counter emf

[jin29_neci]

thanks guys for all the help!!! im still studying doing my best to become a professional electronic technician someday.. thanks

my question is how counter emf occurs?

post pls. thanks

 

[Someone Electro]

You mean how an coil makes that high voltage spike?

This hapends wen you sudenly stop the curent trough an coil.Some of the energy was sotred as an magnetic field around the coil.This field then colapses and is converted back in to eletrcety.This creates an high voltage spike.

This is why there are diodes on releys.If the didode wasent there the voltage spike cod destroy or reduce the liftime of the conected components.

This is also why DC motors make a lot electrical noise in the power suply(Thats why you somtimes see ceramic capacitors on the motors conectors)

 

[ljcox]

thanks guys for all the help!!! im still studying doing my best to become a professional electronic technician someday.. thanks

my question is how counter emf occurs?

post pls. thanks

The voltage across an inductance is v = L di/dt where di/dt is the rate of change of current in the coil. So if the circuit is opened, the current falls rapidly thus creating a high voltage - the back emf or counter emf.

Len

 

[JimB]

The voltage across an inductance is v = L di/dt where di/dt is the rate of change of current in the coil.

To be completely correct the expression is v = - L di/dt.
The -ve is there because the polarity of the voltage is opposite to that of the current.
Probably why it is sometimes referred to as "back EMF".

JimB

 

[Roff]

The voltage across an inductance is v = L di/dt where di/dt is the rate of change of current in the coil.

To be completely correct the expression is v = - L di/dt.
The -ve is there because the polarity of the voltage is opposite to that of the current.
Probably why it is sometimes referred to as "back EMF".

JimB

It depends on the definition of current direction. Below is the definition I have always used. Defining positive current as flowing out of the positive terminal is strange. Using your current convention, if you replaced the inductor with a resistor, Ohm's law would be V=-IR. Damn strange.

 

[eblc1388]

The direction of the counter emf is there to try to maintain the same current flowing. So it is in the same direction of the current and points downwards across the inductor, therefore the negative sign.

 

[Roff]

The direction of the counter emf is there to try to maintain the same current flowing. So it is in the same direction of the current and points downwards across the inductor, therefore the negative sign.

My point is, if current is flowing into the inductor (direction of arrow in my schematic), and you interrupt the current, then di/dt is negative, and the voltage spike is negative. There is no need for a minus sign in the equation.

 

[eblc1388]

I see what you meant. I need to do some thinking on this. I was taught that without even questioning the teacher.

OK. The back emf(e) generated is producing current in the same direction of the load current because it tries to maintain the same current.

Therefore the back emf will produce a voltage drop across the inductor in the direction same as the voltage across it before the current change.

So, both VL = L x di/dt and e= -L x di/dt are correct as they refer to differnt things.

 

[Roff]

Well, as long as you define the direction of the current as being into the positive voltage terminal, I don't see how you can have two definitions.
Below is a simple schematic with current and voltage waveforms. I limited the current slew rate to avoid an extremely high voltage spike.
BTW, if the inductor is ideal, and the current is steady state prior to the transition, there is no voltage across the inductor prior to the current transition (you made reference to that).

 

[eblc1388]

As I understand it, e is the back emf which is generated inside the inductor and is acting as a source so e and current are in the same direction.

voltage across inductor is the result of current flow so the direction is always pointing from the lower potential to the higher potential. Since di/dt is negative, V across inductor has a negative value. So the arrow is pointing up, the value is negative.

BTW, if the inductor is ideal, and the current is steady state prior to the transition, there is no voltage across the inductor prior to the current transition (you made reference to that).

It is like a battery connects to a resistor of zero. There is voltage across the resistor if we draw the circuit, with the direction I mentioned, just that its value is zero as V=IxR and R=0.

 

[Roff]

OK, now I understand the minus sign. It does not apply to the voltage across the inductor - it applies to the back emf which is generated by the changing flux. I think this is what you and Jim meant (and said). I have been referring to the voltage across the inductor, which is what ljcox had also referred to. As you said, eblc1388, we are basically talking about two different ways of looking at the same thing.

 

[ljcox]

This is an interesting point. I looked at the text book I used as a student - Network Analysis & Synthesis by Franklin F Kou.

He states (page 175 in the second edition) that an inductor circuit can be transformed into an inductor in series with a voltage source (pointing in the reverse direction to the current). But this voltage is the initial condition, ie. the voltage at t = 0-.

He starts with v = L di/dt and then integrates both sides and then does the Laplace transform resulting in V(s) = s L I(s) - L i(0-) where i(0-) is the current at t = 0- of course. Hence the term L i(0-) is a voltage pointing in the opposite direction to the current. So I suggest that this is the back emf.

But note that the starting point was L di/dt, not -L di/dt.

Len

 

[JimB]

I seem to have opened up a can of worms here.
Having read my old college textbooks on the subject, it is possible to get very confused, it all depends on what is happening to the inductor at the time, if energy is going in or coming out.

In figure 1, the battery is making a current I flow through the inductor, a magnetic field is generated and stores energy.
If we measure the voltage across the inductor, we see a +ve voltage (measuring from the reference point Ref).


If we now close the switch, (figure 2), the battery no longer drives current through the inductor. The magnetic field collapses, putting energy back into the circuit, trying to maintain the current I. To maintain this current flowing in the same direction the voltage from the inductor is now in the opposite direction from before. On our meter (or oscilloscope) we will see a -ve voltage across the inductor with respect to the reference point.

 

[eblc1388]

I seem to have opened up a can of worms here.

You sure did.

Note emf e is not voltage across the inductor, it is a source inside the inductor. The voltage across the resistor RL remains always positive in your drawing.

 

[JimB]

Ambiguous drawing on my part.

Rl is the resistance of the real world inductor, if we are talking about ideal components Rl = 0.

JimB

 

[Roff]

I seem to have opened up a can of worms here.

You sure did.

Note emf e is not voltage across the inductor, it is a source inside the inductor. The voltage across the resistor RL remains always positive in your drawing.

But for the voltage across RL (Vr) to remain positive (which it does), and for e to be a positive voltage (-L*di/dt), the reference point for e has to be the node between L and RL (which I think you have indicated by the arrow next to e). Why use that point as a reference?

 

[eblc1388]

Sorry Ron, can't understand the question. Why did I choose the junction between RL and L as the reference point for e? I guess simply because only the inductor part can acts as a source when the flux changes.

 

[awright]

OK, jin29_neci, does that clarify everything? jin29_neci? jin29_neci?

Hey. You threw the hunk of meat into this den of hungry lions. Where are you now?

By the way, were you curious about back EMF in a pure inductor, or back EMF in a motor?

awright

 

[JimB]

So much for the theory, I spent some time playing with the following test circuit, and recorded the voltage across the coil with an oscilloscope as the switch was opened.

I tried the inductor (a 24volt relay) on its own, adn with a diode across the coil .

Note that the voltage scales vary between the scope traces.

With the coil on its own, unloaded, there was about a 150volt back EMF from the coil when using a 1.2 volt excitation source.
This is why diodes are essential when the relay is driven by a transistor or IC.