# Maintaining the same current?

1. Jul 13, 2014

### Dash-IQ

Is it possible to maintain, the same current when there is induced back-EMF(that is supposed to reduce it)? I assumed at the cost of more input electrical power it could be done. Am I correct? If I am how can this be done?

2. Jul 13, 2014

### Staff: Mentor

The most direct way of keeping current at some fixed level is to power the circuit via a constant-current source, a circuit that can adjust the voltage to cause current to stay constant. For a more specific answer, you will have to provide the schematic of the circuit you are involved with.

3. Jul 13, 2014

### Dash-IQ

Knowing that the principle is possible, is the most important to me. I assumed such an idea would be impossible to maintain the same current while back-emf is induced.

4. Jul 14, 2014

### Staff: Mentor

An electronic circuit can be constructed to achieve almost anything you desire.

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.

5. Jul 14, 2014

### sophiecentaur

If you were a bit more specific about the particular circuit you have in mind, answers could be more meaningful, I think.

6. Jul 14, 2014

### Staff: Mentor

Constant-current sources are something you can buy off-the-shelf. No magic involved. A very simple implementation just measures the current and adjusts the voltage as necessary.

7. Jul 15, 2014

### Dash-IQ

This is an example to help my understanding, hope you all can contribute!

We have a circuit, that has 1 A of current flowing, at 12Volts, the resistance is 12 ohms power consumed is 12W. Back-emf is induced, and its 5V, so now the current is reduced due to that Back-emf. By using a constant-current source how can this be solved to maintain a current of 1 A even when there is 5v of Back-emf?

This is is a very simple example to help understand how such a circuit would manage the current. I just want to understand the principle more.

8. Jul 15, 2014

### sophiecentaur

If it is a constant source, its output volts will increase (to 17V) to force 1A through the device, against your back emf. If the power supply cannot do this then it is not a content current source. It may be a 'Current Limiting Source', like many commercial power supplies and there can be confusion here..

9. Jul 15, 2014

### Baluncore

If you are considering a situation where inductance is present, the voltage across the inductor will be; V = L * di/dt.
Back EMF suggests that V becomes negative. For that to occur, the current must be falling with time, di/dt is negative.

If you place a diode across the inductor to clamp the back EMF, the inductor voltage will be smaller, but it will still be the forward biassed diode voltage plus the inductor's series resistance, Lr, times the current, I. That will reduce di/dt.

Alternatively you can connect a positive voltage across the inductor, V = I * Lr, which will exactly cancel the resistive component.

If you used a superconducting inductor, you would only need to short it's terminals to elliminate the back EMF and so maintain the current without change.

10. Jul 15, 2014

### sophiecentaur

You would need to produce a circuit diagram so we are all singing from the same hymn sheet, I think.

11. Jul 15, 2014

### sophiecentaur

At switch on of +DC, the back emf will be, as Lenz's law tells us, 'negative' as current is increasing , so it will 'tend to' reduce the current (slow the increase), for a given applied voltage. A 'proper' current source will ramp up the volts to ensure that the current is what was wanted, despite the emf. In an AC situation, the applied voltage (phase and amplitude) will be appropriate to provide the wanted current (phase and amplitude). Not hard in principle and observable with the appropriate probes etc if this is achieved with a feedback system.

12. Jul 15, 2014

### jim hardy

My experiments with feedback-based constant current sources and inductors is:
your design must allow for return of energy to the source.

My classic 7800 series regulator wired for constant one ampere current was immediately destroyed by connection to a substantial inductor. I had to parallel it with a resistor that supplied ~ half the desired current and let the IC trim current to one amp.

Last edited: Jul 15, 2014
13. Jul 15, 2014

### sophiecentaur

It doesn't surprise me that a beefy inductor can take out a weedy bit of silicon. dI/dy can be very high if you aren't careful. A relay coil can cause havock if you don't use a catching diode.

14. Jul 15, 2014

### Baluncore

In the post#7 example given, one amp flows through the inductor because of the 12 ohm series resistance and the long application of +12V. The situation is stable. Energy is being dissipated in the inductor series resistance.

