For a DC coil a diode will work fine but will allow the magnetic field to remain for a longer time, slowing the dropout of whatever is moved by the coil.I thought it was an MOV not a diode?
Thank you.For a DC coil a diode will work fine but will allow the magnetic field to remain for a longer time, slowing the dropout of whatever is moved by the coil.
A well chosen series resistor will maximize the magnetic field decay by dissipating energy.
An MOV will combine the functions of the diode and resistor since it will not conduct at the normal applied voltage.
For an AC coil a diode will not work but an MOV or a pair of zener diodes will.
You can also use a snubber circuit of a capacitor and resistor in series placed across the coil.
Just a capacitor will also limit the voltage spike on either an AC or DC coil, but can produce a resonant ringing instead and will slow the dissipation of the stored energy.
Any snubber circuit has to be in parallel with the control contacts or in parallel with the coil.Thank you.
One other question, I would expect a diode to be in series with the load and a resistor to be in parallel with the load while the MOVs I have seen are in parallel with the load.
So if we dumped the MOV for a run of the mill diode would it be in parallel and series.
This is where a basic understanding is important.
Consider a series circuit of a battery, switch, and an inductor. Assume the inductor inductance is constant. In the real world there will be resistance in series in this circuit. Assume the dominate series resistance is the internal resistance of the inductor coil. Let the reistance be R and the source voltage V.
Also note that the inductor current can not instantaneously change. This means that, if just before time T1 the current is I, then just after T1 the current must = I. From any energy stored in the inductor, if an attempt is made to change that current, then the inductor will generate whatever voltage across its inductor that is required to maintain that same current in the same direction. Theoretically this could mean infinity voltage, but in the real world something will breakdown to limit the voltage.
The loop voltage drop for this circuit is V = i*R + L*di/dt where L is the inductance, and di/dt is the rate of change of current with respect to time. If the starting point is where inductor current is zero, then at closure of the circuit all the voltage drop is across the inductance because at that time the current is 0. After a long time the steady state current builds exponentially to a value of I = V/R. One uses calculus to determine the shape of the current curve.
Theoretically voltage sources have zero internal impedance. This means if the voltage source is reduced to 0 volts, then the voltage source looks like a 0 resistance wire.
Next instantaneously change the voltage source from V to 0 volts. Just before this voltage change the current thru the inductor is flowing from the + end of the inductor thru the inductor and out the negative terminal. Just after the source voltage change the inductor current must remain the same and flow in the same direction. This means the inductor becomes a voltage source and thus the inductor terminal polarity reverses to maintain current flow in the same direction. What was positive is now negative. An exponential current discharge curve occurs that is like the charge curve, but reversed in the vertical axis.
Placing a reversed biased diode across the inductor and the inductor's internal series resistance provides a new series path. When battery voltage is applied the diode is reversed biased and negligible current flows thru the diode. If a switch is in series with the battery loop, then when the switch is opened the current from the battery goes to zero, but the current in the inductor must continue. Since the inductor voltage reverses this forward biases the shunt diode, and it becomes a ahort across the inductor.
The time constant of an L-R series circuit is a function of L/R. The lower R is relative to L the longer is the time constant and the time for the current to increase or decay.
To reduce the time constant to obtain a shorter dropout time one makes R greater, but then the terminal voltage at turn off increases. Non-linear devices such as Transorbs and MOVs become useful as mentioned in other posts.
See http://www.electronics-tutorials.ws/inductor/lr-circuits.html don't be confused by my use of L/R vs R/L you see in the equation. What is in the exponent of the equation is really the ratio of two times --- an independent variable t divided by a time constant (L/R).
Then perhaps a little more direct.Whoosh
That was the sound of your post going right over my head. I do not understand what you are saying. It is to technical for me and I am OK with that.