I've been playing around with logic gates and RAM chips and stuff the past week or so, but I can't figure out how to get a DIP switch to represent a digital '0' while still being able to represent '1' when turned on. Having no current running through doesn't work, and the switch only has one input for each bit. How would I do this?
Octal DIP Switch Digital ‘0’
digital-logicdipswitches
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Even a PLL isn't going to preserve the waveform (including duty cycle) of its reference input, unless you add a second circuit to it to specifically do that. For example, the PLL could drive a one-shot whose pulse width is controlled by a feedback loop that compares the output duty cycle with the input duty cycle.
The best way to do this in an all-digital manner is to have one circuit that measures the frequency and duty cycle of the input signal (if those are the only two parameters you care about), and have another circuit that synthesizes a new signal with the same duty cycle and the desired frequency based on those measured values. (I have done something like this myself when building FPGAs that need to do frame rate doubling/halving on video signals.)
The output signal will in most cases be an approximation of the signal you actually want, and you'll have to decide how good that approximation needs to be.
http://digikey.com/product-detail/en/A8GS-S1305/SW1533-ND/4248837
Maybe try one of these. You could have a cap charged to 5V, a depletion-mode fet that is held off when your power is on and dumps the 5V charge through the coil when power is off? This would allow the switch to manually be reset even if power was lost and not purposefully reset.
edit: If it just needs to pop "ON" when the car turns on, have "ON" be the reset position. When the car turns on, have a circuit that momentarily 'resets' the switch to the "ON" position and then the switch will be toggleable.
Best Answer
You are right, you need something more than a simple single pole single throw switch to make a digital logic signal. Usually that something is a pullup resistor. The resistor goes between the power supply and the logic line, and the switch between ground and the logic line. When the switch is closed, it hard-shorts the line to ground. When the switch is open, the pullup resistor provides a weaker drive to power.
The tradeoff is that you want the pullup stiff enough (low enough resistance) so that it can overcome any current that other parts on the digital line might be leaking to ground. It has to be weak enough (high enough value), so that the current is limited to a reasonable value when the switch is closed.
Most CMOS logic gate inputs leak little enough that a few 100 kΩ to power is enough to guarantee the voltage will be interpreted as a logic high level. However, you don't want the line to pick up stray noise and of course you want some margin. If in doubt, use a 10 kΩ pullup.
This same concept works flipped around too. You can use a pulldown resistor and put the switch between power and the logic line. For various reasons, switches, whether mechanical or transistors, are usually connected to ground, and therefore pullups are more common than pulldowns.