VVV
Joined: 26 Nov 2004
Posts: 1379
Helped: 253
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 02 May 2008 18:28 Re: how to get triangular current in a series RL circuit? |
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You need to apply a trapezoidal voltage.
Here is why:
If the inductor was ideal, then a step voltage would create a current ramp:
i(t)=U*t/L. The voltage across the inductor would then be: uL(t)=L*di(t)/dt
Since you have a resistor in series, the voltage drop across it increases with the current. You then need to compensate for that and make sure the voltage across the inductor stays constant, so the current is still a ramp.
To compensate for the resistive part, you need to add a ramp, equal to uR(t)=R*i(t)
So by adding the two you get the trapezoidal or "ramp-on-a-step" voltage.
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