Electronic – Which circuit best approximates a zener diode

I connected my supply which was 5v. The resistance was chosen to be 220ohms. This will make the max current of the zenor around 7mA.

This means that if your load requires more than 7 mA (I actually calculate 7.7 mA, not 7 mA), the resistor will drop V_out below 3.3 V. In that case the zener will no longer be regulating.

But, when I loaded the zener diode, the voltage across the diode dropped to 2.4v.

Why did that happen?

Your load was taking more than 7.7 mA.

Given that the output voltage dropped to 2.4, we can calculate that the load current was (5-2.4)/220 = 11.8 mA.

You could make this work by reducing your series resistor. If the load is fairly stiff, reducing the resistor to 110 ohms or so should bring you back into regulation. This assumes the load current doesn't increase to more than 15 mA when the supply voltage is raised to 3.3 V.

If it's possible to have an open-circuit condition at the load, you need to choose a zener that can take all the current that would normally go to the load.

You need a resistor between the switch and zener, and a resistor between the Vout point and ground, like so:

^{simulate this circuit – Schematic created using CircuitLab}

With the switch up, the Zener will limit Vout to 4 volts.

With the switch down, the Zener will not conduct, and Vout will be 3 volts (less a bit, due to the current drawn by R2)

Without R1, when the switch is up, infinite current would flow through the Zener diode as V1 tries to deliver 5 volts, and D1 tries to hold the voltage to 4 volts.