Hello...

Using the transformer from the other thread I'm trying to confirm with LTSpice the waveforms of an half wave rectifier using an RL circuit. But LTSpice is not outputting the waveforms I was expecting.

This was what I was expecting to obtain, considering an ideal model for the diode:

But this is what I'm getting with LTSpice


(Ignore the voltage drop due to the diode in LTSpice simulation)

I was expecting the current to be kind of displaced to the right, yet starting at t = 0 due to considering the ideal model for the diode and also that the current would be lower due to the resistor.

So, what is wrong in the circuit?

 

I was expecting the current to be kind of displaced to the right...

Why? Where would the current be going in the meanwhile? There's no capacitor.

 

You need a much larger inductor to see a current phase-shift at that frequency, as shown below:
I had to add the R2C1 damping circuit to suppress the oscillations and absorb the inductor energy when the diode stops conducting.

Did you actually do any calculations when selecting those circuit values or were they just random?

 

@crutschow ok. Big thanks. That exactly that I was expecting.

And yes, it was random value for R1 and L1.



Edited;

What means the value for the cap you used? I tried 0.1uF and it didn't output the same result

 

And yes, it was random value for R1 and L1.

Well random values give random results.
That's obviously not the proper way to design a circuit.

What means the value for the cap you used? I tried 0.1uF and it didn't output the same result

Yes, it's 0.1μF.
So what result did you get?

 

If I use 0.1uF, the current waveform is a bit different close to the point where it gets to 0.


If I use 0,1uF, note the coma instead of the dot in above picture, the current looks better at the same point, close to where it is 0A.

 

0,1uF is interpreted as 1uF in LTspice.

Also you are using 1.5H instead of the 2H I used(?).

 

0,1uF is interpreted as 1uF in LTspice.

Also you are using 1.5H instead of the 2H I used(?).

Hum, ok. Didn't know about 0,1 to be interpreted as 1.

And yes, I was testing different values for the inductor to see the changes!

 

I was testing different values for the inductor to see the changes!

Then why do you seem surprised that the results are different?

the current waveform is a bit different close to the point where it gets to 0.

 

Then why do you seem surprised that the results are different?

The difference I was referring to was about current wave form and it was related to the fact that ',1uf' is interpreted differently from '.1uF' and that I didn't know about!
After fixing that detail, everything was just about the same, at least in terms of wave forms!

 

Hello...

Using the transformer from the other thread I'm trying to confirm with LTSpice the waveforms of an half wave rectifier using an RL circuit. But LTSpice is not outputting the waveforms I was expecting.

This was what I was expecting to obtain, considering an ideal model for the diode:

But this is what I'm getting with LTSpice
View attachment 137120

(Ignore the voltage drop due to the diode in LTSpice simulation)

I was expecting the current to be kind of displaced to the right, yet starting at t = 0 due to considering the ideal model for the diode and also that the current would be lower due to the resistor.

So, what is wrong in the circuit?

Hi,

To get a ball park idea what your phase shift would be you can use this as a very rough estimate:
Ph=-57*atan(0.3*L)

where L is the inductor value in Henries and Ph is the phase shift current to voltage in degrees.
That's for f=50 Hz.

 

Yeah, thank you @MrAl ...

That will be handy for sure! Where can I see where from that formula derives in case I need to change the frequency?


For the circuit I was studying, the RL version of the half-wave rectifier, adjusting the measuring point of the Vout, I could obtain exactly what I was expecting.

 

Hi,

That comes from the 'formula' for the phase shift for an inductor L in series with a resistor R.
The two in series produce the impedance:
Z=R+j*w*L

and so the normalized current is:
I=1/Z

and the simplified phase angle is then:
Ph=-atan((2*pi*f*L)/R)

where Ph is phase angle in rads, f is frequency in Hertz, L is inductance in Henries, R is resistance in Ohms,
and that is the AC phase angle which is an approximation to the phase angle for the circuit you are using.
If you add more components though this will change of course.