Charging inductor

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

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.

SneaKSz said:

The simulator settings where wrong, I have a ramp now with a small resistance. My thoughts were right, just the darn simulator.

Click to expand...

And who set up the "darn simulator"?

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 .

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.