Can Induced Back-EMF Be Maintained with Increased Input Power?

[Dash-IQ]

I think you need to look at this another way - you seem to be putting the cart before the horse.
First, just assume that an actual current source exists. This will (by definition) put the required current through your Inductor (plus its resistance). What Volts are you going to measure across this Inductor? Answer: I(R+jL).
Now all you need to do , to achieve this in practice, is to provide those volts from some other (real circuit), which can be just a very high resistance in series with a high voltage source OR it can involve a supply that will maintain the wanted current by (possibly) varying volts somewhere, by means of a control loop. As the replacement circuit is producing exactly the correct current, the volts will be just what you measured in the original experiment with an ideal source.
There is a small matter of Phase / time lag to be considered here if you want to analyse it completely.

BTW, you keep referring to a steady current - you do mean AC? That's not very good for a electromagnet. Could you make it clearer for me?

Remember that there will be a phase between AC current and voltage waveforms so you need to consider the VA as well as the Watts if you now are wanting to look at the Power situation. This is how Power Factor affects the use of reactive loads.

Of course you will need to change the supply volts if you change the coil resistance but when the current is what you want, the back emf will be the same (unless you are changing the inductance too). Phase will be different again.

This is VERY interesting, providing voltage from another real circuit, that's beyond me!
I know it's difficult to say without any details, but the time lag... is in the rage of seconds or milliseconds normally?

Btw, I always confuse terms... this whole circuits is a DC circuit.

 

[Dash-IQ]

25W

I relied on the calculator to much that I forgot to do a simple multiplication!

 

[sophiecentaur]

This is VERY interesting, providing voltage from another real circuit, that's beyond me!
I know it's difficult to say without any details, but the time lag... is in the rage of seconds or milliseconds normally?

Btw, I always confuse terms... this whole circuits is a DC circuit.

If it's a DC circuit, where is any back emf coming from?

 

[Baluncore]

Consider the general case of a current flowing in a closed circuit.
Any back emf can be modeled by the introduction of an impedance into that circuit.
The resistive component of that introduced impedance may be positive or negative.
The “back” of “back emf” implies a negative resistance has been introduced.

Normally, circuit resistance is positive, so I²R power is released.
But where the introduced resistance is negative, a flow of energy is released into the circuit.
There must be a source of energy available to generate the reverse voltage gradient.
That energy source may be from unaccounted internally stored energy, or from an external source.
An energy flow audit of the circuit must include all energy sources, sinks and stores.


I think it is a mistake to consider an unspecified hypothetical situation here. No matter what solution or explanation is offered, the hypothetical situation may be changed to defy that particular solution or explanation.

A circuit current is maintained constant by introducing a signed, (bipolar), variable resistance into that circuit.

 

[Dash-IQ]

If it's a DC circuit, where is any back emf coming from?

An exterior magnetic field, from a magnet/electromagnet and that field changes.

 

[NascentOxygen]

This is VERY interesting, providing voltage from another real circuit, that's beyond me!
I know it's difficult to say without any details, but the time lag... is in the rage of seconds or milliseconds normally?

Microseconds, if it's desired.

 

[sophiecentaur]

Consider the general case of a current flowing in a closed circuit.
Any back emf can be modeled by the introduction of an impedance into that circuit.
The resistive component of that introduced impedance may be positive or negative.
The “back” of “back emf” implies a negative resistance has been introduced.

Normally, circuit resistance is positive, so I²R power is released.
But where the introduced resistance is negative, a flow of energy is released into the circuit.
There must be a source of energy available to generate the reverse voltage gradient.
That energy source may be from unaccounted internally stored energy, or from an external source.
An energy flow audit of the circuit must include all energy sources, sinks and stores. I think it is a mistake to consider an unspecified hypothetical situation here. No matter what solution or explanation is offered, the hypothetical situation may be changed to defy that particular solution or explanation.

A circuit current is maintained constant by introducing a signed, (bipolar), variable resistance into that circuit.

I have to take issue with introducing a "negative resistance" into this. A negative resistance must introduce energy into a circuit (V/I is negative and so must IV be; implying power input). This must imply another Power source, which doesn't exist in a passive component. Back emf has a mechanical analogue of Mass, which causes a reaction force without introducing any Energy. What happens with an inductor is dealt with perfectly well by having a back emf that is -LdI/dt. This is not resistive and it 'reduces' the Power Flow into the coil.

But Dash-IQ tells us he is considering a DC circuit. So where can any emf come from? I have a feeling that this thread is being carried out on two separate levels.

 

[Baluncore]

If it's a DC circuit, where is any back emf coming from?

An exterior magnetic field, from a magnet/electromagnet and that field changes.

That hypothetical external magnetic field is the source of the energy. Hence the need for negative resistance.

I think it is a mistake to consider an unspecified hypothetical situation here. No matter what solution or explanation is offered, the hypothetical situation may be changed to defy that particular solution or explanation.

I have a feeling that this thread is being carried out on two separate levels.

