i was thinking adding a diode parallel to each relay might be able to solve the issue. Currently, whenever there is a voltage spike, the printed circuit board spoiled easily which made hard for my project.
If there is any other remedies, i am willing to give a shot.

 

This is the current schematic. However, this board design currently has a flaw, hence i was thinking adding a diode parallel to each relay might be able to solve the issue. Currently, whenever there is a voltage spike, the printed circuit board spoiled easily which made hard for my project.
If there is any other remedies, i am willing to give a shot.

Would you please explain what you mean by "...the printed circuit board spoiled easily which made hard for my project." in a little more detail?

 

I recommend you read the paper at Scott's link at #9. It explains the speed advantage of using a zener and some of the other.

A schottky switches faster, but that won't be important here. The lower forward voltage of the schottky actually makes the relay turn-off delay longer. Most "switching" diodes like the 1N4148 switch very fast - in the range of 5 ns. That is faster than the drive transistor turn off time.

If you need the relay to turn off quickly you have to let the flyback pulse amplitude go as high as possible while still protecting the transistor. Your relay operates at 12 volts. If you choose a drive transistor rated at 40 volts (very common rating for small transistors) you should safely be able to allow the flyback pulse to go to 30 volts. If you use a zener, there are different ways to connect it. Let's use this way: cathode of the zener to the transistor collector, anode to circuit common (ground). This would directly limit the voltage across the transistor to the zener voltage. Let's pick a 33 volt zener, just because that is a common standard value. When the relay turns off, the zener current will be identical to the current that was flowing through the transistor - 34 mA. The power dissipation in the zener will be 34 mA x 33 V = 1.12 W. Because the amount of time the zener conducts is short, it would be safe to use a zener rated at 0.4 to 0.5 W, or even a bit less. Pick a zener with that power rating in the package you want to use and check the data sheet carefully to see what the actual voltage would be at 34 mA. It will be higher than 33 V - maybe closer to the 40 V rating of the transistor than you want to allow. You might decide to use a 30 V zener, instead, or maybe use one with a 1 W rating.

With an ordinary diode across the coil, the voltage across the coil will be about 0.7 volts while the stored energy is being discharged. With the zener, the voltage across the coil will be 33 V - 12 V = 21 V. By using the zener, the current will drop about 30 times faster (21 V / 0.7 V)
δi/δt = V/L i in amperes, t in seconds, V in volts and L in henries; again this ignores the coil resitance
That doesn't mean the relay will open 30 times faster, since there is a mechanical process, too.

If you use the simple diode method, which will cost the least, almost any small "signal" or "switching" diode will have adequate current rating. Most will be rated at at least 30 volts, but be sure you don't pick one rated for less than 15 volts. There are many many to choose from. If you use a small diode elsewhere in your circuit, there is a good chance it will be suitable.

 

Would you please explain what you mean by "...the printed circuit board spoiled easily which made hard for my project." in a little more detail?

I meant by whenever there is a voltage spike, the relay chip of RTE 24012 would be burned and the board could not be used anymore.

 

hi nike,
I have used a series combination of a 1N400x diode and Zener diode in order to reduce the 'hold on' time of the relay.
Choose the Vz voltage lower than the Vce breakdown of the transistor.
E

EDIT:
This LTS sim shows two relays, one with a diode and the other with a diode and zener.
Note:
the shorter current decay time of relay2 with the zener, but note the higher Vc swing.

 

Thank you everyone for the help.

 

I meant by whenever there is a voltage spike, the relay chip of RTE 24012 would be burned and the board could not be used anymore.

We may be getting closer...

The RTE-21012 does not have a semiconductor chip inside. That leaves me wondering what you mean by "chip". Is the chip you mentioned on the schematic you uploaded in post #21?

Also, when you referred to a voltage spike, do you think this spike is generated on the board or is external to the board?

Where does the +12V to drive the relays come from?

 

Is it possible for me to edit my reply post ? as i made some type and errors.

MOD:
There is a short time limit to Edit while you have less than 10 posts.

 

Is it possible for me to edit my reply post ? as i made some type and errors.

