Coil Voltage/Current: Understand Resistance in Relay Circuits

[erobz]

Maybe a not well thought out question:

https://cdn.automationdirect.com/static/specs/78relays.pdf

781-1C-24A Relay is listed as having $177 \Omega$ Coil Resistance. I'm assuming that figure will give the Real Power dissipated by the coil (entire relay circuitry) for some current $I$, or some voltage $V_{ac}$ across its terminals?

Just an ME here, my circuits knowledge is a little rusty, small, and minimally applied. Be Gentle!

I'm confused, because I was under the impression that coils didn't consume real power. They consume reactive power. So I'm assuming the manufacturer is referring to the resistance in the circuitry of the entire unit., not the "coil" per se?

 

[Guineafowl]

Ideal coils (inductors) only draw reactive power. Real-world coils have some resistance and so dissipate some real power.

 

[erobz]

Ideal coils (inductors) only draw reactive power. Real-world coils have some resistance and so dissipate some real power.

The whole circuit seems to be acting like this:

I'm taking $R$ to be $177 \, \rm{\Omega}$, and $r \approx 0.83 \, \rm{\Omega}$, $V_{ac} = 24.6 \, \rm{V}$ ( I don't know the supply phase angle).

So I'm thinking I don't need the inductance of the coil here to calculate the magnitude of the current in the circuit. The voltage across the relay is $24.5 \,\rm{V}$, so that leaves $0.1 \, \rm{V}$ across the $r$ (the resistance of the transmission lines), yielding $119 \, \rm{mA}$.

The inductance only effects the phase angle here I suppose.

 

[erobz]

Ideal coils (inductors) only draw reactive power. Real-world coils have some resistance and so dissipate some real power.

Ok. So, it is from the coil. I guess I never really gave it much thought. Take a straight piece of wire it has a resistance. Coil it up you have an inductor...it still has resistance. I don't ever remember modeling it this way in introductory circuits.

 

[Guineafowl]

( I don't know the supply phase angle).

The $V_{ac}$ is normally set to $0^o$, against which the other phase angles are measured.

 

[erobz]

The whole circuit seems to be acting like this:

View attachment 314845

I'm taking $R$ to be $177 \, \rm{\Omega}$, and $r \approx 0.83 \, \rm{\Omega}$, $V_{ac} = 24.6 \, \rm{V}$ ( I don't know the supply phase angle).

So I'm thinking I don't need the inductance of the coil here to calculate the magnitude of the current in the circuit. The voltage across the relay is $24.5 \,\rm{V}$, so that leaves $0.1 \, \rm{V}$ across the $r$ (the resistance of the transmission lines), yielding $119 \, \rm{mA}$.

The inductance only effects the phase angle here I suppose.

Actually, I think I'm only getting away with this because the inductance of this coil must be small in comparison to its resistance. The inductance does affect the magnitude of the current in the circuit.

$$ |\mathbf{Z}| = \sqrt{ \left(177+0.83\right)^2 + X_L^2} \quad \Omega $$

But I believe its safe to say that worst case the highest current in the circuit is $138 \, \rm{mA}$, since any appreciable inductance will drop the current?

Is the inductance of this coil "computationally hidden" in some way in the data sheet? As far as I can see it's not listed anywhere. Maybe by omission the manufacturer is saying to just ignore it in computations?

 

[Guineafowl]

I think you’re confusing:
Inductance, $L$, measured in henrys, $=\frac{\Phi}{I}$ with
Reactance, $X_L$, measured in ohms, $=j\omega L$

Since the latter is frequency dependent, you can’t calculate the complex impedance without knowing that and the inductance.

 

[erobz]

I think you’re confusing:
Inductance, measured in henrys, $=\frac{\Phi}{I}$ with
Reactance, measured in ohms, $=j\omega L$

Since the latter is frequency dependent, you can’t calculate the complex impedance.

Everything seems to be ok with what I said though because the reactance depends on the inductance. The inductance $L$ is what I don't see on the datasheet for the coil. I believe the frequency $\omega = 60 \, \rm{Hz}$.

However, when put that way I now notice the reactance will actually decrease with increasing inductance ( since $j^2 = -1$), so the current must be larger than $138 \, \rm{mA}$, not smaller.

 

[hutchphd]

There is one more part of this that must be included. The harmonic analysis is for steady state and does not include the transients that happen when you switch the relay coil. These excursions are often the important considerations.
It is very good that you realize your analysis is far from correct. You need to slow down and actually learn how to do this analysis (or not...your call). Get a textbook and do the work!
You asking a series of ill-formed questions is not the Socratic Method, is very inefficient, and personally I will not be involved. (That was the gentle version. Just FYI.)

