How much current limiting resistor should this LED circuit be connected to? At present, a 5V / 4A power supply is connected to six ordinary LED lamp beads. The lamp beads need to be connected in parallel. How large is the current limiting resistance of each path?

How much current limiting resistor should this LED circuit be connected to? At present, a 5V / 4A power supply is connected to six ordinary LED lamp beads. The lamp beads need to be connected in parallel. How large is the current limiting resistance of each path?


Is your lamp led? White light or red, green, blue? White light has a larger voltage drop. Subtract the voltage drop from 5V, and then set the brightness of the lamp bead to 15mA. (10-20) if you want to light up a little, set it higher, if you want to dark down a little, then use the formula r = u / I
(for example, if only one LED red light bead is lit, the voltage drop of the red light is 1.8V, and the preset brightness is 15mA, then (5v-1.8v) / 0.015 = 213.3 Ω can be obtained according to the formula.)



How to know how much resistance to add in the circuit


If the added resistor is used as load, it can not overload the device and itself; if it is used as voltage divider, it is required to obtain the required voltage and not overload itself



In a circuit, the resistance is increased by 4 Ω and the current is reduced to half of the original value. How big is the resistance in the original circuit





In a 12V circuit to connect two 3V led. Ask. I want to use how much resistance? Thank you
In a 12V circuit to connect two 3V led. Ask. I want to use how much resistance?
I don't quite understand the circuit, please give some advice. Thank you. Speed reply


The resistance depends on your connection method and LED power. If the power is small, it's OK to use the resistance to reduce the voltage, but it's not recommended to use the resistance in this way. If two LEDs are connected in series with a current of 20mA, r = u / I = (12v-3v * 2) / 0.02A = 300 Ω. If two LEDs are connected in parallel, each LED is connected in series with a resistance, r = u / I = (12v-3v) / 0.02A = 450 Ω