Sepic di/dt(on) not affected by "leakage" inductor sizing?

[Flyback]

Hello,
In the (below) ‘coupled-inductor’ SEPIC converter schematic, why does the sizing of the “leakage” inductor (L4) , have no effect on the di/dt(ON), or the di (ON), of the FET current?
Whether the "leakage" inductor is 47uH , or 470nH, it makes no difference to the FET current.

Also, its a constant-off-time converter, but the sizing of the "Leakage" inductor makes no difference to the frequency of operation..why is this?

(L4 is an extra “leakage” inductor added in so as to reduce the input ripple current, and also to reduce the high frequency ringing currents which can occur in the coupled inductor if the leakage is not high enough in value)
SEPIC schematic:
https://i43.tinypic.com/28hg8i1.jpg

 

[MrAl]

Hi,

I had trouble viewing the image, but in general inductance does not affect the turn on of a transistor because the turn on period itself is mostly of high harmonic content and one simple view of an inductor is that is acts like an open circuit for high frequencies.

As you probably know, when you first apply a voltage across an inductor, nothing much happens. With no resistance the slope depends on the inductance, but the absolute current level depends on time. So over short time periods there is very little current change regardless of the size of the inductance. With a small resistance the slope is even less, so takes a little more time to build up the current.

So with a smaller inductor the current slope is steeper, but the initial current change is still zero over a zero time length. So with any size inductor the transistor is still turning on with the same current level. What happens after that though does depend on the inductor size.

 

[Flyback]

OK thanks, When transistor M1 turns on, the input voltage (vin) is superimposed across L4 and L1. That gives a di/dt(on) of X Amps/us {from V=L. (di/dt) }

The fascinating thing is, that if I change L4 from say 1uH to 47uH , then the fascinating thing is that the di/dt seen in that current loop is the same. Its as if the coupled inductor is showing a different inductance to the FET current loop so as to keep (L1 + L4)uH's value the same.

The below simulation shows it in LTspice, and if you run it, and change L4 from 47uH to 1uH then you can see what I mean.

(ltspice is free from linear.com, you may run it by changing it to .asc and opening it then hit the running man icon)

 

[ericgibbs]

hi F,
Tried the asc in LTS, what running time do you find for the sim to complete.?

E

 

[Flyback]

..you only need about 20 ms total, -the load current gets to its desired value and settles after about 15ms (as you know, remember to note where you've put the sim (.asc) file and delete the waveform file when youre done. (as its very big)

 

[simonbramble]

Mine changes drastically. Looking at a timeslice at 8ms, with 1uH inductor I get 52,573A/s and at 8ms with a 10uH inductor I get 13,193A/s

 

[Flyback]

..what about peak current?