understanding a relay

[SughiRam P.]

within this discussion, can anyone tell me how the flywheel diode dissipate the surge EMF?? will it flow to the source or it will flow through the coil??? If the surge EMF is less than Source, will the source oppose the EMF to flow through the diode and coil?? I have searched in Forums. someone says, this EMF will flow through the source and dissipated via decoupling capacitor and other guys said, it won't flow to the source, it will flow to coil as circular loop. I can't come to a solution.

 

[goldsmith]

Dear SughiRam P.
Hi
Do you know about lenz law ?
Consider that we have an inductor . and we gave it a current . and after some time , we want cut that current .
What will happen ? we have a fundamentally formula for inductor : Vind ( the voltage across ind ) = L*di/dt ----> if time is low , and current is a bit large and the L value is a little large , so the voltage across the coil will be very high .
As you know , we used a transistor as a switch to drive the ralay , thus when the transistor is on , we haven't any special problem. but at the time that transistor want to be off , the inductor , can create the high voltage ( remember that this voltage is in opposite the last voltage across ind ) and in reverse polarity , that can destroy the transistor simply . and the duty of fly back diode is , to restrain that instantaneous pulse to prevent damages .( it will short circuit that opposite voltage .).
Best Wishes
Goldsmith

 

[kak111]

Read post #19
https://www.edaboard.com/threads/235544/#post1005529

 

[enjunear]

within this discussion, can anyone tell me how the flywheel diode dissipate the surge EMF?? will it flow to the source or it will flow through the coil??? If the surge EMF is less than Source, will the source oppose the EMF to flow through the diode and coil?? I have searched in Forums. someone says, this EMF will flow through the source and dissipated via decoupling capacitor and other guys said, it won't flow to the source, it will flow to coil as circular loop. I can't come to a solution.

See goldsmith's and kak111's descriptions for how the reverse voltage is created to forward bias the clamping (freewheeling, flywheel, etc) diode.

Depending on the exact situation, power dissipation can be both of the mechanisms that you mentioned. If there is enough energy to bias the diode and create a node voltage higher than the source potential, then some energy will certainly flow back to the power source. The decoupling caps won't really dissipate much energy, but they will reflect the high frequency content back toward the relay (if you design filters, you'll know that they work by signal reflection due to impedance mismatch). The power that makes it into the source can be dissipated by the circuits inside the power supply (switching transistors, iron core transformers, etc).

Also, because the magnetic field of the relay's coil is collapsing, some of the current going through the clamp diode can travel back into the coil in order to "close the loop", at which point, that energy is dissipated by the resistance of the coil. So, both responses are (mostly) correct.