Here's my question, the instructor said that the layout is important to reduce the number of components. I've already figured out that I need to connect my switch to the load side of the circuit, but he said there was another trick to it. This is a pic of what he drew on the board.
Does anyone know what this means?
I've tried thinking it out but I couldnt come up with anything.
Looks like the switch will select various zener (shunt) diodes. There will be some series dropping resistor before the switch to the left. The problem comes with the lowest voltage zener, to prevent it from burning when no load is connected to the output and the zener have to shunt all the extra current. One way around this is to select the series resistor as well using a dual selector switch.