How to Use 74HC595 Shift Registers for LED Display with Arduino

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To set up a 13x16 LED display using four 74HC595 shift registers with Arduino, ensure correct wiring and power connections, including a 0.1uF capacitor across the power pins. The process involves setting the data pin (pin 14) to a desired value, pulsing the shift clock (pin 11) high and low eight times for each data value, and then pulsing the latch clock (pin 12) to transfer the data to the output. It's important to keep the clear pin (pin 10) high and the output enable pin (pin 13) low for proper operation. If the LEDs are not lighting up, double-check the connections and ensure the registers are correctly configured. Proper setup and timing are crucial for successful LED display operation.
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I have a 13x16 led screen I'm making(leds are put together), but now I'm trying to set up 4x 74hc595's. I've looked online and lots of the site seam similar but don't work and I do not know/understand how I can get my arduino to send (for example have the shift registers do "1010101010101010").
 
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here is a site
http://arduino.cc/en/tutorial/ShiftOut#.UyKNM7TubD4

But I don't understand the capacitor on the latch pin. In the text they say 0.1uF. If you use 1uF you will need to have a very long latch pulse. You shouldn't need a capacitor at all on the latch pin, but you need a capacitor (0.1uF ceramic) across the power pins.

Set the data (pin 14) to a value.
pulse the shift clock high then low (pin 11)
do that 8 times, each time with a different data value.
Then pulse the latch clock high then low (pin 12). That will transfer the data to the output.

Make sure you have correct power, ground, have clear (pin 10) high and output enable (pin 13) low.
 
That's what I originally used but I never worked right... I have. Currently 4 connected together like how it shows on there with an led only on the first of each. I set my arduino so it sets them all to high but leds didn't turn on
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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