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I am making a kick solenoid lock with AC.

I designed the circuit to run on 220 V, 50 Hz (household plug.)

The inductor (coil) is air core 500 turns, 150 meter (500 ft) in length with inductive reactance of 55 Ω.

DC resistance of the coil is 8.6 Ω

According to Ohm's law, the current should be 4 A so the solenoid will be 880 watt.

This the circuit.

enter image description here

I have some questions:

  1. Is the design OK? Will the circuit draw 4 A?
  2. Will there be a loss in current or power? If so how to find it (calculation)?
  3. Will 880 W, 220 V and 4 A create a strong electromagnetic field to be used as a kick solenoid?
JRE
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Marco David
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  • Dc resistance should also be known ... and some ... about wire used. – Antonio51 Jul 23 '22 at 13:29
  • Why dc resistance should be know?, 18awg – Marco David Jul 23 '22 at 13:38
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    Because current drawn will depend on the complex impedance and power dissipated will depend on resistance. – Transistor Jul 23 '22 at 13:58
  • 8.6 ohm dc resistance – Marco David Jul 23 '22 at 14:01
  • Magnetic field can be calculated by simple formulae: https://en.wikipedia.org/wiki/Solenoid

    Current will be limited (roughly) by AC and DC resistance in series. Magnetic field, and thus, force will be sinusoidal. Required force depends on what you want to kick.

     – Vladimir Jul 23 '22 at 14:07
  • @Transistor It is pure inductor no any other resistance or component – Marco David Jul 23 '22 at 14:40
  • "It is pure inductor ..." There's no such thing in the real world. A pure inductor would have zero resistance. Yours has 8.6 ohms. – Transistor Jul 23 '22 at 16:22
  • EE&O, current is ~ 4 A, real power dissipated is ~125 W, and reactive power is ~ 860 VA (not W) for a total of ~ 880 VA ... Heating a "lot" ... so, "temperature" would be calculated. – Antonio51 Jul 23 '22 at 16:40
  • How you calculated this? This solenoid will work by push button just for 1s or less – Marco David Jul 23 '22 at 22:00
  • These are "steady" conditions giving an idea of "magnitude". If the button is pushed 1s (which was not defined) then the calculus is a "transient" analysis depending on all the variables in the system and the "time starting" (random) in the sinusoidal wave ... Does the inductance remains "constant" in a "kick" solenoid? ... – Antonio51 Jul 23 '22 at 22:07
  • Yes inductance is constant, according to your words, the current, voltage will depends on sinusoidal wave, so if i push button when the voltage wave is nealy zero will cause the current decrease so much too? – Marco David Jul 23 '22 at 22:12

1 Answers1

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Here is a simulation where you can see what happens when you push a button in a "inductive" circuit. Equations can also be written ...

enter image description here

There is 3 phases in the behavior.

1- When you just push the "button" (here, it is a "clean" push ...). The starting current (~ 9 A peak) can be somewhat higher then in steady state because of the switching transient. It is a function relative to the voltage "phase" at starting "push".

2- Steady state (current calculated as usual in "sinusoidal" circuits).

3- When button is released, energy in inductor (if not equal to zero) will try "continue" the current, but it can't. As there is always a parasitic capacitor in parallel, energy is released (\$1/2*L*I^2\$) and create an "impressive" impulse voltage (\$1/2*L*I^2 = 1/2*C*Umax^2\$ >> \$ Umax = sqrt(L/C)*I \$ ) that can disrupt "air", it is an "arcing" that we can see.
Here (Umax is only some kV). Voltage is reduced because of the "big" capacitor that was added .

Examples :
if I=5 ; L= 170 mH ; C = 0.1 uF ; then Umax = 6.5 kV.
if I=5 ; L= 170 mH ; C = 1.0 nF ; then Umax = 65.0 kV.

EDIT:
To prevent "high voltages" transient, use a VDR (MOV).

enter image description here

See this for a correct choice of VDR

How to choose chart in the just above note ...

enter image description here

enter image description here

Antonio51
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  • Thanks for this great analysis, so after push the button the current will be about 9A for some milliseconds?, the third phase did not understant what is (Vps) refer to and can voltage up to kilo voltage in this circuit? – Marco David Jul 24 '22 at 03:59
  • Not always 9A. Function of the "phase" voltage input when button is pressed. It can be also - 9 A or anything else (perhaps 0 sometimes). Third phase (when button just released) is a "transfer" of "magnetic" energy of inductor (if any, because current is not equal to zero at that time) to electrostatic energy. Vps is the voltage across inductor at any time. When button pressed, Vps=220*sqrt(2) sine Voltage. – Antonio51 Jul 24 '22 at 07:26
  • Do i need to use snubber in this circuit? – Marco David Jul 25 '22 at 21:26
  • As it is used with AC, something as a VDR is needed. Added in answer. – Antonio51 Jul 26 '22 at 07:08
  • The problem with vdr, it work for once one protection the will be needed to replace and with every push to button will very high voltage (kv) ( current not equal zero) as per your analysis need to be discharge – Marco David Jul 26 '22 at 12:47
  • Ok. Did not know that it was for once work only). Snubber as I used (R-C-) added also ... If I remember well, also TRANSIL diode+serial Resistor could do the some job ? – Antonio51 Jul 26 '22 at 12:59
  • You should read this https://eepower.com/resistor-guide/resistor-types/varistor/# and this https://www.electronics-tutorials.ws/resistor/varistor.html#:~:text=Varistor%20Summary,generated%20by%20inductive%20switching%20transients. If VDR chosen carefully, it can be used because, here, transient is "short" (disrupt energy to be calculated) and solenoid does not function very often. Technical note : https://www.manualshelf.com/manual/vishay/vdrs07h275bse/1-brochure-english/page-12.html – Antonio51 Jul 26 '22 at 21:00
  • https://www.vishay.com/search/?searchChoice=part&query=technical%20note%20vdr%20resistor https://www.vishay.com/docs/29079/varintro.pdf How to use/choose these VDR. – Antonio51 Jul 26 '22 at 21:14