Yes. This thread is quite out of hand, because it has never been nailed down..

 

[sophiecentaur]

That hypothetical external magnetic field is the source of the energy. Hence the need for negative resistance.



Yes. This thread is quite out of hand, because it has never been nailed down..

OK but that isn't a 'back emf', surely; it's just an emf and could be in either sense, depending in the direction of the field that's suddenly switched on. It doesn't constitute a negative resistance either - it's just another added emf that isn't related to the existing current flowing at all. You are right; this thread has no direction. It needs to be split up and each part could concentrate on one of the random ideas.

 

[Dash-IQ]

Maintaining a fixed current in the face of changing conditions is not difficult, in principle. Of course, this can not happen over an unlimited range. But you could readily devise a fast-acting circuit that varies its voltage over a range of, say, 1V to 200V, in order to try to keep a constant current flowing through some path.

So going from 1V to 100 or 0.01 to 10V etc..., in a very short duration (less than a second) is possible?! How did I miss that.

 

[Dash-IQ]

Now all you need to do , to achieve this in practice, is to provide those volts from some other (real circuit), which can be just a very high resistance in series with a high voltage source OR it can involve a supply that will maintain the wanted current by (possibly) varying volts somewhere, by means of a control loop.

How can this be done? We can just add ONLY voltage from another circuit and current from another? And just total out to be new voltage and the same current all around the circuit? What about resistance etc...?

I mean, for me right now I'm struggling with the idea of "voltage" and resistance in a constant current source circuit, So I'm going to break down my problems one by one.

1) The back-emf cancels the source-emf or the source-emf is still there? By that I mean in general when we have a source-emf that's 10V with back-emf(exterior magnetic field changing flux) being 5V, the voltage is now 5V right what happened to that other 5V?

2) By increasing the resistance, aren't we causing a problem? If you look at my example previously, we increased the voltage to "oppose" back-emf but out of all that comes out a value, I have to use an example (I struggle explaining it while confused):
V_source = 10V,
I = 2A, V_emf =5V
R_initial =5Ohms

In order to maintain that same current that is 2A, we need to increase V_source to 15V, that will require resistance to be what 7.5 Ohms instead of 5, if the 15V now drops to 10V total how can current be the same with the resistance being higher?




I know I'm repeating this over and over, but I'm struggling a bit, sorry :shy:.

 

[Baluncore]

The component network must be precisely specified at the start of the game.
You may specify flipping a switch in the network at some time.
If you really need to, you can specify a change to the value of one component at a time.
Once the game has started, you may NOT introduce an external component into the network.
External sources of energy, current or EMF are strictly NOT permitted.

Now, to start, define your circuit precisely.

 

[jim hardy]

If one follows the rules
performs his "walk around the loop writing Kirchoff's Voltage Law"
and rigourously solving the resultant equations
these questions disappear.

That's why the subject of electric circuits is taught the way it is. "What" before "why".

 

[MrSparkle]

1) The back-emf cancels the source-emf or the source-emf is still there? By that I mean in general when we have a source-emf that's 10V with back-emf(exterior magnetic field changing flux) being 5V, the voltage is now 5V right what happened to that other 5V?

The back-emf takes away from the source-emf. When you apply a voltage to a motor, it starts spinning faster and faster. Now, what causes it to eventually stop accelerating and reach a steady state? Well, the spinning rotor induces the back-emf that opposes for source-emf. Its basically acting like a generator. When the source-emf equals the back-emf, the motor stops accelerating (in reality, it stops accelerating a little before this because of losses). Now, depending on the load of the motor, the current at this point could be very very small. When you use a constant current source however, you are forcing this current to be significant. The source-emf will be higher than the back-emf, and the motor will keep accelerating.

2) By increasing the resistance, aren't we causing a problem? If you look at my example previously, we increased the voltage to "oppose" back-emf but out of all that comes out a value, I have to use an example (I struggle explaining it while confused):

I'm not sure you are getting this. A current source raises the voltage to whatever is necessary to apply the set current. It is that simple. If it needs to put out 17 volts so that 2 amps are pushed through, then it puts out 17 volts. If it needs 30 volts so that 2 amps are pushed through, then it pushes out 30 volts.

 

[mfb]

So going from 1V to 100 or 0.01 to 10V etc..., in a very short duration (less than a second) is possible?! How did I miss that.

Good function generators can go from 0.01 V to 10 V in nanoseconds. I guess they won't do that in your setup as the current measurement and internal electronics probably needs more time to react, but at least it is possible.
I never saw function generators for 100 V, but this is still easy to switch in less than a microsecond.

 

[Baluncore]

70 volt pulses with picosecond risetimes are available from avalanch breakdown of some 1$ NPN transistors.
But it takes a very expensive oscilloscope with a non-linear transmission line sampler to see them.

 

[sophiecentaur]

Chasing and controlling a variation in emf would be a matter of the whole feedback loop phase shift. The real delay in level change of a control signal would be much longer than this. Remember one foot per nanosecond - on a good day with the wind behind you.