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## Main Question or Discussion Point

Take as an example a 120VAC 60 Hz primary, soft iron core transformer.

Fed it with 120VAC with a 2VDC offset, what will happen?

Fed it with 120VAC with a 2VDC offset, what will happen?

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Take as an example a 120VAC 60 Hz primary, soft iron core transformer.

Fed it with 120VAC with a 2VDC offset, what will happen?

Fed it with 120VAC with a 2VDC offset, what will happen?

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berkeman

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The offset pushes you to one side of the hysteresis curve. If you push too far with DC, you will start to saturate the core at one end of the input current swing, which will distort the transfer function.Take as an example a 120VAC 60 Hz primary, soft iron core transformer.

Fed it with 120VAC with a 2VDC offset, what will happen?

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berkeman

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I'm not sure I understand what you're asking, but the best way to think of it (for me) is to look at the B-H curve, and picture the excitation waveform going in on the horizontal axis, and the resulting flux (coupling to the secondary) on the vertical axis. If you offset the excitation waveform with a DC voltage (say, pushing the excitation to the right on the x-axis for H), then you see how you will start to approach the saturation of the core for postive values of the AC excitation. That will cut down the coupling to the secondary at that point, affecting the output waveform, and will modulate the impedance seen by the driving circuit.

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That helps. To solve, I should start with the excitation current rather than the voltage.

Dropping constants that are particular to a given transformer:

H = i(t)

B = Hu

Phi = B

e(t) = d(Phi)/dt

This is some background to my musings. Model a transformer primary as an inductor with series resistance due to wire resistance.

If you assume that mu is constant, B/H=constant, then any DC voltage will end up across the resistance, with zero volts across the inductive part.

Without the resistance, d(Phi)/dt is constant, and the current ramps to infinity (where mu has been assumed constant, of course).

But where the B-H curve is other than a straight line this may not happen because the equation e=f(i) is no longer a linear equation.

This is an issue that's always bothered me when driving transformers in the kilohertz with an H, or half wave bridge where the primary resistance is very small. The slightest imbalance in duty cycle is an DC voltage offset. the flux should unavoidably ratchet into saturation in a matter of seconds if not milliseconds.

Dropping constants that are particular to a given transformer:

H = i(t)

B = Hu

Phi = B

e(t) = d(Phi)/dt

This is some background to my musings. Model a transformer primary as an inductor with series resistance due to wire resistance.

If you assume that mu is constant, B/H=constant, then any DC voltage will end up across the resistance, with zero volts across the inductive part.

Without the resistance, d(Phi)/dt is constant, and the current ramps to infinity (where mu has been assumed constant, of course).

But where the B-H curve is other than a straight line this may not happen because the equation e=f(i) is no longer a linear equation.

This is an issue that's always bothered me when driving transformers in the kilohertz with an H, or half wave bridge where the primary resistance is very small. The slightest imbalance in duty cycle is an DC voltage offset. the flux should unavoidably ratchet into saturation in a matter of seconds if not milliseconds.

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berkeman

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I don't see the "ratchet" effect that you are worried about. What causes ratcheting?

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sophiecentaur

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I have in mind a square wave input across a transformer. For a region on the B-H curve where mu is fairly constant, the unloaded current is a triangular wave. The peak value might be picked to be about half the saturation current as the contributions to flux density, B from the additional primary and secondary currents due to secondary loading, cancel.I don't see the "ratchet" effect that you are worried about. What causes ratcheting?

We can introduce two equivalent offsets errors to the primary voltage. One is to add a DC offset voltage to the square wave input. The equivalent is to bias the duty cycle by a small amount, say 49% push and 51% pull.

So we can examine one offset and and it looks like the other where both are small. An additional positive DC offset voltage is a constantly increasing current curve superimposed on the triangular wave. Given enough cycles, without compensation, the offset current would be of comparable to the triangular peak current. This is what I meant by ratcheting; the current waveform is a triangular waveform superimposed on a ramp.

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sophiecentaur

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berkeman

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EDIT -- Ack, aced out by the centaur again!

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sophiecentaur

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Yes - I don't see where he's coming from on this one. DC volts cause DC current afaik.

EDIT -- Ack, aced out by the centaur again!

(Early Worm again.)

My earlier point was that you could eliminate even this current if the overal dc volts are allowed to float, to compensate.

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I was considering an ideal transformer for simplicity. In this case, the current waveform is a triangular waveform superimposed on a ramp.For the DC offset voltage in your 3rd paragraph, there is no "constantly increasing current". For DC there is no inductance, only the winding resistance. The DC currrent that results from the DC offset voltage is determined by the winding resistance.

