Why does the back emf effect increase with frequency?

[Just some guy]

Hi,

If I remember correctly, if one applies an alternating voltage across an inductor then the back emf produced by the inductor increases with the frequency of the voltage. However when I decrease the frequency of the voltage to less than 100hz and I plot a trace of voltage against time across the inductor the waveform becomes seriously distorted. This isn't a problem with my signal generator and when I turn the frequency up the distortion dissapears. I'm just wondering why this happens only at low frequencies, and I would have thought the back emf would be larger at high frequencies, distorting the waveform to an even greater degree?

Cheers,
Zachary.

 

[Astronuc]

Please describe the distortion. Is it clipping of the waveform?

Given that $E_L = L\,\frac{di}{dt}$, and one assumes the inductance, L, is constant, then the back emf should decrease with frequency.

 

[Just some guy]

http://www.blorktronics.com/blorky.gif a plot of the current and voltage.

I thought the rate of change of current would get larger with increasing frequency though, as I = sin(wt+phase shift) dt/dt=wsin(wt+phase shift). As f increases so does w and so the back emf gets larger?

That aside, I'm still completely stumped as to what's happening. I'm not using a laminated iron core so there should be a lot of eddy currents, but iirc Lenz's law says that these should be opposite in sign to the applied voltage, and so would act as a resistor and not actually distort the waveform.

Could hysteresis cause this? I don't quite see how but I'm a bit lost for ideas...

 

[Astronuc]

Is the applied voltage sinusoidal? And this doesn't happen at higher frequency?

It certainly looks like a saturation, but the spiking is interesting. Have you made a diagram of the circuit?

Have you tried to calculate the field in the inductor?

The voltage, 2V, is not terribly high.

In eddy current systems, the penetration of the magnetic field increases as frequency decreases. Perhaps there is some mutual inductance?

I'd have to think about the hysteresis effect.

 

[Just some guy]

Is the applied voltage sinusoidal? And this doesn't happen at higher frequency?

It certainly looks like a saturation, but the spiking is interesting. Have you made a diagram of the circuit?

Have you tried to calculate the field in the inductor?

The voltage, 2V, is not terribly high.

In eddy current systems, the penetration of the magnetic field increases as frequency decreases. Perhaps there is some mutual inductance?

I'd have to think about the hysteresis effect.

The applied voltage was sinusoidal when I measured with without the inductor. I haven't drawn the circuit yet but it's just an inductor and an ammeter connected in series to an alternative voltage and a voltmeter across the inductor.

I should also add that the iron core isn't solid - it's in three segments which bend at right angles to form a U, however this effect occurs when I only wind the wire around one of the segments.

I haven't measured the field of the inductor yet (I can work out the field from the flux, right?)

The effect diminishes as I increase the frequency (or decrease the voltage) until it doesn't become noticeable at around 500Hz.

Cheers,
Zachary.

 

[Astronuc]

I should also add that the iron core isn't solid - it's in three segments which bend at right angles to form a U, however this effect occurs when I only wind the wire around one of the segments.

Can you move the inductor to the middle part of the U? If you wrap the wire around the whole U, you don't see this effect?

Try to find a straight segment of iron, and repeat, and you probably won't find this effect.

I am thinking that perhaps when one is using one leg of the U, somehow the other leg becomes involved.

I have asked berkeman to look into this.

 

[Just some guy]

Can you move the inductor to the middle part of the U? If you wrap the wire around the whole U, you don't see this effect?

Try to find a straight segment of iron, and repeat, and you probably won't find this effect.

I am thinking that perhaps when one is using one leg of the U, somehow the other leg becomes involved.

I have asked berkeman to look into this.

Sorry, I dismantled my experiment last week, so I can't rewind the inductor

The effect did persist when it was around the whole U though, so I'm not sure if the shape of the inductor has anything to do with it, though I could always be horribly wrong :uhh:

p.s. I also had it wound around half the inductor (one leg and half the middle of the U) and the effect was still there.

 

[Astronuc]

I believe the shape of the core, U, does have an effect. It's like a transformer core with a large air gap. The voltage induced in one part will act against the voltage in another part, depending on frequency, IIRC.

The only way to be sure is to test the inductor with a straight core.

 

[Just some guy]

ah well, there's no way I can test that. How come it has such a large effect though with such a large air gap? And how come it still has an effect when the wire is would around the entire inductor?

I also decided to plot a current against voltage graph to see if anything interesting shows up. (http://www.blorktronics.com/IVgraph.gif). Anyway, it gives a very strange curve (not least those bumps at the ends), and looks sort of like a madly distorted hysteresis curve, but it's current against voltage and isn't comparing magnetic fields...

 

[Just some guy]

Here's what I get at higher frequencies: http://www.blorktronics.com/curve.gif

 

[Bystander]

(snip)... it's just an inductor and an ammeter connected in series to an alternative voltage ...(snip)

Any other questions? Consider everything when analyzing circuits.

 

[Just some guy]

Any other questions? Consider everything when analyzing circuits.

what? it has a resistance of an ohm, that's around a quarter of the resistance of the wire in the inductor, I don't see how that could cause this...

 

[Bystander]

... and and inductance of ?

 

[Just some guy]

oh, ok, I think I see what you mean - it has an inductor coil in it. I should add this isn't a moving-coil ammeter but something linked up to a digital sensing system. Iirc those things work by passing the current through an inductor and measuring the generated emf. However I

a) don't see why that would affect this, two inductors in series just makes a larger inductor iirc
b) I assume it's designed to have as small an impact on the circuit as possible. i.e. it's inductive reactance is tiny.
c) why would that affect my circuits more at low frequencies?

