I am building a circuit from a 4.2 volt battery and I cannot use resistors to lower the current, and I cannot calculate the load resistance but I want to protect the switch from over current, and short circuit. I was thinking of using a diode. To the order of positive terminal, load, diode, switch, negative terminal. I need max amps to pass through the load, and minimum amps to pass through the switch. Will that work?
I need max amps to pass through the load, and minimum amps to pass through the switch. Will that work?
Since all your elements are in series, the same current flows through all elements. You need your switch to be rated for the same current as your load (for safety reasons you would use a switch rated for higher than the expected current).
A forward-biased diode will drop the voltage by a small amount (e.g. 0.7 V) but will not create a difference in the current either side of the diode. A reverse biased diode would stop any current flowing through your load.
To allow a small switch to control a large current you would connect the switch to a relay (or solid-state equivalent).
You want overcurrent protection. A series diode won't accomplish that.
If your load is passive, I would suggest trying to find out your load's resistance so you could use something like a basic current limiter which is just two BJTs and several resistors. You can't use it without knowing your load.
You can also use a Low Dropout Regulator which is like a miniature power supply. These provide a flat voltage and have thermal and short circuit protection. You can get them in single voltage versions (78XX) or if you want something adjustable or oddball, you can use an LM317.
There are also PTC (Positive Thermal Coefficient) resistors, but these are often too slow to save anything sensitive.