Electrical – Zener Diode Voltage and Curve Equation

The current from the relay opening doesn't go into the Vcc rail at all. It follows the path shown here:

The stored energy is dissipated in the diode drop and the coil resistance of the relay.

In the Zener diode configuration, the stored energy is dissipated in the full Zener voltage of the diode. V*I is a lot higher power, so the current will fall faster and the relay might open a little faster:

MOVs are different than Zeners, but fulfill a similar circuit function: They absorb energy when the voltage exceeds a certain level. They are used for overvoltage protection, not for precision things like voltage regulators.

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.