Cheers,
Zachary.

 

[Just some guy]

Ok, maybe not - here's the current sensor I was using: http://store.pasco.com/pascostore/showdetl.cfm?&DID=9&Product_ID=1357&Detail=1

I doesn't even have an inductor coil, just a 1ohm resistor, so that's that idea out of the window

 

[Astronuc]

Ok, maybe not - here's the current sensor I was using: http://store.pasco.com/pascostore/showdetl.cfm?&DID=9&Product_ID=1357&Detail=1

I doesn't even have an inductor coil, just a 1ohm resistor, so that's that idea out of the window

It could be a Hall effect probe, in which case it might affect the circuit, or be affected by other parts of the circuit if the inductor is nearby.
http://en.wikipedia.org/wiki/Hall_effect
http://hyperphysics.phy-astr.gsu.edu/HBASE/magnetic/hall.html

Please contact PASCO Technical Support for assistance at 1-800-772-8700.

You might wish to ask these guys if the probe is affected by a magnetic field.

 

[Just some guy]

It could be a Hall effect probe, in which case it might affect the circuit, or be affected by other parts of the circuit if the inductor is nearby.
http://en.wikipedia.org/wiki/Hall_effect
http://hyperphysics.phy-astr.gsu.edu/HBASE/magnetic/hall.html

Please contact PASCO Technical Support for assistance at 1-800-772-8700.

You might wish to ask these guys if the probe is affected by a magnetic field.

It's not a hall-effect probe. I know this because their high current sensor is a Hall-effect probe and considerably more expensive ;)

The product description says it just measures the voltage across a 1ohm resistor, so I don't see why they would include an inductor (there's hardly room anyways, the thing is tiny).

Cheers,
Zachary.

 

[Bystander]

http://www.blorktronics.com/blorky.gif a plot of the current and voltage.(snip)

That is slow clipping --- you might be looking at a capacitor breakdown in the output stage of your oscillator; no load, everything looks fine, draw current and charge a bypass or isolation capacitor to 2 V where it breaks down. However, once capacitors start "leaking," they usually go ahead and burn to a short or wide open state very quickly. At higher frequencies you don't get sufficient charging to break down, and everything looks normal.

 

[berkeman]

http://www.blorktronics.com/blorky.gif a plot of the current and voltage.

I thought the rate of change of current would get larger with increasing frequency though, as I = sin(wt+phase shift) dt/dt=wsin(wt+phase shift). As f increases so does w and so the back emf gets larger?

That aside, I'm still completely stumped as to what's happening. I'm not using a laminated iron core so there should be a lot of eddy currents, but iirc Lenz's law says that these should be opposite in sign to the applied voltage, and so would act as a resistor and not actually distort the waveform.

Could hysteresis cause this? I don't quite see how but I'm a bit lost for ideas...

The back emf of an inductor is what gives it a rising impedance characteristic with frequency. Look at the equation for the impedance of an inductor versus frequency, and look at the output impedance of your signal source to understand the amplitude of the voltage across the inductor.

And as the impedance of the inductor drops enough at low frequencies, you will be pushing enough current through the inductor to cause saturation, and the incremental inductance will drop and you'll get closer to an impedance equal to the winding resistance. I think that's consistent with the 'scope shots that you've posted.

Does that make sense?

 

[Just some guy]

The back emf of an inductor is what gives it a rising impedance characteristic with frequency. Look at the equation for the impedance of an inductor versus frequency, and look at the output impedance of your signal source to understand the amplitude of the voltage across the inductor.

And as the impedance of the inductor drops enough at low frequencies, you will be pushing enough current through the inductor to cause saturation, and the incremental inductance will drop and you'll get closer to an impedance equal to the winding resistance. I think that's consistent with the 'scope shots that you've posted.

Does that make sense?

Umm, sort of. Are you saying that when the voltage peak collapses to practically a straight line the inductor has saturated?

Sorry if I'm being dense, you've been a tremendous help

 

[berkeman]

Umm, sort of. Are you saying that when the voltage peak collapses to practically a straight line the inductor has saturated?

Sorry if I'm being dense, you've been a tremendous help

Don't worry, you aren't being dense. There are just several things going on at at the same time. Looking at your original 'scope trace, the voltage rises quickly because the excitation frequency is low enough that the back emf effect is small, and the impedance of the inductor is low. When the output voltage snaps back down to a constant level (with the corresponding constant current) for the rest of each half-cycle, it appears that the voltage source of the signal source has gone into a fold-back current limit. It may not be an explicit fold-back circuit in the output stage of the signal generator -- it could be something as simple as the outputs saturating, and the voltage regulator on the output stage going into fold-back.

Often to understand the effects of saturation on an inductor, you'll use a square wave generator instead of a sine wave. When you apply a square wave voltage excitation to an inductor, the current through the inductor ramps up linearly (do you know why? what's the equation?) at first because of the constant voltage across it. But as the current gets close to the saturation current of the inductor (where the B versus H curver starts to tilt over), the current starts rising at a faster rate than earlier. Eventually you will have to hit the limit of something somewhere, and it often results in a current limited by the output resistance of the voltage source and the DC resistance of the coil widing. When you are testing an inductor to see what its saturation current is, you will start with a high-frequency voltage square wave, and turn down the frequency until you start to see just the beginnings of a non-linear current ramp starting at the end of each half of the square wave.