On Tue, Oct 14, 2008 at 6:21 PM, dudleyhd78 <> wrote:
> hello im trying to run 4 optocouples of a 7404 and dont understand how
> to sink the current in to power them
>
> i have pin 7grounded and pin14 to 5volt
>
> pin 1 is my squarewave input
>
> pin 6,8,10,12 are my optocouple out puts
>
> i have pins 13,11,9,5 connected to each other
>
> i was trying to sink the current into pin 3 with an led hooked to
> geoud of pin 4
>
> but im sure something isnt right some help would really be
> appreciated thank you james
>
>
A 7404 is not an optocouple, it is a hex inverter. The signal that is
input into an inverter, is only opposite
in polarity. The high level output current is 400 microamps (uA), the
low level is 16 milliamps (mA). So it
is said to be able to sink 16mA. Usually only about 10mA is needed to
"drive" an LED.
The forward diode drop of an LED is about 1.2 volts (V). The maximum
Volts out at a low level for the 7404 is 0.4V
If 5V is used to power the LED, then to calculate the current limit
resistor needed you would
5-1.2-0.4=3.4
That means the resistor needs to drop 3.4V when the current is at
10mA. Using ohms law
3.4/0.01=340
So you need about 340 ohms of resistance to limit the current at 10mA.
Connect your input signal to pin 1 of the 7404. Connect pin 2 to one
side of your resistor. Connect the other side of the resistor to
the cathode of your LED. Connect the anode of the LED to 5V. Connect
one end of a 1 Kohm resistor to pins 3,5,9,11, and 13, and
the other end to 5V. Pin 14 should be tied directly to 5V. Pin 7
should be tied directly to ground.
Hope that helps,
Steve
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