# What affect peak current in an RL load rectifier circuit?

so I'm running a half wave RL rectifier in simulink. I don't have much knowledge. May I know what affect the peak current in the half wave rectifier circuit (the theory behind it) and how do I increase it?

I'm using 240V RMS, 50Hz with 25ohm and 0.3H. Below is the waveform simulation of my circuit, I would like to increase the peak current so that it looks more visible

Orange - Current source Blue - Voltage source Yellow - Voltage output

The DC resistance of any coil, L is called DCR. This limits the current according to Ohm's Law. I=V(R)/R.

The impedance of a coil L+DCR is Z(ω)=DCR+ωL

Thus for a voltage source V(ω), Ohm's Law may be applied to complex impedance here.

• I(ω)=V(ω)/Z(ω) for ω=2πf

Current rectification with a diode causes a rapid discontinuity with V=LdI/dt if there is no capacitance. What is not shown on schematics, that ought to be , are component non ideal critical parameters like capacitance in the diode. When the diode is "OFF" it has a voltage controlled capacitance and leakage resistance. Thus when in series between a voltage source and ground , they appear to be a parallel resonant circuit.

See simulation below;

Max C is when diode voltage rated at V = 0 then reduces in a log function of -ve voltage. I'll leave that for another question as it is not simulated in the above power diode. Also know that the larger the area/gap and power rating of the diode, the larger the capacitance. This is also true of FETs but due to physical semi. differences is only consistent within the same family. Thus RdsOn Coss =T is a figure of merit.

Notice that the left plot shows a cutoff voltage is extended past zero crossing because of the phase lag in the L/R=T time constant relative to 1/f. The right plot shows the voltage and current just for the resistor.

If you looked up the datasheet for C(0v) of a diode, you can estimate the off-state resonant frequency from $$\ω=\sqrt{\dfrac{1}{LC}}\$$