In this tutorial we will learn How to Use Diode as As Series Noise Clippers. Often, signal noise can cause unwanted explosions of sensitive circuits. When the sound voltage (Vn) is smaller than the diode forward voltage drop (Vf) and the voltage signal (Vs) is greater, a pair of parallel connected, opposite polarity (single cathode connected to anode of another) diode in a series with a suitable resistor (R) can be used to eliminate an unwanted part of the input signal. Since the sound voltage is not large enough to transmit bias or diode between signal layers the result will be zero, producing a dead +/- Vf band worldwide. When Vs forward bias or diode the output voltage will be (Vs – Vf)

If the Vn is too large for normal diodes, two Zener diodes placed cathode at the cathode are usually replaced. Zener voltage (Vz) has been selected to be greater than the audio voltage. When the input signal runs smoothly, one diode moves forward, while the other enters the degradation mode. In the absence of a signal, roles are reversed. The deadline is +/- (Vf + Vz), only larger signals will be transferred to the output.

As the voltage drop across both diodes is output to the input signal, the output value is (Vs – Vf – Vz).

For example, suppose we have a +/- 6V input signal with +/- 2V amplitude sound:
1) Find the correct Zener diary and calculate the value of R.
2) Calculate the height of the output signals.

Given that Vz> 2V, the manufacturer’s specification update sheets inform 1N746A with Vz = 3.3V.

Output signal (Vo) therefore:
= +/- (Vs – Vf – Vz)
= +/- (6V – 0.7V – 3.3V)

= +/- 2V.

In the absence of a load across the R, the series resistor must exceed the current sufficient to keep the diode running where the signal is. From the data sheet, the current Zener test (Izt) is given as 20mA in Vz.

1) To ensure that the current through the resistor (Ir) is greater than the Zener or knee current break, set to about 1/4 X Izt = 1/4 X 20mA = 5mA.
2) Make the voltage drop across the Resistor (Vr) equal to +/- 2V.

Now, the resistance value (R) will be equal to Vr / Ir = 2 / 5mA = 400 ohms. The nearest normal value will be 390 ohms. Therefore, the power distribution in R will be equal to (Vo X Vo) / R = (2V X 2V) / 390 ohms = 10.3mW.