# Ohms Law and Zener Resistor. Simple Design 4 Zener is a kind of Diode widely used in Voltage regulation and Switching applications. Zener is similar to other signal diodes but it conducts only if a specific voltage is provided to it based on its voltage rate. For example, a 5 volt Zener conducts only if it gets more than 5 volts. When the voltage to the zener rises above its value, its p-n junction undergoes breakdown and conducts. This breakdown point is known as Avalanche point. Unlike other diodes, current through the zener must be controlled, otherwise it will fail. So a series resistor must be connected to control the current through it. Here comes the importance of Ohms law .

Like other diodes, Zener diode also has Anode and Cathode ends. The Rectifier and Signal diodes are connected with Anode to the positive line and Cathode to the negative line. That means, the current flow is in the Anode – Cathode direction. But Zener is usually connected in the “Reverse bias” mode. That is, Cathode to Positive and Anode to Negative. It is easy to identify the Anode and Cathode of Zener. The method is same as that of other diodes. A white or black ring will be present at one end of the diode. It is the Cathode side. How to select the correct Zener resistor?

Generally, the Zener requires minimum 5 Milli Ampere current for its breakdown. The resistor in series with the zener will protect it from excessive current.

The diagram below shows the typical zener application to produce a regulated voltage. The supply voltage used is 9 volts 500 mA. From this 9 volt, we have to produce a fixed 5 volts. So we need a 5 volt (or 5.1V) zener and a correct series resistor. The Cathode of zener is connected to the resistor and it goes to the positive rail. The Anode of zener is connected to the negative rail. 5V regulated DC is tapped from the junction of the resistor and the cathode of zener.  Suppose we need 5 volt DC and 25 mA current to the load. So let us check.
Input voltage – 9V (V in)
Voltage for the load – 5.1 V (V out)
Load current – 25 mA (Iin)
Zener current – 5 mA

The formula for calculating the resistor value is
V in – V out / I in = 9 – 5.1 / 30 mA x 1000 = 120 Ohms
So the current flowing through the zener is
9 – 5.1 V / 120 Ohms = 32.5 mA.
So 32.5 mA passes through the zener. Out of it, minimum 7.5 mA passes through the zener – resistor chain ( 32.5 – 25 = 7.5 ). So the output current to the load will be 25 mA at 5.1 volt.
Power dissipation in Zener
Power rating of zener is very important. When excess current flows through the zener, it heats up and burn. Zener is available in different power ratings like 400 Milli Watt, 1 Watt, 2 Watts , 5 Watts etc. So it is important to calculate the power dissipation of the zener before selecting it.
Power dissipation in Zener
Zener voltage x Zener current

In the above example, 7 .5 mA current passes through the zener (total current – load current). Let us round up it as 10 mA. So,
5.1 V x 10 mA = 51 Milli watts.
So up to 350 Milli Watts, you can use 400 mW zener. Above that, use 1 W or 2 W based on the power dissipation.
Special note – For the proper breakdown of the zener, the input voltage to it should be 1-2 volts excess than its voltage rating. For example, a 5 volt zener requires more than 6 volts.

Special note – For the proper breakdown of the zener, the input voltage to it should be 1-2 volts excess than its voltage rating. For example,  a 5 volt zener requires more than 6 volts.