Input voltage for calculation of transformer pirmary number of turns

[kender]

Colleagues,

I'm making calculations for a push-pull transformer and following equations in Microchip application note AN1207 (pp.51ff).

\[{\Delta} B = \frac{V_P T_{ON}}{N_P A_{core}}\] eq.175

\[N_P=\frac{V_{DC,min}-V_{Q,on}}{A_{core}f_{PWM} {\Delta} B}D_{max}\] eq.176

where
\[V_{DC}\] is the input voltage
\[D_{max}\] is the max PWM duty cycle (between 0 and 0.5 for push-pull topology)

Why do they use \[V_{DC,min}\] and not \[V_{DC,max}\] in this calculation?
I though that higher voltage across the primary will cause higher magnetic flux. Higher magnetic flux can saturate the core.

Any suggestion, insight or reference is really appreciated!

Cheers,
- Nick

 

[chuckey]

With the lowest DC input, to get the correct output, the current input will be the highest, so this is the worst case, as the DC input rises, the PWM changes and the input current drops.
Frank

 

[kender]

Frank,

Thank you for your response.

With the lowest DC input, to get the correct output, the current input will be the highest, so this is the worst case [...]

Worst case. All right.
Suppose, the transformer, which is designed for \[V_{DC,min}\] is fed with \[V_{DC,toolow}\] such that \[V_{DC,toolow}<V_{DC,min}\]. This is worse than the worst case. What would be the failure mechanism?
I can see why I should use \[I_{P,max}\] (which occurs at \[V_{DC,min}\]) for estimating the copper losses and sizing the wire gauge for the primary. But the equations 175 and 176 in the O.P. don't have current in them.

As far as I understand, the core goes into saturation when \[{\Delta}B\] is too high. Eq. 175 in the O.P. shows the direct relationship between \[{\Delta}B\] and \[V_{DC}\] (or\[V_P\]). Wouldn't \[V_{DC,max}\] be the worst case for core saturation :?: (While \[V_{DC,min}\] is the worst case for primary copper.)

- Nick

 

[rohitkhanna]

..... I can see why I should use \[I_{P,max}\] (which occurs at \[V_{DC,min}\]) for estimating the copper losses and sizing the wire gauge for the primary. But the equations 175 and 176 in the O.P. don't have current in them.

As far as I understand, the core goes into saturation when \[{\Delta}B\] is too high. Eq. 175 in the O.P. shows the direct relationship between \[{\Delta}B\] and \[V_{DC}\] (or\[V_P\]). Wouldn't \[V_{DC,max}\] be the worst case for core saturation :?: (While \[V_{DC,min}\] is the worst case for primary copper.)

- Nick

Magnetic flux & saturation are of course functions of current, and not voltage.
However since we are "charging" and inductor, the equations (#175, 176) inherently have hidden within them this current - as a function of Ton & Vdc !! Notice the dependancy on T-on and see what that tells you.

cheers!

 

[kender]

@rohitkhanna

All right. I see the current.

\[L {\Delta}I=VT\]

Eq.175 can be rewritten as

\[{\Delta}B = \frac{V_P T_{ON}}{N_P A_{core}} = \frac{L {\Delta}I_P}{N_P A_{core}\]

But there still is a direct relationship: \[V_P\uparrow\] causes \[I_P\uparrow\], which causes \[{\Delta}B\uparrow\]. I still don't see the answer to the original question: "Why \[V_{DC,min}\] and not \[V_{DC,max}\] ?"
I ought to be missing a piece somewhere.

- Nick

P.S. By the way, here's a parallel thread with the same question.

 

[rohitkhanna]

...original question: "Why \[V_{DC,min}\] and not \[V_{DC,max}\] ?"
....

I thought that had already been answered in #2

 

[kender]

I thought that had already been answered in #2

It's answered when I say that it's answered.

 

[mtwieg]

These equations have only to do with magnetizing current and magnetizing flux, so the load current does not matter for these equations. Ideally it should not matter whether you use the minimum or maximum input voltage, since the converter's feedback will automatically scale D in order to maintain a roughly constant volt-time product (neglecting rectifier voltage drop). If you factor in voltage drop due to the secondary rectifier diodes, then it should turn out that at lower input voltages, this is not quite the case, and you will need a slightly greater volt-time product on the primary side. That is likely why they use minimum input voltage for the equation. But the difference should be very small.

 

[FvM]

I think, you are reading too much into the rather basic equation 176.

Suppose a given maximum duty cycle (e.g. defined by the controller), a designed input voltage range and a fixed output voltage. Then the maximum duty cycle will be only used with minimum input voltage. At higher input voltages D will be reduced by the controller.

This means that the number of primary turns will be calculated based on Dmax, respective input voltage, core data, PWM frequency. That's all.