Designing a switch for a shunt trip coil - Page 1
Forum section: electronics / projects

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 07, 2020, 04:10:19 pm
content: Hello,In the circuit below L1 is a shunt coil that was tested at 115 VDC. It has a DC resistance of 300 ohms. I need to use this coil in an application with wide voltage 130-800VDC from a bridge rectifier. I will apply a 30 ms pulse to the MOSFET. This means the coil won't see the high voltage for a long time. My thought is that if the pulse is that quickly the coil wouldn't get damaged. Do you think this approach would work? Any thoughts?

author: duak
author_rank: Super Contributor
timestamp: March 07, 2020, 04:31:16 pm
content: Well, if the voltage across the FET is low then the remaining voltage must be across the coil, no? I think you mean current, which is what actually determines how much heat is generated in the resistance of the coil. vis: P = I^2 x RThe current in an inductor is easy to determine given circuit values, the applied voltage and the time. Here's a link to a description of how an inductor behaves when a voltage is applied to it:https://www.electronics-tutorials.ws/inductor/lr-circuits.htmlAn inductor stores energy by creating a magnetic field. When the FET goes open circuit, the current continues to flow through D1. Most of the energy will be dissipated in the resistance of the coil.I think what you may really want to do is apply the voltage to the coil while monitoring the current, and then shut the FET off when the current reaches a certain level. The current will continue to flow and do the work to trip the switch.

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2020, 04:54:01 pm
content: At 800V, your coil is overloaded by a factor of nearly 50 - not good!If one ignores the unknown inductance of the coil and treat it as a purely resistive load (justifiable for a ball-park estimate if its core saturates in a time that's short compared to the pulse length), the pulse will dissipate 64 Joules in the copper of the coil. Assume the heat loss during a 30 ms pulse is negligible. To determine if the coil has any chance to survive, you'd need to calculate the expected temperature rise from the mass of copper present and the specific heat capacity of copper, then, assuming max. ambient temperature, see if that exceeds the max. temperature rating of the coil insulation. If you cant weigh just the coil, estimate it from the coil volume assuming close packing of the turns leaves 10% waste space.I suspect that the results wont be favourable and you will need to implement some sort of current control as Duak suggests, to maintain a safe current in the coil for long enough for the mechanism to trip.I'd also be concerned about layer to layer insulation breakdown at seven times its rated voltage.

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 07, 2020, 05:21:45 pm
content: This is all in the datasheet of the coil:1. Winding: 3750 turns +/-10% of 40 awg double build polyurethane wire with Min. Of 130 °c temperature rating.2. Resistance: 290 +/-10% ohms at 25 °c.3. Force: must provide a min. Of 2.8N. At 0.059 striking distance at a supply voltage of 115 VDC.

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2020, 05:52:29 pm
content: That's not all in the datasheet, as you don't know the length of the 40 AWG wire comprising the 3750 turns (less 10% for worst case), so cant calculate its mass and thus the worst case temperature rise.While you could estimate the length from the coil resistance, wire CSA and the resistivity of electrical grade copper, the result will have a high degree of uncertainty vs estimating the copper volume from the coil dimensions, allowing for 10% waste space due to the packing fraction, and additionally for the thickness of the "double build polyurethane" coating on the wire.