Hi Dr.K! Thank you for the video! Everything else was very clear but I have one question. Can you tell me how did you get average current on secondary side of the transformer as Is = (pi*I0)/2?
That's actually the peak value of the current on the secondary side. This comes from the computation of the average value of a fully rectified sine wave. The average value of |Apk*sin(wt)| is 2*Apk/pi, where Apk is the peak value. In this case it Is. Solving for Is using the average, one gets Io*pi/2. If only 1/2 wave rectification is used, then the equation would change to Io*pi. Hope this helps. Best wishes on your design. -Dr. K
@@powerelectronicswithdr.k1017Thank you Dr. K! That makes more sense. I think you mispoke in the video by mistake and I took it literally😅😂. Thank you for your lecture too! Have a great day!
If one switch is connected to ground and the other switch connected to Vds shouldn't it be a voltage between 0 and Vdc ? Like a square wave with an offset of Vdc/2. But you showed it as a -Vdc/2 and +Vdc/2 how it can be -Vdc/2
Hi Dr.K! Thank you for the video! Everything else was very clear but I have one question. Can you tell me how did you get average current on secondary side of the transformer as Is = (pi*I0)/2?
That's actually the peak value of the current on the secondary side. This comes from the computation of the average value of a fully rectified sine wave. The average value of |Apk*sin(wt)| is 2*Apk/pi, where Apk is the peak value. In this case it Is. Solving for Is using the average, one gets Io*pi/2. If only 1/2 wave rectification is used, then the equation would change to Io*pi. Hope this helps. Best wishes on your design. -Dr. K
@@powerelectronicswithdr.k1017Thank you Dr. K! That makes more sense. I think you mispoke in the video by mistake and I took it literally😅😂. Thank you for your lecture too! Have a great day!
If one switch is connected to ground and the other switch connected to Vds shouldn't it be a voltage between 0 and Vdc ? Like a square wave with an offset of Vdc/2. But you showed it as a -Vdc/2 and +Vdc/2 how it can be -Vdc/2
Hi Utku, the capacitor blocks the DC portion of the signal. The result is shifting the waveform. Hope this helps. - Dr. K
Thank you - nice explanation!
You're welcome. -Dr. K
Thank you
You're welcome. Hope it helped with your design.