12 V to 5 V (500 mA) using Zener Diode

This circuit converts 12 V from a car battery to 5 V and provides up to 500 mA current output, to power a USB device. If a 5.1 V Zener diode is used, and it requires 5 mA to maintain the 5.1 V Zener voltage, then calculate the value of the voltage drop resistor R. Use the E24 series to provide a suitable value of the resistor. Since the Zener diode requires 5 mA, and our USB power supply project requires at least 500 mA to operate, then the total current flowing through the resistor R has to be 505 mA, when we close the switch.

The car battery provides 12 V and the Zener diode regulates the voltage to 5.1 V, therefore the voltage drop across the resistor R must be (12 V - 5.1 V), which is 6.9 V.

Since we now have the voltage across the resistor R and the current flowing through it, we can calculate its value using Ohm's Law where R=V/I, and therefore the resistance R is given by 6.9/0.505, which is 13.66 Ω. We therefore choose 13 Ω from the E24 series because it will ensure that the Zener current is not less than 5 mA.

With the switch open, all the current flows through the resistor and the Zener diode. The voltage across the Zener diode remains at 5.1 V, and the voltage across the resistor remains at 6.9 V. With a 13 Ω resistor, the actual current flowing through it will be given by 6.9/13, which is 0.53 A.

Since the switch is open, all the current (0.53 A) also flows through the Zener diode, therefore we have all the parameters required to calculate the Zener diode power requirement.

P=I × V

P=0.53 × 5.1

P=2.703-watts

Therefore, a Zener diode with a suitable wattage rating would be required for this application.

Since the current through the resistor R is 0.53 A, and the voltage across it is 6.9 V, we can also calculate the power dissipation of the resistor.

P=I × V

P=0.53 × 6.9

P=3.657-watts