Search Results

Connect the collector via a R (e.g. 10k) suitable for LED brightness to the LED to GND. When the 12 V falls below 9V-2*0.7, the PNP will will turn on and light the LED. …

Select outputs 1, 3, 6,7,8 to activate the LED. …

You can get a single white LED that is rated for that amount of current. 2) If you put multiple LEDs directly in parallel, small variations in the LEDs themselves will cause them to run different currents … in each LED. …

The LED itself would only be sensitive to the total lifetime, but reliability is measured in 10's of years. … Most likely failure for a self-made system is the solder joints, or wires between the Led and PCB, or PCB and power source. …

The inductance of the wires will create voltage spikes -- be careful for your drivers. In any case, they may not be able to regulate the current in a stable way with large series inductances. Likely …

Also, the order of the LED and resistor doesn't generally matter, but it is slightly safer to put the R at the top because if you short the junction to ground (e.g. with a DMM lead or a scope probe), you … wouldn't burn out the LED. …

The internal resistance of a 2450 coin cell is about 10 Ω when fresh (see https://data.energizer.com/pdfs/cr2450.pdf). If you consumed 300 mA, the voltage would drop by 3 V -- and you wouldn't be able …

It is only necessary if you want to modify the LED current in a continuous way. …

For example, the efficiency of the LED, the temperature, the ambient light level etc. … Part of the design process is to find a reasonable compromise between unsafe (above spec. for the LED) high current level, and too low brightness (at low supply voltage and perhaps an LED with extreme …

simulate this circuit – Schematic created using CircuitLab When Q1 turns on, the red LED will short across the green+diode and turn on. …

That IC is not fully protected in this case. Basically, it wants the BOOST supply to be around 5 V. You can supply from VIN with some zeners if needed, or as you are doing from VOUT. But, if VOUT beco …

When the driver turns off, the current will stop, and the LED voltage will become zero. …

Therefore the LED current is defined by its series R1 and V1. With ~ 2.2 VF needed for the LED, the current is (5-2.2)/54 = 51 mA. In circuit 2, the NPN is not saturated. … If the LED requires 2.2 V, this leaves only 0.4 V across the 54 Ω, for a current of 0.4/54 = 7 mA. …

A (red) LED requires about 1.7 V to start to light. So, if you apply 2 V to the motor, it will (internally) generate about 1.6 V -- nearly enough to light the LED. …

The reason to use resistors is because a) the 5 V supply will vary over time and temperature; b) the LED forward voltage required to obtain a certain brightness will vary significantly over temperature … Thus to have any control over the LED brightness, you use a resistor to 'swamp' the other variations. …