DC coils

[Gibson486]

So, how does a DC coil actually work? Under DC, in school, you learn that inductors are shorts when DC is applied. How, then would a DC coil on a relay work? What causes the magnetic field? If I were to put a DC oil in parallel with a resistor, how would you calculate how much current would go through each branch since a DC coil would only have impedance with no resistance (atleast in the AC world)?

 

[PottedMeat]

So, how does a DC coil actually work? Under DC, in school, you learn that inductors are shorts when DC is applied. How, then would a DC coil on a relay work? What causes the magnetic field? If I were to put a DC oil in parallel with a resistor, how would you calculate how much current would go through each branch since a DC coil would only have impedance with no resistance (atleast in the AC world)?

Remember real inductors are always at the very least an ideal resistor + ideal inductor. Any relay datasheet should specify coil resistance, i.e. a '12V relay' will say something like 'coil resistance - 24 Ohms' so when 12V is applied 500mA is generating the magnetic field necessary to close the contact.

 

[Born2bwire]

Any DC current radiates a magnetic field and a long series of coils produces a strong magnetic field in what is typically referred to as a solenoid.

 

[futuristicmonkey]

So, how does a DC coil actually work? Under DC, in school, you learn that inductors are shorts when DC is applied. How, then would a DC coil on a relay work? What causes the magnetic field? If I were to put a DC oil in parallel with a resistor, how would you calculate how much current would go through each branch since a DC coil would only have impedance with no resistance (atleast in the AC world)?

It looks like your last sentence hasn't been addressed. If an IDEAL coil is put in parallel with the resistor the current from the DC source will be equal to V/R of the resistor plus the time-integral of voltage across the inductor divided by its inductance. Recall that the voltage across the inductor is: Vl = L * di/dt. If you divide both sides of the equation by L and then take the integral with respect to time you will see that is the case.

Since the coil is a real, physical component it will have resistance. The current through the coil will build up exponentially until it equals V/Req. Consult a textbook or google search for the details.

 

[MrDudeMan]

To add to what was already said, the field still exists, but it isn't changing direction the way it would be with AC current.

 

[Gibson486]

Well, I guess what leads me to asking this is putting a DC coil in parallel with A 47.5K load. My first instinct is that the relay coil is going to take most of this load. According to the spec sheet, the load is 1 watt, so that means under 12V, it would be the equivalent to a 144 Ohm load. However, since it's an inductor, you cannot simply combine them like two resistors since the inductor stores energy and the resistor dissipates. The energy between the two branches has to equal. So to evaluate what the current will be on the 47.5K branch, I will have to see what the current will be when it dissipates 1W, correct?

 

[TecHNooB]

Can you draw the setup?

 

[PottedMeat]

Well, I guess what leads me to asking this is putting a DC coil in parallel with A 47.5K load. My first instinct is that the relay coil is going to take most of this load. According to the spec sheet, the load is 1 watt, so that means under 12V, it would be the equivalent to a 144 Ohm load. However, since it's an inductor, you cannot simply combine them like two resistors since the inductor stores energy and the resistor dissipates. The energy between the two branches has to equal. So to evaluate what the current will be on the 47.5K branch, I will have to see what the current will be when it dissipates 1W, correct?

You're applying 12V across a parallel combination of 47.5K and 144 Ohms, so the current in the 47.5k branch is just 12V/47.5KOhms no matter what. The current in the coil is 12V/144Ohms when energized. At steady state the inductor dissipates no energy and stores no energy, but the resistive portion dissipates so it's all that matters.

 

[MrDudeMan]

Well, I guess what leads me to asking this is putting a DC coil in parallel with A 47.5K load. My first instinct is that the relay coil is going to take most of this load. According to the spec sheet, the load is 1 watt, so that means under 12V, it would be the equivalent to a 144 Ohm load. However, since it's an inductor, you cannot simply combine them like two resistors since the inductor stores energy and the resistor dissipates. The energy between the two branches has to equal. So to evaluate what the current will be on the 47.5K branch, I will have to see what the current will be when it dissipates 1W, correct?

Why does the energy have to be equal? The voltage drop across them has to be equal, but the current does not. P=V*I, so the power could be very different.

Assume you have a 12V source driving this circuit. When you first turn that circuit on, the inductor is an open so no current will flow through it. Energy begins to be stored in the inductor and current will flow following an exponential curve, but the voltage across it is always 12V, so the power dissipation will continue to increase until it hits steady state. What you have to know is the DC resistance of the coil to find the steady state current value. If the coil is, say, 100 ohms, then you know at steady state it will be sinking 12V / 100 ohms = 120mA, which also means it is dissipating 12V * 120mA = 1.44W. The current through the resistor is a separate issue as the principle of super position applies. The voltage across the resistor is always 12V, so the current through it is always 12V / 47.5k ohms = 0.25mA and the power is always 12V * 0.25mA = 3mW.

The transient period is a different story. The charging of the inductor follows this equation:

V = 12V
R = 100 ohms
t = time
L = inductance
tau = L/R

I = V/R * (1 - e^(-t/tau))

The current through and voltage across the inductor look like this:

The internal resistance is in series with the inductance, and the voltage across them changes exactly inversely of how the current charges in the inductor. Another way to say that is the voltage across the inductor when it is open is 12V and as it begins to charge the voltage tends to 0V. The voltage across the resistor is exactly the opposite. When the inductor is open, there is no current flowing and the resistor has no voltage drop (V = IR, I = 0 so V = 0). As a small amount of current begins to flow, a little less than 12V is across the inductor and a little more than 0V is across the resistor, but the total is always 12V. Eventually, at steady state, the voltage drop across the resistor is 12V and the drop across the inductor is 0V.

 

[C1]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

For DC, inductors can be considered pure resistors. It is in the process of becoming energized/de-energized (and the frequency thereof) that is interesting. Initially the build up of the magnetic field in the inductor resists the direction of current flow. Once stabilized, the coil can be represented as a resistor. When the current ceases, the magnetic field collapses and generates current flow in the direction the of the external current that was responsible for generating the field.

 

[Mark R]

A perfect inductor would ramp up to an infinite current, if connected to a DC source. In reality, inductors aren't perfect, as they have resistance. So, it's easiest to think of your DC coil as an inductor and resistor in series. This allows you to model the behavior - even if you can't physically separate the inductor and resistor components.

So, while a DC voltage is applied, the inductor component has zero impedance, and stores or releases zero energy - so it can be completely ignored in the analysis.

Things get a lot more exciting when non-DC is applied - and that instants when power is applied and removed - and the energy stored and released needs to be considered.

For example, the instant you apply power to a DC coil - the current is 0, but then ramps up to the steady state value. As the current is ramping up, energy is being stored in the inductor.

When the power is disconnected, the current *must* ramp down by releasing energy from the inductor - the energy and current cannot suddenly disappear. As a result, the inductor will force current into the rest of your circuit. If your circuit has electronic components, they may fry as the coil forces the current in.

I've drawn a diagram:

 

[Colt45]

steady-state current on DC relays and solenoids is entirely a function of the real resistance of the coil.

The reason your relay doesn't melt is because the coil has a mile of cunthair thin wire, which has a fairly large real resistance.This keeps power draw reasonable.

Inductors are only interesting with changes. be that ac, pulsed dc, the first bit a dc supply is switched on, a little ripple riding on some dc, etc.