MOD:
There is a short time limit to Edit while you have less than 10 posts.

Oh oh alright thanks so much for the reply

 

I recommend you read the paper at Scott's link at #9. It explains the speed advantage of using a zener and some of the other.

A schottky switches faster, but that won't be important here. The lower forward voltage of the schottky actually makes the relay turn-off delay longer. Most "switching" diodes like the 1N4148 switch very fast - in the range of 5 ns. That is faster than the drive transistor turn off time.

If you need the relay to turn off quickly you have to let the flyback pulse amplitude go as high as possible while still protecting the transistor. Your relay operates at 12 volts. If you choose a drive transistor rated at 40 volts (very common rating for small transistors) you should safely be able to allow the flyback pulse to go to 30 volts. If you use a zener, there are different ways to connect it. Let's use this way: cathode of the zener to the transistor collector, anode to circuit common (ground). This would directly limit the voltage across the transistor to the zener voltage. Let's pick a 33 volt zener, just because that is a common standard value. When the relay turns off, the zener current will be identical to the current that was flowing through the transistor - 34 mA. The power dissipation in the zener will be 34 mA x 33 V = 1.12 W. Because the amount of time the zener conducts is short, it would be safe to use a zener rated at 0.4 to 0.5 W, or even a bit less. Pick a zener with that power rating in the package you want to use and check the data sheet carefully to see what the actual voltage would be at 34 mA. It will be higher than 33 V - maybe closer to the 40 V rating of the transistor than you want to allow. You might decide to use a 30 V zener, instead, or maybe use one with a 1 W rating.

With an ordinary diode across the coil, the voltage across the coil will be about 0.7 volts while the stored energy is being discharged. With the zener, the voltage across the coil will be 33 V - 12 V = 21 V. By using the zener, the current will drop about 30 times faster (21 V / 0.7 V)
δi/δt = V/L i in amperes, t in seconds, V in volts and L in henries; again this ignores the coil resitance
That doesn't mean the relay will open 30 times faster, since there is a mechanical process, too.

If you use the simple diode method, which will cost the least, almost any small "signal" or "switching" diode will have adequate current rating. Most will be rated at at least 30 volts, but be sure you don't pick one rated for less than 15 volts. There are many many to choose from. If you use a small diode elsewhere in your circuit, there is a good chance it will be suitable.

Thank you very much for the heads up. I will try my best

 

hi nike,
I have used a series combination of a 1N400x diode and Zener diode in order to reduce the 'hold on' time of the relay.
Choose the Vz voltage lower than the Vce breakdown of the transistor.
E

EDIT:
This LTS sim shows two relays, one with a diode and the other with a diode and zener.
Note:
the shorter current decay time of relay2 with the zener, but note the higher Vc swing.

Would you recommend using a SMD diode ?

 

We may be getting closer...

The RTE-21012 does not have a semiconductor chip inside. That leaves me wondering what you mean by "chip". Is the chip you mentioned on the schematic you uploaded in post #21?

Also, when you referred to a voltage spike, do you think this spike is generated on the board or is external to the board?

Where does the +12V to drive the relays come from?

Sorry for using the wrong term. I was referring to the RTE21012.
I would think that it is external to the board.
In the design, there is two different voltage. 115V and 12V. There is a dip switch whereby i could toggle it. However, i am currently facing the voltage spike in 115V.
Please pardon me if i could not explain clearly as it is my first time trying a pcb design.

 

I think this means that the problem is that the relays are being damaged by voltage spikes on the 115V. You removed the schematic that you posted in post #21 but from memory this is 115VAC and this is switched by the relays but is not related to the relay coils.

The advice you have been receiving is about how to deal with voltage spikes that develop across the relay coil when you de-energise the coil.

The reason for the failure of the relays and the burning of the circuit board is not immediately evident.

Are the relay contacts being burned from high voltage arcing? What is the load -is it something like a motor that is inductive?

If you can post a schematic and is possible a photograph of the damaged relay and board we would have a good chance of properly diagnosing the root problem and recommending a solution. The application (what is the load?) needs to be understood before a solution can be offered.