push pull / half and full bridge waveforms

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.

The last line should be ( a^2 + ab + b^2 ) / 3

And that would be sqrt( (a^2 + ab + b^2) / 3)

Unless I've made a mistake too. :-)

-- John

Good tech data on SMPS design, w/ voltage & current waveforms, at:

W Letendre

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.

Cheers Terry