Jan 26 2010, 12:57 AM
Ok, so I want to launch multiple model rockets individually from the same board. I have devised a scheme that would work using CAT6 wire, so I should be able to connect 5 ignitors to a common ground (the 6th wire). Each of the 5 firing circuits has its button or momentary switch. Each has a continuity light to let me know when the circuit is firing. However, I don't know how to make the ENTIRE setup have a continuity light. Using my current diagram, there is a master switch that gives power to all the circuits. I want a light to illuminate when I turn this switch on. How do I do that without stealing the power from the sub-circuits? BTW, I will only be firing one circuit at a time.
Feb 21 2010, 07:38 PM
If you don't want to steal power from the subcircuits (even though an LED uses VERY little power, maybe around 20mA), I would use something called a Driver chip.
Now, I am assuming you are using a separate power source of power supply to drive the LED (otherwise you would be "stealing power from the subcircuits"). This seems quite unnescessary, but you know more about your project than I do.
A Driver chip comes in a DIP package (breadboard) or surface mount so you can put it wherever you want. It essentially has a bunch of inputs and a bunch of outputs. The inputs take a logical high or low. There is one ground wire as well. So what happens is, if I apply a logic high to input pin 1, it will ground output pin 1.
So, what does this mean? Usually, they are used in the following way:
The VCC leg of the LED is connected to VCC through a resistor (use an LED resistor calculator to find out what value you need based on the voltage and current of you LED). The other leg, (GND) is connected to the output pin 1 of the Driver chip. The input pin of the driver chip is connected to your VCC for your firing circuit somewhere after the switch. Now, when the ALL POWER switch is not switched on, the Driver chip input will receive a logic 0 (technically it is an indeterminate state, but let's not worry about that for an application like this) and the LED will not be grounded, so it will be off. When you turn the ALL POWER switch on, the chip will receive logic 1, ground pin 1, and complete a circuit for the ALL POWER LED.
Again, this seems like massive overkill and I would just wire a single LED with a large resistor in series to the right of that purple wire, but best of luck. If this seems like a good solution, I can help you find a suitable driver.
Hope this helps!