Buck-Boost transformer model conversion ratio

Thread Starter

BitsNBytes

Joined Mar 22, 2021
10
I am tasked to calculate the conversion ratio; I presume it is the duty cycle of a Buck-Boost. The transformer model of the original circuit is shown in my work.
The turns ratios are correct: 1:D and 1:1/(1-D). The effective resistance is also correct (RL+Ron*D), and the effective voltage is also correct (Vd*(1-D)).
Pushing the voltages and resistances through the transformer I arrive a the last circuit so it becomes a simple voltage divider.
I do believe my voltage divider equation that I boxed is correct. If RL and Ron were to be 0 ohms, we are left with 1/1 which reduces to 1. If the diode effective voltage was 0 then we are left with the ideal buck-boost equation:

(Vg*D)/(1-D) = V, where Vg is input and V is output.

The problem is, how in the bloody hell can this be manipulated to achieve V/Vg?
 

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Thread Starter

BitsNBytes

Joined Mar 22, 2021
10
I figured it out. There is a section of worked problems, and the solution is to divide by Vg. This still leaves the (Vd/Vg) term, but the statement in the similar worked problem said that it is ok. I presume because Vg is considered a "constant".
 
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