Quote:
|
Originally Posted by Caleb Fulton
i(t) = {0, t < 0
{Vm / R (1 - exp(-R / L * t)), 0 <= t < Tw
{i1(Tw) exp(-R / L * (t - Tw)), t >= Tw
[/code]
So this is what the waveform for the current looks like when there is a resistor in series with an inductor as a response to a pulse.
|
Caleb,
Nice work on the math derivation. The equation should simplify to
I(t)= (V*(1-e^(-(t*R)/L)))/R
Where t is the time the voltage has been applied
L is the inductance of the coil
R is the series resistance of the coil + all other resistive series losses
e is the natural log
V is the applied voltage
If you look closely you will see Ohm's Law (I=V/R) In the case of a motor at speed, the true applied voltage is the supply voltage minus the counter EMF and the inductance is reduced to the average inductance of the windings as they turn.
For all as rules of thumb, the time constant of an L,R circuit is approx. L/R and the current will be about 66% of maximum in one time constant. Current will reach very near maximum between 5 and 6 time constants. So for Greg's example motor above, L/R=40 exp -6/.0859=.465mSec. A 10% duty cycle at 120 Hz is .833mSec so the current at this duty cycle is not quite at 100% but greater than 66% of maximum. Backing into the math you can see that 6 time constants is a little over 2.4mSec or about 30% duty cycle. The graphs that Greg has referenced show this.
Quote:
|
Originally Posted by Manoel
Actually, I think you are confusing the fact that the Victor will output a green light (full forward) in the 240-255 range, but its output won't necessarily be the supply voltage.
|
Can you elaborate? At full up the Victor should output DC at power supply input minus the series "on" resistance of the FETs.