Voltage across inductor

[elec123]

Hi,

Please find the circuit attached.

I know the voltage seen across the inductor will be a triangle wave and the average voltage across it is zero volts.

Could anyone explain to me why this is so.

Thanks

p.s Vs is 15-30V, 2A
p.p.s R=0.2ohm
p.p.p.s The mosfet used is IRF511

 

[Styx]

one thing hte diode is in the wrong place. As it stands the FET will blow up. The diode needs to contain the Inductor in its free-wheel path as well

 

[ljcox]

one thing hte diode is in the wrong place. As it stands the FET will blow up. The diode needs to contain the Inductor in its free-wheel path as well

I agree. However, in order to clarify this point, he means to move the anode of the diode from the top of the inductor to the bottom. As drawn, the diode does nothing. It must be across the inductor in order to protect it from the back EMF of the inductor when the FET is switched off. Otherwise, it will destroy the FET.

I assume that Vs is DC and there is a squarewave signal applied to the Gate.

If so, the voltage across the inductor will be exponential. The fall time (ie. when the FET is switched on) will be similar to that of the rise time since the lamp is in series with the inductor in both cases.

If the frequency of the signal applied to the Gate is high enough, the waveform will be approximately triangular since the FET is switching faster than the time constant (L/R) where R is the resistance of the lamp.

But if you reduce the frequency, the expontential curve will be evident on both the rise and fall.

Hope this helps

 

[Optikon]

Hi,

Please find the circuit attached.

I know the voltage seen across the inductor will be a triangle wave and the average voltage across it is zero volts.

Could anyone explain to me why this is so.

Thanks

p.s Vs is 15-30V, 2A
p.p.s R=0.2ohm
p.p.p.s The mosfet used is IRF511

Whatever the waveform is you create across your inductor (determined by your circuit and operating parameters), the average voltage will be zero across the inductor because the inductor cannot support a non-zero DC voltage (DC = Average in this switching case). Why because an inductor looks like a short at DC. That's the theoretical argument, in practice the inductor has a DC resistance which, will support a DC voltage across it.