I agree with you both about the effects of series, or winding resistance.

While still assuming a relatively constant mu, inclusion of resistance means, to first approximation at least, a triangular wave is superimposed on an L/R time constant curve.

After all the expaining of the problem, it's become obvious to me how series resistance operates. I_offset = V_offset / R_series. H=I_offset*turns/effective_magnetic_path_length. So we can know how much effective offset flux density we can expect.

But I'm still curious as to whether with resistance or not, the AC component corrects the DC component. I haven't had time to look at it properly between work and a large fraction of required down time.

I'm going to try the family of curves B = c1[atan(c2 H)]^k to approximate core material (no hysteresis) --and hope it's analytical.

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sophiecentaur

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Isn't it just an example of a DC cut (high pass) filter function? If the voltage drive to the primary is decoupled to DC (very easy to achieve with some feedback) then the DC component is, by definition, eliminated. This will bring the part of the cycle that lasts longer will be 'nearer, the zero line than the shorter part. (i.e. it will drift so that the mean value is zero)But I'm still curious as to whether with resistance or not, the AC component corrects the DC component. I haven't had time to look at it properly between work and a large fraction of required down time.

I'm going to try the family of curves B = c1[atan(c2 H)]^k to approximate core material (no hysteresis) --and hope it's analytical.

If you don't eliminate the DC volts in the primary circuit then there will be a DC current. Isn't this trivial in the ideal case?

If the dc resistance of the winding and also the source are very low then a huge / embarrassing dc current could flow.

I don't think the hysteresis of the core can affect this, basically, because the core is ac coupled to the source and winding - you can't force dc to flow in the windings due to the core (can you?). I would agree that you could imagine an 'effective current' to be flowing somewhere due to the non linearity - as with a rectifier circuit following a transformer - but where? Isn't that 'effective current ' just another way of describing the loss effects of the hysteresis (i.e. in - phase currents in the core domains)

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I dunno, sophiecentaur. I always end up thinking is circles about this, so I'm trying to resort to mathematical rigor, if possible.

I got this far:

Assuming that B=atan(H) for simplicity then dB/dt = 2(dH/dt)/(1+H^2). This is a "bump function" and it's a polynomial which is nice.

Normalizing all transformer constants like core area and magnetic path length and anthropic units, then I get

[tex]\left( \frac{1}{1+i^2 (t)} \right) \frac{di}{dt} = e(t)\ .[/tex]

e(t) is supposed to be a sine wave with some comparably small DC offset so,

[tex]e(t) = a\ sin(\omega t) + b \ \ and \ \ b<<a\ .[/tex]

The solution is not obvious to me.

A hysteresis curve looks like the integral of a smooth bump function. There are other bump functions that might be easier to use in a differential equation like

[tex]y = e^{\frac{-1}{(1-x^2)}\ ,[/tex]

[There's supposed to be a negative sign in front of the numerator of the fraction so it's negative one instead of one, but for some reason, it won't edit no matter how many times I resend.]

that should also approximate a BH curve with all the general characteristics that might actually be easier to use in a differential equation. Any ideas?

-------

I'm afraid I jumped way too many steps, and I should rewrite the whole thing, but this really does all follow from using the equations for inductors with closed cores.

I got this far:

Assuming that B=atan(H) for simplicity then dB/dt = 2(dH/dt)/(1+H^2). This is a "bump function" and it's a polynomial which is nice.

Normalizing all transformer constants like core area and magnetic path length and anthropic units, then I get

[tex]\left( \frac{1}{1+i^2 (t)} \right) \frac{di}{dt} = e(t)\ .[/tex]

e(t) is supposed to be a sine wave with some comparably small DC offset so,

[tex]e(t) = a\ sin(\omega t) + b \ \ and \ \ b<<a\ .[/tex]

The solution is not obvious to me.

A hysteresis curve looks like the integral of a smooth bump function. There are other bump functions that might be easier to use in a differential equation like

[tex]y = e^{\frac{-1}{(1-x^2)}\ ,[/tex]

[There's supposed to be a negative sign in front of the numerator of the fraction so it's negative one instead of one, but for some reason, it won't edit no matter how many times I resend.]

that should also approximate a BH curve with all the general characteristics that might actually be easier to use in a differential equation. Any ideas?

-------

I'm afraid I jumped way too many steps, and I should rewrite the whole thing, but this really does all follow from using the equations for inductors with closed cores.

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