Hi,
Im trying to get the LED on my PC to be ON all the time and cut when there is HDD activity. Well I found out that the motherboard controls the cathode to blink the led. I have included a schematic of what I built, for some reason I can't get it to work right. If anyone has any ideas why please let me know!
all you want is just invert the LED to light up when there is no HDD activity.
You might choose either way shown in the attachment.
Inverting the signal using a NAND-chip leaves three gates unconnected and requires some space. Additionally CMOS-gates won't source 20mA at 5V. Therefor I used a low current LED (2mA)
With a simple PNP-transistor the result is the same, requiring less space and you might use any standard LED (20mA).
On their datasheets, Texas Instruments show curves of typical and minimum output current of their CD40xx Cmos gates and inverters at 5V, 10V and 15V supply voltages and with various output voltages down to a shorted output.
With a 5V supply the typical current into a 2V red LED (without a current-limiting resistor) is only 3.5mA. With a 10V supply the typical current into a 2V red LED (without a current-limiting resistor) is 14mA to 18mA.
The inputs on 7400 series ICs pull up quite strongly. You have to pull them down very near to ground to make the input low. It is usual to have a pull up resistor and a switch or transistor to ground.
The outputs are also asymmetrical. They can sink a lot more current than they can source, so you should connect the LED between +5 V and the gate output, via a resistor.
https://www.electro-tech-online.com/custompdfs/2009/05/sn74ls00.pdf gives the specifications. A low input should be less than 0.8 V and 1.6 mA will flow to ground from the output.
A high output can only provide 0.4 mA and may only rise to 2.4 V, so not enough to light an LED. A low output can take in 16 mA and it will fall to 0.4 V (leaving 4.6 V to the power supply) so it can light an LED.