Nick designed into Balloon 2’s board a handy feature – a jumper that disconnects power from the batteries to the board when open. This would allow us to package the balloon and only plug in a jumper when ready to launch. While nice, we were a bit concerned about still messing around with the internals of the balloon minutes before launch. To make launching easier and less of a hassle, a small SPST (Single Pole Single Throw) Switch was purchased from radio shack, and wired up to a small pcb, along with a power LED.
This allows us to completley seal the box, and simply throw a switch before launch.
It’s a simple circuit. The switch connects the two pins of the jumper when thrown and leaves them open when switched off. The led draws power from the jumper’s circuit when powered on. There was no ground locally available though, so an extra header pin was required to bring in a common ground from the board. A 160 ohm resistor was used to protect the LED from the high current coming straight from the board.
To calculate resistor values for an LED in a circuit, we can use Ohm’s Law.
It states that:
V = I * R
Where V = voltage, I = current, and R = resistance.
Since we’re solving to find a resistor value, a little algebra manipulates the equation into R = V / I. Now we can just plug in the values and solve for R! If the LED specification is unknown, generally safe values to use are 2V at .02A. Since we want 2V, and our circuit is providing 5, we subtract 2 from 5 to get 3 – the voltage drop we want to accomplish with our resistor. Now our equation is:
R = (5 – 2) / .02
Where R = 150. Since all we had in the lab were 160 ohm, I used one of those. It doesn’t hurt to use a higher resistance, it will just shine a little less bright. Wire it in series with the power, or ground of your LED. Either one is fine, and everything should work!