 

[Guineafowl]

I see your question is about characterising the current through the relay coil, which as you point out, requires knowing the inductance. I think it would be reasonable to calculate the max current using the resistance only.

A few more points:
$\omega$ is the angular frequency, in radians/sec, calculated from $2\pi f$
No, reactance increases with inductance.

Rather than have me nitpick your posts, how about this link, which explains what you’re dealing with:
https://en.wikipedia.org/wiki/RL_circuit

 

[Guineafowl]

This one is more approachable, and deals with transients:
https://www.electronics-tutorials.ws/inductor/lr-circuits.html

 

[erobz]

I see your question is about characterising the current through the relay coil, which as you point out, requires knowing the inductance. I think it would be reasonable to calculate the max current using the resistance only.

A few more points:
$\omega$ is the angular frequency, in radians/sec, calculated from $2\pi f$
No, reactance increases with inductance.

Rather than have me nitpick your posts, how about this link, which explains what you’re dealing with:
https://en.wikipedia.org/wiki/RL_circuit

Yeah, I haven’t solved any problems using these concepts for well over a decade. I’ve only ever taken a single course, Introductory Circuits. I’m not trying to annoy anyone.

Thank you for your help.

 

[DaveE]

Larger AC relays (contactors, mostly) will have a dramatically different inductance between the open and closed positions. This is because the magnetic path (i.e. core) has a significant air gap when open and moves to a small (zero?) gap when closed. The consequences are that there is often a large current surge when closing them, compared to to holding current to maintain closure because of the increased inductance. This probably doesn't apply for your relay, but, I didn't look into it very carefully.

In any case, I don't ever recall a relay or contactor manufacturer specifying the coil inductance. People mostly ignore that part and just use the specs. given. The fact that you quote coil resistance instead of operating currents leads me to believe that the inductance isn't a significant factor.

I have also seen manufacturers say "coil resistance", when they actually mean "impedance magnitude"; $|Z|=\sqrt{R^2 + (\omega L)^2}$. It's not evil, they just don't want to confuse electricians and such.

The bottom line is if you need a good model (which I suspect you don't), then you'll probably have to test one yourself. All you really need is a current probe and an oscilloscope.

Anyway, to answer your question, all relays and contactors consume power because of the winding resistance; it is the dominant factor. This is because to generate the large magnetic field they need, they are made with many turns of small wire.

 

[erobz]

In any case, I don't ever recall a relay or contactor manufacturer specifying the coil inductance.

To me that seems funny, because if you designed a circuit you might want to know it...I say "might" because I already designed and implemented this circuit a year ago, and it works fine...without knowing it... . I think that's because they kind of engineer the engineering out of it. I'm asking about it because it occurred to me while I was checking the system over, probing voltages, that I really didn't understand the effect of the inductor on the current. The answer now seems to be while I was probing it (probably steady state conditions), there was no (measurable with my very standard AmProbe 60 Multimeter) effect of the inductor ( other than it’s internal resistance).

 

[Averagesupernova]

Considering a relay coil is designed to pull in the contacts as quickly as possible, the manufacturer will give you a voltage they want it to be driven with and a current that you can expect it to draw. It's not like it's going to be part of some impedance matching network. Knowing the inductance isn't all that important.
-
Now that being said, I have worked with large contactors for 12 volt DC winches. Those contactors can develope significant flyback when de-energized. Also, the motor they are switching can introduce noise back through the contactor coil. My application needed some suppression of this noise while also sensing the moment that the coil was de-energized. This required a bit an idea what the source impedance was (relay coil in this case). Sure it's easy to use a brute force filter of some kind but it caused significant delay in sensing when the coil was de-energized. More finesse was needed.

 

[Tom.G]

What was overlloked in the discussion so far is that the 177 Ohm coil resistance that the OP referred to is for a 12V DC relay, in which case the AC reactance does not exist (unless you are driving it with AC).
<page 3 Coil Specifications of datasheet:
https://cdn.automationdirect.com/static/specs/78relays.pdf>

The coil resistance, voltage, and power do not exactly match the theoretical unless you assume the coil resistance tolerance is -12% AND the voltage tolerance is +12%.

