Ponke
Newbie level 3
Hi all!
I've got some questions about the forward converter.
Now, since it is derived from the Buck converter, the output to input voltage ratio is:
\[\frac{Vout}{Vin} = \frac{Ns}{Np}\cdot {D}\]
However, my question is whether the current ratio is simply, assuming Pin = Pout to be:
\[\frac{Iout}{Iin} = \frac{Np}{Ns}\cdot {\frac{1}{D}}\] ?
Also, another question I have is about Imax on the load side.
Consider that Imax is 10 Amps. If the input VOLTAGE drops by say 30%, how will that affect output current?
My thought is that it shouldn't matter, since Imax on the secondary is constant. An input voltage drop doesn't imply a necessary input current drop, or am I on the wrong trail here?
Thankful for any points in the right direction!
Cheers!
I've got some questions about the forward converter.
Now, since it is derived from the Buck converter, the output to input voltage ratio is:
\[\frac{Vout}{Vin} = \frac{Ns}{Np}\cdot {D}\]
However, my question is whether the current ratio is simply, assuming Pin = Pout to be:
\[\frac{Iout}{Iin} = \frac{Np}{Ns}\cdot {\frac{1}{D}}\] ?
Also, another question I have is about Imax on the load side.
Consider that Imax is 10 Amps. If the input VOLTAGE drops by say 30%, how will that affect output current?
My thought is that it shouldn't matter, since Imax on the secondary is constant. An input voltage drop doesn't imply a necessary input current drop, or am I on the wrong trail here?
Thankful for any points in the right direction!
Cheers!