I wasn't recommending 1000-fold increase, just noting that the inductor size purchased by and mentioned in one post was 1000 times the size I estimated would have 8A ripple.
Ah, that explains the confusion. I circled back and found the post where jimbob32 mentions he bought 100mH inductors but then goes on to say he experimented with 100uH inductors, so perhaps he can clarify whether the reference to 100mH was a typo or not.
As voltage on capacitor changes, voltage across inductor changes and di/dt changes. But I avoid math involving waveforms and calculus as much as possible. I either calculate with DC applied or steady-state AC sine wave applied.
For a resistor, of course steady DC voltage means steady current.
RC circuit, exponential decay, so I either calculate peak current at beginning or RC time constant to reach 1/e of final voltage.
Inductor, steady voltage is steady ramp in current. Sine wave of some frequency sees reactive impedance, so RMS current from RMS voltage.
Everything you state above summarizes my rudimentary level of understanding.
RL circuit, AC and DC impedance add orthogonally, and Pythagorean theorem gives hypotenuse, magnitude of impedance so current can be calculated. Trig gives phase angle.
Understand the concept but have never made use of the principle.
LC has natural resonant frequency, and amplitude with forced sine wave is attenuated or boosted, to infinity at resonance. Resistance reduces and broadens the peak. For the PWM circuit, I figure it could be over/under/critically damped but I didn't try to analyze that.
If we’re adding a 100uH inductor in series between a switched current source and an input capacitance, does that constitute LC?
I just tried, ignoring change in voltage across capacitor (assume infinite capacitance) to see what inductance would make current ripple the 8A you specified.
Ah, so I think you are saying the inductor feeding inout capacitance is LC and the minimum inductor needed to keep ripple under 8A is ~100uH. I’m not clearly understanding why the capacitance does not enter into the calculation but that’s OK.
The point is you’ve confirmed that addition of 300uH as jimbob32 has done has a good chance of keeping ripple current under 8A…
You mentioned frequency of switching. A square wave could be made up of sine wave fundamental and odd harmonics. Instead of trying to do that hairy math I just considered "on" time causing current ramp. "off" time was different.
I believe the PWM controllers have diode protection to prevent reverse current, so ‘OFF’ is probably more akin to output floating rather than grounded.
Whether current drops to zero each cycle, pumps up to reach a steady state current with voltage ratio determined by switching duty ratio, or voltage rises above that ratio, depends on load and inductor value. This was part of SMPS theory class I took.
When switch is "off", in SMPS circuit the current circulates through a diode and keeps flowing into load as energy stored in inductor decays. I don't know if your PWM motor speed control has components which let that happen or just cause inductor's energy to dissipate. I was imagining the basic "buck" architecture.
I’m not sure either but I think there are some schematics floating around on YouTube…
Just for my understanding, if a transistor is being used to drive current from a 25V battery through an inductor feeding a capacitor, what will happen in the two cases with and without a diode in series?
Inductor keeps pushing current into capacitor so capacitor voltage continues to rise.
Inductor is a decaying resistance so continued forward current means floating output voltage rises above 25VDC battery voltsge.
With a diode present, that output voltage will be able to whip up as far above 25V as needed and when the transistor is next turned on, current through the diode will quickly drive output voltage back to 25V.
Without a diode present, voltage rise on output of off transistor can reach the level the transistor can be damaged.
Is that about right?
en.wikipedia.org
With PWM connecting battery to capacitor, problem was the transistor was acting as a resistor between them and dissipating power as the voltages were brought to same value.
Purpose of inductor is to let transistor switch as fast as possible, voltage drop across inductor instead. Power is then stored in inductor, later transferred to load. Similar to having square wave mechanical motion transferred to a load with spring instead of shock absorber.
Yes, all of this I understand.
100-300uH inductor allows transistor to switch quickly without driving much current and without generating much heat.
Input capacitance charging up more slowly with exponential input current rise for the full ON time (20uS for 25kHz square wave).
Output current to the Microinverter is average DC current, let’s say 10A for 250W of power, so input capacitance voltage is actually decreasing until input current through inductor increases to 10A, at which point it begins to increase.
Let’s assume inductor current increases to 14A or less by the end of ON time.
Inductor keeps supplying now-decaying level of current and as long as that current exceeds 10A, capacitor voltage will continue to rise (along with output voltage protected by diode).
Once inductor current drops below 10A, input capacitor voltage will begin to drop and output voltage along with it.
If we assume inductor current can drop to 0A before the end of OFF time, this will mean capacitor is discharging at 10A and is linearly decreasing both it’s voltage and the output voltage from whatever peak overshoot voltage it reached.
In the ideal scenario where the input capacitor discharged to exactly 25V just as inductor current decayed to 0A and transistor turns back on, this would mean both capacitor and output are both at 25V when transistor turns on, so no instantaneous current through the transistor when it is switched ON and about as efficient as possible.
In the more likely scenario where input capacitor has discharged to less than 25V (likely because of constant 10A discharge current), instantaneous step-up in output voltage (and inductor voltage) will be less than 25V which will ultimately translate to less-than-maximum 20A output current through the full cycle.
And in the most-likely scenario that current through inductor has decayed but not all the way to 0A, the inductor is essentially smoothing current through the OFF phase so that average current of 10A is being maintained with exponential increase to 14A or less during the ON phase and exponential decrease to 6A or more during the OFF phase.
So yes, since average current is irrelevant to the equation, we just need an inductor sized to maintain current below 50% of max ripple over a time which is 50% of overall period…
There is some reality-check needed in terms of max voltage swing of the input capacitor, but am I understanding thus correctly to the first-order?