Using those worst-case tolerances yields a worst-case power dissipation of 1.16W, close to the datasheet value of 1.15W. (-15%R and +10%V also work)

The image in post #3...
https://www.physicsforums.com/attachments/1664562671186-png.314845/
...would apply to the 12VAC relay with a coil resistance of 44 Ohms and the power drain (1.4W) specified for 50/60Hz. Those numbers supply enough information to work out the inductance for those willing to make the effort. As a note, the AC power dissipated would be due to the 44 Ohm coil resistance.

As others have pointed out, the inductance will be lower, and the current higher, until the relay armature is fully pulled in.

Cheers,
Tom

 

[erobz]

What was overlloked in the discussion so far is that the 177 Ohm coil resistance that the OP referred to is for a 12V DC relay, in which case the AC reactance does not exist (unless you are driving it with AC).

Hi Tom,

I think you got your lines crossed a bit

I have 781-1C-24A

 

[erobz]

Anyhow, looking over some of the material I realize that in my circuit (24V ac) the effect of the coil does not go away in steady state conditions, correct (otherwise the whole phasor business doesn't make any sense)?

I was looking at:

$$ L\frac{dI}{dt} + IR = V \implies I = \frac{V}{R}\left( 1 - e^{-\frac{R}{L}t} \right)$$

as presented in post #11 https://www.electronics-tutorials.ws/inductor/lr-circuits.html

However, after some apparent contradictions surfaced while fiddling around with the math, I realized that solution is constant voltage DC.

The equation I need to solve to get the characteristics (both transient and steady state) is given by:

$$ L\frac{dI}{dt} + IR = V \cos ( \omega t) $$

 

[erobz]

If I were going to solve the ODE above, how would I know the initial condition $i(0)$? I think it should be 0, but I can’t reason why it must be. I read that the coil initially chokes the circuit with back emf, but why does it seem like I must know about this property to get an initial condition for the current? For instance, if I had instead chose a sine wave for the voltage, at $t=0$ we have 0 applied voltage, so I would expect no current there as well. But it seems that both of the conditions for the voltage (sine or cosine) cannot be true. Any guidance on this appreciated.

 

[hutchphd]

However, after some apparent contradictions surfaced while fiddling around with the math, I realized that solution is constant voltage DC.

The equation I need to solve to get the characteristics (both transient and steady state) is given by:

$$ L\frac{dI}{dt} + IR = V \cos ( \omega t) $$

First I would solve $$ L\frac{dI}{dt} + IR = V_0 \sin ( \omega t) $$ instead of $$ L\frac{dI}{dt} + IR = V_0 \cos ( \omega t) $$ because V crosses zero at t=0 and the algebra is easier maybe. Guess a solution of the form $$ I(t)=I_0 \sin(\omega t+\phi )$$ and try to solve for $I_0$ and $\phi$. You will get very reasonable values.

 

[erobz]

I used the method of undetermined coefficients:

$$ i(t) = C_1 e^{-\frac{R}{L}t} + \frac{V_p}{( \omega L)^2 + R^2 } \left( \omega L \sin ( \omega t ) +R \cos ( \omega t ) \right) $$

If I use $i(0) = 0$ ( which I don't understand how to justify the prior knowledge of the currents behavior, before I know how the current behaves) I get:

$$C_1 = -\frac{V_p R}{(\omega L)^2 + R^2}$$

$$ i(t) = -\frac{V_p R}{(\omega L)^2 + R^2} e^{-\frac{R}{L}t} + \frac{V_p}{(\omega L)^2 + R^2 } \bigg( \omega L \sin ( \omega t ) +R \cos ( \omega t ) \bigg) $$

Seems reasonable? But still concerned about that initial condition.

 

[hutchphd]

Your Latex did not display for me

 

[erobz]

Your Latex did not display for me

I always forget about the preview. Miss a brace and it all goes to crap

 

[hutchphd]

The initial condition that you have chosen i(0)=0 means that t=0 is not when the external voltage is applied. Remember that this is a first order inhomogeneous equation. To match the boundary value desired you can always add or subtract any solution to to the homogeneous equation (which is actually the $e^{-\frac {-R} L t}$ part of your solution) This is why I recommended sine. But the steady state current is indeed the voltage over the magnitude of the impedance etc etc.

 

[erobz]

The initial condition that you have chosen i(0)=0 means that t=0 is not when the external voltage is applied.

I don't understand. At $t= 0$, the RHS of the ODE I solved is $V_p$?

 

[erobz]

If we were to pick $V_p \sin ( \omega t )$ as the RHS. at $t=0$, the ODE is $0$, and I would expect again $i(0) = 0$. Not because of this "choking" effect, but because the applied voltage is $0$ at $t=0$. The problem there is that its the wrong initial condition for the sine wave ODE, because the current would be in phase with the voltage. It seems like the sine wave initial condition (whatever non-zero current it is at $t= 0$) would be a hard pill to swallow given no applied voltage?

