Current through inductor - Page 1
Forum section: electronics / beginners

author: npelov
author_rank: Frequent Contributor
timestamp: June 26, 2018, 10:36:59 am
content: 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 allgraphs of inductor current I seeare 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 foundthis calculatorthat says Irms = Ipk/sqrt(3). and Ipk = Vind*Ton/L. I took values:L=68uHILsat =0.77AF=62.5kHzTon(70%max) = 0.7*(1/125kHz) = 5.6msVbatt=4.2Vcesat=1VVL= 3.2VIpk = 0.527AIrms = 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.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 26, 2018, 05:06:27 pm
content: 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.

author: TimNJ
author_rank: Super Contributor
timestamp: June 26, 2018, 05:43:48 pm
content: 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 26, 2018, 06:51:06 pm
content: Quote from: npelov on June 26, 2018, 10:36:59 amYes.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

author: npelov
author_rank: Frequent Contributor
timestamp: June 27, 2018, 12:47:35 am
content: Quote from: T3sl4co1l on June 26, 2018, 06:51:06 pmSo, 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 pmWell 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 27, 2018, 01:13:02 am
content: Quote from: npelov on June 27, 2018, 12:47:35 amNot the input or output filter cap, the switch capacitance.But those too: the output ripple of a buck (or the input ripple of a boost, same thing) is a small segment of the (much slower) resonance between the switched inductor and the filter capacitor. It's reversed every half cycle so you see a small segment of this going up, then going down, forever.So, the output ripple is approximated as a parabolic wave, i.e., the integral of a triangle wave (plus switching noise, passed in through the inductor's parasitic capacitance and the filter cap's ESL). Which is close enough to a sine wave not to care, so it's further approximated as that.The other side (buck input / boost output) has 100% current ripple (i.e., current goes to full, then to zero), so is approximated as a square (at low current ripple %) or trapezoid wave. The voltage waveform, then, is a triangle wave, or a triangle slope one side and a bit of a curve the other side (a hybrid parabola + triangle wave).The triangle segment of that, assumes a constant current source/load. This is unrealistic; more often, it's a resistance of some sort, in which case the triangle side is actually a small segment of an RC exponential wave. An example would be local ceramic bypass caps charged by electrolytics (through their ESR).QuoteWell... presumably one of those is the DC (thermal) rating, and the other is the saturation.I'd suggest shopping at places that, y'know... actually know what they're selling.What is that, RS? Think I've seen crap like that there...Tim

author: npelov
author_rank: Frequent Contributor
timestamp: June 27, 2018, 08:00:55 am
content: You get what you payed forhttps://store.comet.bg/en/Catalogue/Product/22944/https://store.comet.bg/download-file.php?id=445I just picked the cheapest one. Here is an example of another datasheet that specify what DC1 and DC2 are:https://store.comet.bg/download-file.php?id=444It is 90% inductance decrease, not 80%. And DC2 is the temperature rating. But I don't care that much. It probably won't get close to those currents.

author: npelov
author_rank: Frequent Contributor
timestamp: June 27, 2018, 10:41:45 am
content: So, I want to know the rms current through the inductor. I wasn't happy enough with Ipk*sqrt(3), so I dig and I foundthis question:So We don't care about time when we want the current through the inductor. The rms is sqrt( (I1*I1 + I1*I2 + I2*I2) / 3 ), where I1 - lowest current, I2 is peak current. And if I1 = 0 we get Irms = I2*sqrt(3) - discontinuous mode.So (correct me if I'm wrong):Duty cycle for minimum input voltage (Vinmin = 3.8 because I don't want to balance below 3.8V)D = 1-Vinmin*eff/Vout = 1-3.8*0.8/7.6 = 0.6 (60%) - that's the estimated maximum duty cycleI2 or Ipk = ((Vin-Vsw) *Ton)/L = 2.8*0.6 = about 200mA at 125kHzI1 = I2 - (Vin-Vd)*Toff/L = 0.2 - (3.8-0.5)*(0.6*8us)/68uH = 0.2 - 3.3*0.6*8/68 = 0.2-0.03 = 0.17ASo delta I is somewhere between 0.17 and 0.2AWell I actually need to reverse these to calculate Duty cicle for required maximum input rms current ... Or I could just build it, measure the input current, leave some margins and stop pretending I can do math