Zero-Cross Detector Circuit using Transistor
The figure below shows the circuit diagram of the Zero-Cross Detector Circuit using Transistor. The 230V AC mains is stepped down by transforme to deliver the secondary output of 6V, 500 mA. The transformer output is rectified by a full-wave bridge rectifier comprising diodes D1 through D4 and then regulated by IC 7805.
You can learn more about zero-cross detector here: Zero-Cross Detector Theory
Electrolytic capacitors C2= 470uF and ceramic disc C3=0.1 uF are used for bypassing the ripples present in the regulated 5V power supply. A capacitor above 10μF is connected across the output of the regulator IC, while diode D9 protects the regulator IC in case their input is short to ground. LED acts as the power-on indicator and resistor R1 limits the current through LED.
About the Circuit
The components required for the circuit are:
1. 230V AC Primary to 9V, 500 mA Secondary Transformer
2. D1-D8 1N4007 Diode
3. 7805 Voltage Regulator IC
4. Q1,Q2 – BC547 Transistor
5. LED
6. Electrolytic Capacitor- 1uF,25V & 470uF,16 V
7. Ceramic Disk – 1uF
8. Resistors – 1K, 4.7K, 470E, 330E, 1K
Assemble the circuit as seen in the circuit diagram or you can try in the Proteus for simulation.
Working of the Zero-Cross Detector Circuit
This regulated 5V is also used as biasing voltage for both transistors Q1 & Q2. A pulsating DC voltage is applied to the base of transistor Q1 through diode D5 and resistors R3 & R2.
When the pulsating voltage goes to zero, the collector of transistor Q1 goes high. This is used for detecting the pulse when the voltage is zero. Finally, the detected pulse from the collector of transistor Q2.
The complete operation can be well understood with the help of wave-forms in figure below:
1. The waveform before feeding into transistor Q1 is a fully rectified wave that is fed to its base.
- When the base voltage falls below transistor Q1 is switched off, pulling the output higher. This
results in a very short positive pulse, which is available at the collector taken from the base of transistor Q2 as shown in the second waveform. -
As this positive pulse is inverted by transistor Q2, it produces one negative pulse of the same width
at from collector of Q2 as shown in the third waveform.
1 Comment
Admin, there are a certain inaccuracies in the article. You wrote that:
“The components required for the circuit are: … D1-D8 1N4007 Diode”. But on the schematic there is: D7 LED-GREEN diode (according to the description of D1-D8, these are the 1N4007 diodes).
On the schematic there is D9 1N4007 diode but not exist in the required components.
So it might be confusing to an electronics beginner 🙂