 

[hutchphd]

I think the solution you have obtained in #28 is correct (for $t\geq 0$).

The problem there is that its the wrong initial condition for the sine wave ODE, because the current would be in phase with the voltage.

At t=0 the current has no defined phase because the transient part is not a simple sine wave. Look at the values for t=0 and t large and see that the solution is good. For large t the phase relationship between input V and i(t) is as expected, and i(0)=0 . (Ignore my initial look at the solution in#24). You good?

 

[erobz]

You good?

Good enough. I've plotted the function for what I thought could be reasonable values for the inductance of that small coil $L = ( 0.1 - 1 ) \, \rm{Henry}$.

Conclusion: There doesn't seem to be any cause for concern about transient behavior in this one (as others have already pointed out early on), nor did the inductance have any dramatic effect on the peak current, everything I checked was under $190 \, \rm{mA}$

Thanks to everyone for your help. A special thanks to you @hutchphd for goading me to go down this path. I think it's been a fun and rewarding exercise.

 

[erobz]

Just to bring it back full circle and tie up loose ends - I realized that I can estimate the inductance $L$ of the relay after all without all the ODE stuff. I approximately know the transmission line resistance $r \approx 0.8 \, \rm{\Omega}$. The internal resistance of the coil $R = 177 \,\rm{\Omega}$ is given by the manufacturer. I measure the voltage drop ( indirectly ) across $Vr = 0.1 \, \rm{V}$ by measuring the voltage across the coil terminals ( $V_c = 24.5 \, \rm{V}$ ). That does fix the current at $I = 125 \, \rm{mA}$.

That leaves a voltage drop across the "ideal" inductor as:

$$ \begin{align} V_L &= V_s - V_r - R I \\ &= 24.6 \, \rm{V} - 0.1 \, \rm{V} -177 \, \Omega \cdot 0.125 \, \rm{A} = 2.4 \, \rm{V} \end{align}$$

If I assume $f = 60 \, \rm{Hz}$ then it follows that:

$$ V_L = X_L I $$

$$ L = \frac{V_L}{I \omega} = \frac{1}{2 \pi f}\frac{V_L}{I } \approx 0.05 \rm{H} $$

So... not a bad guess for the range before, but it turns out an unnecessary one (I know I'm estimating several values - and not including uncertainty... just a ballpark figure).

If there are issues with this, please let me know.

 

[hutchphd]

Sorry where did the 26.5V come from? And why are you adding it in phase ?

 

[erobz]

Sorry where did the 26.5V come from? And why are you adding it in phase ?

The 26.5 volts was a goof - me jumbling numbers up. The source voltage is 24.6 V.

And why are you adding it in phase ?

I must still be confused. If I measure a voltage drop across a resistor, and I know the resistance, can't I calculate the current through resistor? Or is it that the voltage I measure across the resistor is only representative of the portion of current that is in phase with the resistor at $0^{\circ}$?

I'm probably being too loose with my analysis.

What I'm asking: Is the voltage I'm measuring on the multimeter actually $V_r = |I |r \cos \theta$, where $\theta$ is the phase angle?

 

[erobz]

I'm guessing Kirchoff's is actually this:

$$ 0 = V_s \angle 0^{\circ} - I \angle \theta^{\circ} r \angle 0^{\circ} - I \angle{ \theta^{\circ} } R \angle 0^{\circ} - I \angle \theta^{\circ} \omega L \angle 90^{\circ} $$

This AC electricity stuff is tricky business. My hat is off to you guys!

 

[anorlunda]

This AC electricity stuff is tricky business. My hat is off to you guys!

If you could learn to work with complex numbers, you will find that the equations are exactly the same as the DC case. Of course, scalar is simpler than complex [no pun intended], but those trigonometric forms you showed are much more difficult than complex.

 

[erobz]

If you could learn to work with complex numbers, you will find that the equations are exactly the same as the DC case. Of course, scalar is simpler than complex [no pun intended], but those trigonometric forms you showed are much more difficult than complex.

I think my main problem here is I don't know what I'm actually measuring with my multimeter. ( I'm rusty on complex number tricks - but conceptually they aren't a problem).

 

[anorlunda]

hink my main problem here is I don't know what I'm actually measuring with my multimeter.

I searched this thread. You didn't say whether the multimeter is set to an AC scale or DC scale.