I built a flyback two-way output circuit in matlab. The inductance and turns of the flyback primary side and all secondary sides found that when R1 is equal to R2, the rate of change of the two secondary side currents is the same (green line and blue line ) But when R1 is not equal to R2, the current rate of change between the two is different. Isn’t the formula for the rate of change of inductor current U=L*(di/dt)? Since the voltage values of the secondary windings are equal, the two secondary Shouldn't the rate of change of current of the sides always be equal?
Dayton ポストする December 18, 2020
With current, the load affects the voltage. Don't you even consider this?
Ashanti ポストする December 18, 2020
The current change does affect the voltage, but isn't the voltage induced on the coupled inductor always equal?
Dayton ポストする December 18, 2020
The voltage drop of the current on the resistance should be calculated in the U of your formula.
Kellan ポストする December 18, 2020
The change rate of the secondary inductor current is directly related to the load, and has nothing to do with the output voltage. The only relationship is the load of each road. The heavier the load, the greater the rate of change until DCM. The lighter the load, the more it tends to 0.
