Electronic – Coilgun capacitor reverse voltage in the schematic

It is useful to provide the component name in the question if known (here = TRU-12VDC-FB-CL) and a datasheet reference - if Olin had had that when you first asked he would have been able to explain better.

After reading the following you should, in future, be able to find and read a data sheet of this sort and be able to find and understand all the information I have provided below.

Manufacturer is Taiwanese Tai-Shing Electronics Components Corp http://www.tai-shing.com.tw

A better copy of the datasheet can be found here - this is on the manufacturer's site and does not have the annoying frames and ads which try and kidnap you at every turn. I foiund this by reading the manufacturer's website address from the captive and annoying data sheet.

The meaning of the relay patrt numbver is clearly explained under "ordering information" at the top of the data sheet.

TRU-12VDC-FB-CL

• TRU General family

  12 - nominal coil voltage

  DC - DC operated

  F - Flux free type

  B - AgCdO contacts - ie NOT ROHS compliant (which they note)

  C - Form C contact layout - diagram supplied

  L - high sensitivity

The data sheet shows that at 12 VDC

• Nominal operate current is 30 mA

• Pull in (operate) voltage = 75% (=9V)

• Drop out voltage is 5% minimum (MAY be as low as 0.6V !!!)

• Coil resistance is 400 ohms.

• Maximum allowable voltage is 130% (15.6V) As Olin said, the ratings on the label are the contact current ratings at various AC and DC voltages. THese are listed about hald way down "page 52" (2nd provided page) of the daty sheet.

As Olin notes, there are TWO 28 V DC contact ratings.
The data sheet explains why on page 52.

• 28 VDC 12 Amp is the "UL" rating

• 28 VDC 15A is "TUV" rating

UL is a US approvals authority.
TUV is a European (German) approvals authority.

The reverse image stylised "UR" logo on the relay is the symbol of the UL organisation (Underwriters Laboratory).

TTI on the relay is the manufacturers brand name.

Since you are making the coil yourself, you can wind it to various voltage versus current tradeoffs. It seems you used too many turns of too thin wire, which therefore has a high DC resistance and needs high voltage to generate the large magnetic field you want.

Use less turns of a thicker wire. That will require more current to generate the same magnetic field, but a lower voltage to cause the current.

You can maybe achieve this with the wire you have by breaking down the "several thousand turns" into several winding and connecting them in parallel. For example, let's say you have one winding of 3000 turns now, which a DC resistance of 300 Ω. 10 A would require 3 kV, and provide 30 kAmp-turns of magnetic field. If you make that into three windings of 1000 turns each, then each winding has 100 Ω DC resistance. Put in parallel, the combined coil has 33 Ω resistance. It would require 1 kV and draw 30 A to produce the same 30 kAmp-turns magnetic field. This is actually still unrealistic. You want to go lower resistance and higher current overall. I just used these numbers as example to show how to calculate equivalent magnetic field.

I would find the largest capacitors you can get your hands on, and design the coil for that. This means putting enough current thru the wire for a short time to heat it just below damage level. The time will be so short that there won't be opportunity to conduct the heat away. Therefore, figure all the energy goes into heating the copper. Figure the maximum copper temperature you are willing to tolerate, use the specific heat of copper to find the energy per unit length of wire, then figure how much of that wire you need to use to absorb all the energy of the capacitor.