Does anyone know where to get actual or accurate voltage and current waveforms for push pull, half bridge and full bridge PSU topologies with centre tapped secondary's.

I would like voltage and current waveforms for: transformer primary, both secondary diodes, output inductor and load.

Also how to derive the voltage transfer fn from first principles in each case

In response to your other question (no need to post multiples), the average of a trapezoidal shape is (a + b) / 2, where a and b are the lengths of the two bases (parallel sides). Average this over your duty cycle to find the overall average (= DC).

If you want RMS, that's a different matter. RMS = sqrt( [Integral (from t0 to t1) f(x)^2] / [t1 - t0] ) Let f(x) = m*x + c, t0 = 0, t1 = 1. For f(x) = a at t0 and b at t1, c = a and m = (b - a), so f(x) = (b - a)*x + a, a trapezoidal shape. The integrand is then: ([b - a]*x + a)^2 = (b - a)^2 * x^2 + 2*a*(b - a)*x + a^2 = (b^2 - 2*a*b + a^2)*x^2 + 2*(a*b + a^2)*x + a^2 So the integral is: (b^2 - 2*a*b + a^2)*x^3 / 3 + 2*(a*b + a^2)*x^2 / 2 + a^2*x Evaluated from 0 to 1 is: (b^2 - 2*a*b + a^2)*1 / 3 + 2*(a*b + a^2)*1 / 2 + a^2*1 + 0 = b^2 / 3 - (2/3)*a*b + a^2 / 3 + a*b + a^2 + a^2 = b^2 / 3 + a*b / 3 + (7/3)*a^2 = (b^2 + a*b + 7*a^2) / 3

This is divided by (t1 - t0) = 1, and the root is: RMS = sqrt([b^2 + a*b + 7*a^2] / 3)

I don't think there is a real solution to this expression that simplifies this further (e.g., finding the perfect square).

I didn't write this down, so someone may want to check my algebra.

Tim

-- Deep Fryer: A very philosophical monk. Website @

This line should be (allowing implied multiplies) = (b^2 - 2ab + a^2) x^2 + 2 (ab - a^2) x + a^2 i.e., the sign of the a^2 in the second term is negative.

nope, thats right. an interesting exercise is to plot RMS vs (max-min)/avg. for even large amounts of slope, the rms is still pretty close to that of the rectangular approximation.

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