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push pull / half and full bridge waveforms

R

reggie

Jan 1, 1970
0
Hi all,

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

Thanks

reggie
 
T

Tim Williams

Jan 1, 1970
0
http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html

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
 
J

John O'Flaherty

Jan 1, 1970
0
http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html

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

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.
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

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

And that would be sqrt( (a^2 + ab + b^2) / 3)
I don't think there is a real solution to this expression that simplifies
this further (e.g., finding the perfect square).

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

WJLServo

Jan 1, 1970
0
Hi all,

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

Thanks

reggie

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

http://www.smps.us/Unitrode.html

W Letendre
 
T

Terry Given

Jan 1, 1970
0
John said:
http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html

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


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.

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


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

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


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

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


Unless I've made a mistake too. :)

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
 
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