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Charging inductor

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SneaKSz

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Hi all,

Well in SMPS we have the coil the is being charged when a voltage stands across it ( eg boost). Thereby the current ramps di/dt.

If I use an inductor with a resistance in series with it, τ=L/R and the inductor will charge so Vl = 0V and all the voltage stands across R. The current flows exponential.

My question is now with SMPS, the mosfets ( in series with the inductor) RDSon is low (<1ohm), why doesn't the current flow exponential but it ramps up when the mosfet conducts?

Sometimes there's a current sense resistor in series with the inductor, and still the current ramps up, I'm a little confused.

Hopefully someone can help me out!

Edit : posted the question in the wrong section of this forum, my bad.

Kind regards
 
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It does have an exponential ramp due to the resistance. It's just that the time-constant (L/R) is so long, due to the small resistance and the relatively large inductance, as compared to the on-time of the MOSFET that the ramp appears linear.
 
The simulator settings where wrong, I have a ramp now with a small resistance. My thoughts were right, just the darn simulator.
 
The simulator settings where wrong, I have a ramp now with a small resistance. My thoughts were right, just the darn simulator.
And who set up the "darn simulator"? :D
 
hahaha, me off course, but the problem with multisim is that it initially, automatically calculates the DC operating point ( default), so you'll never see a capacitor/ inductor charge if this setting is on. If you set the initial conditions to zero, you'll see the cap/ inductor charge, and that's what I did :D.
 
I understand. It can take awhile to learn all the subtleties of a simulator. Couldn't resist giving you a little tweak about it though. ;)
 
Hi,

The observed ramp for an inductor is I=t*E/L where E is the applied step voltage and L is the inductance and t is time.
If we analyze the RL circuit for current we get an equation with R and L and t, and if we take the limit as R goes to zero we get I=t*E/L, the very same equation. Thus for small R the inductor looks like a pure inductance. It does depend on the application though whether or not this holds true.
 
Hi,

ye I know, I found it out with the formule Ul =E(1- e^(-t/tau)), where E is the applied voltage, and t=L/R, if r=0, Ul=E so Ul=L.di/dt and the current ramps up, and for low R values or high inductor values ( tau), this equation will be correct.
 
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