0
\$\begingroup\$

I have the following circuit:

I basically have 2 transformer, one is 230V-8V the other one has a ratio of 1:1, so 230V on the secondary.

I'm rectifying the two signals with two full bridge rectifier, and I'm controlling the two signals through the two switches, which are solid state relays.

In principle I need to feed the coil (which is a big mechanical switch) with the 230V half sine waves signal in order to close the contact of the switch. The pulse of 230V has to be very short, in fact once the switch is closed, I feed the coil with the 8V half sine waves in order to mantain the contact of the big switch closed.

My idea is to activate the SSRs simultaneusly, in order to have the shortest time possible between the commutation 230V --> 8V once I turn off the SSR SW2. The two transformers share the same AC source at the primary. The diode and varistor in parallel with the coil are used to dump the energy of the coil when a high spike of voltage occures.

My question is:

  • Due to the very large inductance of the coil, once I want to turn off the system, or simply switch from 230V to 8V, I'm afraid that the big voltage spike could damage the SSRs and the transformers. How can I improve the circuit in order to prevent breaking down of the SSRs or of the transformers due to induced high voltage, and how can I be sure that my varistor network will dump all the energy stored without any leakage?

Thank you.

Edit:

I forgot to mention that I couldn't find the varistor symbol, so I used the resistor's one. The varistor I'm using is the SIOV-B32K275 from Siemens

\$\endgroup\$
0
1
\$\begingroup\$

You haven't drawn a varistor network on the coil, you've drawn a resistor and diode.

Sizing the resistor and diode properly is the key to having them do as you want.

At the point you stop supplying the coil, what is the current? This is the current that will initially flow through the D3 and Rvar. D3 needs to be big enough to handle the current. Although a 1N4007 is rated at 1A continuous, it can take an 8mS surge of 30A. The figures for the 540x series are 3A and 200A respectively.

When this current flows through Rvar, it generates a voltage. Rvar must be low enough resistance so that this voltage stays within your safe level. It must be physically large enough to absorb this pulse of energy without overheating. Not many resistors have pulse handling specified.

If in fact you've drawn a resistor because you didn't have a varistor symbol, then things are a little easier to specify, because a varistor will have pulse handling and maximum voltage specified. Make sure it's big enough to pass the current, absorb the power, and limit the voltage to what you want.

\$\endgroup\$
1
  • \$\begingroup\$ Yes, I used the resistor symbol because I couldn't fine the varistor's one. I edited the question. Thank you for pointing out. I will also check on the varistor's datasheet what you mentioned \$\endgroup\$ – Luca Daidone May 23 '18 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.