Why Zener getting heat at no load in Capacitor power supply arrangement..

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saj

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Why Zener getting heat at no load in Capacitor power supply arrangement..
October 1, 2019 - 2:52am

Hi,
With this PS, trying to operate 2 relays(5v, spdt, rated coil current 80mA for each) with pic mcu.
 It is noticed that at no load(when the relays not being operated) the zener(5.1v, 1.3watt) is geeing heat.
And if the relays are kept turned on the zener is ok.
Can you suggest what should i do.IMG-0642.JPG

tnx

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between 230v and diode bride
October 1, 2019 - 2:27pm

between 230v and diode bride you need  transformer to down voltaga from 230v   to 9v

between the zener and capacitor need resistor to split the voltage and limit the current

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saj

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CHARLI, 1ST GO TO SCHOOL
October 3, 2019 - 6:01pm

CHARLI, 1ST GO TO SCHOOL BEFORE REPLY...

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Do not try this circuit
October 4, 2019 - 1:05pm

Hi saj,

I am not sure where you found this circuit from, but it is has some serious mistakes.

You are converting a 230V AC mains voltage to DC using a full bridge rectifer. In that case your output voltage will be Vout = root 2 * 230 which is around 325V DC. Then you are connecting this 325V DC to a 25V capacitor and 5.1V zener which in most cases will fail immediatly or even burn completly.

So, add a step down transformer as charli suggested. This transformer will bring down your input voltage from 230V to 6V or 9V depending on the transformer you are using. Then you can proceed with the same circuit.

Also using a zener diode to regulate voltage is not a good idea. Consider using a voltage regulator like 7805 to regulate the required 5V to drive your relays. Zener diodes are not meant for voltage regulation 

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zener hot without load
November 18, 2019 - 2:32am

Ignore the other entries, they don't understand what they are looking at.  First, the 1uF 400V capacitors right after the 230V A/C power are there to limit current, based on your 50Hz frequency.  They do this using field (capacitive reactance), instead of thermal, because a resistor would be big, expensive and/or get hot.  So the circuit above, using a 1uF currnet limiting capacitor is probably dropping the voltage to about 10V.

The full-wave bridge rectifier converts the rest to a DC pulse, and the following 1000uF (1F) capacitor smooths it and is 25V for safety.  The zener then dumps current to drop the voltage to 5.1V.

The reason the zener gets hot without a load, is because no load beyond the zener, means the zener itself must pass all the current (the zener becomes the load).  You shouldn't run this circuit without a load.  Remember this for other things you do.

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Transformer less power supply - CAUTION!
November 18, 2019 - 7:32am

Hey the circuit that you are working on is called a transformer less power supply circuit. As Jim pointed out it uses a X-rated capacitor to drop the AC mains voltage to low level. You can read this article on how it works 

https://circuitdigest.com/electronic-circuits/transformerless-power-supply

But be heavily advised, since you are working on AC mains do not try this circuit unless you know what you are doing. Also for the zener probelm, if you want to operate this circuit at no-load you can use a voltage regulator like 7805.