i am driving a inductive load (2 mH) with steady state of 10 amps with
30 amp relay at 12 Volts DC. no flyback diodes or any other supression of the inductive kickbacki also have a 100nf capacitor from the contract of the relay to ground.
what i am seeing is that after repeated turn off on the load, my cap is getting shorted.
i understand the current has to keep flowing in the load so it must be going thru my cap
has someone else seen this and/or can someone explain why its shorting
What is the voltage rating of the capacitor?
You have an L-C-R circuit when the relay opens. This can generate very high voltages that can pop the cap.
Is there a reason why you aren't using a reversed biased rectifier to dump the inductive spike?
too expensive and too big...need cap for esd protection
can i reduce the capacitance to prevent this?
"no flyback diodes or any other supression of the inductive kickback" Why not? It's killing your cap. It is failing because your circuit repeatedly hits the cap with a huge spike: V = L*di/dt. The cap needs to be protected by a series resistance and needs to be rated at a much higher voltage.
Ed
You are very likely getting much bigger spikes than that off of a 12 volt inductor. A reverse diode across the inductor will eliminate the problem.
Ben Miller
As others have said, the voltage exceeds your cap rating. 100 ohms in series would help but do you have a good reason not to put a diode across the points? That will slow down the drop out but it will save the points.
Just a quick calculation... Energy stored in inductor = Energy transferred to cap
Z=squreroot(L/C)=141 ohm
10A makes the voltage on the cap =1410V assuming the cap start from 0V which is not, it may be at 12V but that is nothing compared to 1410V.Assuming no energy is lost in the coil resistance and core. Still is easy to see >500V across that poor 50V cap.
Reducing the cap value makes things worse, the contact may arc. Like other suggested place the diode. Cost and size do not mean a thing if the circuit fail.
MG
Try using a cap that's meant for use accross the points in a car distributer, these are designed for this type of use.
Pip
The problem is the cap is too small or the voltage is too low. A 2 mH inductor at 10 amps will store 100mJ of energy and a capacitor of
100nF will charge well above 50 volts with 100mJ. Work it out and you will find the required cap voltage j=1/2 ce^2. 100mJ = 1/2 (100nF) e^2 e=1414 volts.So, you need a 100nF, 2KV cap.
Maybe a larger cap is the answer?
-Bill
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.