At the instant the +12V is reduced, inductor current starts to fall because, the inductive component will produce a back EMF. The 1A current will still be dropping +12V across the series resistance. Something must be preventing a huge reverse excursion of the EMF and holding it to 5V. If the current remains stable at 1 amp then there will be no back EMF to counter, the external constant current circuit will be a +12V supply.

If you want to turn off the current quickly, then it pays to install a dump resistor in series with the flyback diode. That allows more back EMF, a larger di/dt, so less time. The possibilities are limited by the voltage rating of the switch.

My experience is that you must prevent a damaging back EMF from destroying the switch. The joules should preferably go somewhere else, such as the source if it is rechargeable, or a resistive dump load.

I prefer to dump inductive energy into a resistive load to quickly reduce the current. But in one example, when the dump load often failed to high resistance, the voltage exceeded –1200V. I installed a crowbar TRIAC that fired at –600V to short out the inductor when the dump load failed. It then took significantly longer for the current in the inductor to fall because there was less back EMF while the current is being circulated through the inductor series resistance alone, without the additional external series resistive load.

My problem was that I needed one of those newfangled digital storage oscilloscopes to see what was happening. The image on the old oscilloscope was useless, because while the inductor was conducting more than half an amp, the image slid sideways clear off the CRT. It was only a 10 henry inductor, but it had a huge air gap. Without my crowbar circuit, when the dump load failed, a green plasma genii would rise from the switchgear. The green colour confirming that the previous contactor set had once been made from copper.

15. Jul 16, 2014

### Dash-IQ

Hows fast is this process? Of stabilizing current with the presences of back-emf?

16. Jul 16, 2014

### Baluncore

@ Dash-IQ.
You have not told us what produces the back EMF.

If it is an inductor with series resistance, then maintain the DC voltage to keep the current constant.

If it is a DC motor, then maintaining the current will maintain the torque no matter what the RPM. If the motor is powered from a current regulator then the current will remain the same. You can regulate current very rapidly, 10 usec is easy, 10 nsec is possible if you use a series RF inductor in the circuit.

17. Jul 16, 2014

### sophiecentaur

To be sure that we are all talking about the same circuit, the OP should post a circuit diagram and a precise description of the situation. It has confused me because there will not be any 'back emf' if there is a 'Constant" current (i.e. DC).
We are talking in terms of an ideal (or spend as much money as necessary on it) source. I am assuming that the source will react as fast as necessary (instantly), which is not a problem because we are not talking about a fast change of current; just a modification of supply PD. without the inductance.

If a back emf value is being quoted then, either this must refer to an RMS voltage with AC or the result of a steadily increasing current supplied - DC does not apply here (of course). If you want a current ramp of steady slope, the required supply voltage will need to be a ramp also and will just need to be increased by the stated 5V of back emf that was quoted. If it's AC, then the AC supply needs to be 5V RMS higher than without the inductance.

People seem to have assumed that the thread is about Power distribution but isn't that just adding extra complexity when initially discussing a basic idea? Practical details can mask the basic principle.

18. Jul 16, 2014

### Dash-IQ

Im sorry for creating a confusing matter, I didn't attach a diagram because this is more of a general problem.
I wanted to know how to deal with back-emf that's induced by the change in magnetic flux. I wanted to know if it was ever possible to maintain the same current where there is change in the flux(and back-emf is induced).

@sophiecentaur, and @NascentOxygen perfectly answered that general question for me.
I thank the rest for all for their contributions, I don't think I need to attach a diagram and go on with this subject, because what I understood is there is indeed a way to maintain the same current even when back-emf is induced.

19. Jul 16, 2014

### sophiecentaur

I think we started off on the wrong foot here. 'Back emf' is, afaik, a term used to describe the effect of varying current in a component. Varying magnetic flux by some other means is hardly 'back emf'. It is just an emf, induced by an outside agency. Like I said: a back emf will not be there for an unchanging current.
Kirchoff 2 shows that emfs around a circuit to add together and will equal the sums of the IRs. That implies you can add or subtract volts to make up for the existence of 'back (or any other) emf'.

20. Jul 16, 2014

### Staff: Mentor

What will we observe if current is held fixed but L is made to vary with time? This could be achieved by tap-changing, or by partly withdrawing the solenoid's core.