\begingroup\$ "no voltage drop across LED (all voltage will drop across resistor)" is a fundamental misunderstanding of what happens when you mix passive and active components. The voltage drop across a resistor is given by Ohm's Law. There is no voltage-divider law. The voltage-divider calculation comes from Kirchoff's Laws applied to two resistors, both obeying Ohm's Law. The LED isn't covered by Ohm's Law. \$\endgroup\$
    – Ben Voigt
    Sep 18, 2023 at 20:54

2 Answers 2

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IV curvature shows that for 40 mA it is required forward voltage of 3V, which means that in such a case there is no need for additional resistors, right?

That is correct, but temperature changes or manufacturing tolerances cause that curve to differ and the curve is steep. So a small increase in ambient temperature would cause more current -> more internal heating -> ...

Also, if LED have nearly zero resistance, then in case if we add resistor in series with an LED, there will be no voltage drop across LED (all voltage will drop across resistor). As result you cannot reach voltage drop across LED that is greater than forward voltage and current won't flow through it, isn't it so?

LED does have a voltage drop, but the drop doesn't behave as a resistive load. Part of the behaviour can be considered as a resistance. Particularly the linear part of the IV curve.

So you need to pick a value to on the IV curve, use that as the voltage drop, size the external resistor and as a result you'll be somewhere near to what you wanted, but temperature change and manufacturing tolerances didn't let you hit the mark exactly.

If there was no voltage, there would be no power, so the LED couldn't radiate any energy (light up).

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The steep I-V curve of LED indicates that it is like a voltage source with low internal resistance. For proper operation, you should power the LED with a current source. Multiple LEDs should either be connected in series, or in series using resistors to make up for that small internal resistance.

This is completely contrary to the old incandescant bulbs. The resistance of the metal filament increases with temperature, so its I-V curve is mainly flat. When increasing voltage, the current hardly increases. It is more like a current source with high internal resistance. For proper operation, the incandescant bulb is connected to a voltage source. Multiple incandescant bulbs can be connected in parallel rather than in series. When connected in series, or with a large resistor to connect to a higher voltage, the lamp will burn out much earlier.

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