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Current through inductor
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npelov:
I have a question about inductors. Does the current through an ideal inductor rise linearly? If resistance of power supply is 0 and inductor DC resistance is 0. Because all graphs of inductor current I see are exponential. Does the inductor DC resistance make it go from linear to exponential?

I ask because they always use linear equation in boost converters.

I also tried simulation and I saw that decreasing the inductor and supply resistance close to 0 makes the current rise linearly. So I guess if the time is low enough and the current doesn't go close to inductor saturation current then the current is very close to linear (like sin(x) when x -> 0)


Background:
I'm trying to balance a li-ion 3-cell battery and I want to use inductor instead of resistor and a diode to charge the whole pack Toff - just like boost converter. I used boost converter formulas to calculate frequency and max duty cycle. However I also need to know the average current through the inductor so I can calculate how much current I draw from the battery cell. I found this calculator that says Irms = Ipk/sqrt(3). and Ipk = Vind*Ton/L. I took values:
L=68uH
ILsat =0.77A
F=62.5kHz
Ton(70%max) = 0.7*(1/125kHz) = 5.6ms
Vbatt=4.2
Vcesat=1V
VL = 3.2V
Ipk = 0.527A
Irms = 0.304A (max balancing current)
(I did not yet properly calculate Toff - should be enough for the inductor to dissipate all the stored energy)


P.S. the question is about change of current through inductors in time, but feel free to comment on my balance circuit.
Doctorandus_P:
If you put 1Vdc over an inductor of 1H, then the current will increase with 1V/s.
1A after 1s, 2A after 2s, and after a day you will have 86400Amp through the inductor.

The decreasing exponent in your picture is because of the series circuit of a resistor and inductor. At the start of the meaurement there is no current yet (Inductors need time before they will pass current) and all the voltage is over the inductor. For the resitor: (U = I*R = 0*R = 0V) after the circuit has stabilized, all the voltage is over the resistor, and can be calculated with I = U/R.
TimNJ:
Mathematically, the current through an inductor is equal to the integral of the voltage applied across it.

So, for a DC voltage, v(t) = C (a constant).

The integral of C is C*t, that is a linear ramp with a slope dictated by C.

The difference is that you are looking at an RL circuit attached to a voltage source, not just an inductor on is own. In effect, when you first "flip the switch", a large current "wants" to flow through the inductor, but cannot, because it is opposed by the back-EMF of the inductor. This is what causes the current to ramp up logarithmically, and prevents the current from jumping up instantaneously.

Eventually, the current through the inductor tapers off at a value of I = V/R. That is, the inductor essentially becomes a short circuit and does not oppose the flow of current.
T3sl4co1l:

--- Quote from: npelov on June 26, 2018, 10:36:59 am ---I have a question about inductors. Does the current through an ideal inductor rise linearly? If resistance of power supply is 0 and inductor DC resistance is 0. Because all graphs of inductor current I see are exponential. Does the inductor DC resistance make it go from linear to exponential?

I ask because they always use linear equation in boost converters.
--- End quote ---

Yes.

The point of a converter is to be efficient.  It follows that you need a low loss inductor.  As close to ideal as possible*.  In that case, the exponent is so tall that, over the duration of a cycle, it doesn't seem to curve at all, and can be approximated as a linear slope.

*More specifically, ideal enough that other losses (conduction, switching and bypass) dominate.

It may also be interesting to consider the waveform if you didn't have a diode in a buck/boost converter (and unlimited switch voltage rating).  In that case, when the switch turns off, the voltage shoots up, and the inductor rings against whatever capacitance is in the circuit.  You get a decaying oscillation waveform.  You see this normally in a discontinuous mode converter, at very short pulse widths and light load, where the diode hardly conducts, and the inductor just rings down on its own.  Well, at longer pulse widths, it would do the same thing, except for the diode.  What relevance is this?  The rising edge, when the switch turns off and before the diode turns on, is not a straight linear rise -- it's just approximated that way -- but actually a small segment of a very large ringing waveform!

Also, to be perfectly correct, the rise is ONLY linear when the applied voltage is square.  For other waveforms, current is proportional to flux (flux is the integral of voltage with respect to time, the area under the curve).

Also also, assuming the inductance is linear.  If not, it varies with flux (saturation, inductance drops at high flux / current), and the waveforms change accordingly.  In a typical switching circuit, the inductor is selected to remain reasonably linear.  However, transistor and diode capacitance also vary with voltage, so the ringing waveform will typically be distorted.

Tim
npelov:

--- Quote from: T3sl4co1l on June 26, 2018, 06:51:06 pm ---It may also be interesting to consider the waveform if you didn't have a diode in a buck/boost converter (and unlimited switch voltage rating).  In that case, when the switch turns off, the voltage shoots up, and the inductor rings against whatever capacitance is in the circuit.  You get a decaying oscillation waveform.

--- End quote ---

So, technically in a SMPS it's not (only) the inductor resistance that gives the non-linearity. It's the output capacitor  that makes the ringing together with the inductor. So if the inductor is large enough the portion of the wave form will be linear enough for the purpose of calculation.


--- Quote from: T3sl4co1l on June 26, 2018, 06:51:06 pm ---Also also, assuming the inductance is linear.  If not, it varies with flux (saturation, inductance drops at high flux / current), and the waveforms change accordingly.  In a typical switching circuit, the inductor is selected to remain reasonably linear.  However, transistor and diode capacitance also vary with voltage, so the ringing waveform will typically be distorted.

--- End quote ---

Well you have a point there. The inductor is rated at DC1=0.77A and DC2=1.2A. They don't mention which current which is but I remember seeing another datasheet where one of the two currents where induction drops to 80%, so it's probably not constant in the whole current range.

Technically in this case I don't really care about efficiency. I just want to dissipate as less power as heat as possible to avoid heat problems. It's not impossible to dissipate 0.3A(max)*4.2V = 1.26W * 2 cells (you never load all the 3 cells), but I don't want to have temperature rise. Well it's actually 0.3*3.2 = 0.96W in the resistor and 0.3W in the transistor.

And with new batteries it will work in discontinuous mode with very short pulses. Is the ringing going to be a problem. Maybe it's good to put low-esr capacitors for each cell, close to the  switching elements.
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