cURRENT SENSE TRANSFORMER: core area?

[grizedale]

Hello,

I calculate that this current sense transformer has a core with a cross sectional area of 2.66mm^2.

**broken link removed**

Is this right?

 

[FvM]

No idea how you "calculated" it, but it looks like an E5.3 core which has a specified Ae of 2.66 mm².

 

[mtwieg]

Yeah I can't see how you could possibly calculate core area from that datasheet. And I don't see why you would need to know that either. It says that Bmax is 2000G (200mT), along with the secondary inductance/ESR, so that is enough info by itself to derive its max Vt and avoid saturation.

 

[grizedale]

i calculated it by using the equation given in the datasheet.

Knowing the inductance and the maximum allowable flux density wont help you to know whether or not you are going into saturation.

-to know if saturating or not, you need the core area aswell.

 

[mtwieg]

-to know if saturating or not, you need the core area aswell.

Okay I see what you mean. Yes I also calculate 2.66mm^2 (assuming in their equation D is a percent, so 50% would mean D=50). That seems like a reasonable number given their mechanical drawings.

 

[neonwarrior]

Hello everyone can you explain how do you calculate cross sectional area from that datasheet equation??

Bpk = 37.59 * Vref * (Duty_Cycle_Max) * 10e5 / (N * Freq_kHz)

And which equation do you use to know if saturating or not??
Is it this equation?
Bmax = Ls x Imax x 10e4 / ( Ns x Ae)
Where Ls : Secondary inductance. in Henry
Imax : the secondary current in A
Ns : secondary turns
Ae : effective cross sectional Area in cm2
Bmax : Maximum flux density in Tesla

 

[FvM]

There isn't much to calculate. The "formula" tells how flux translates to B for the given core.
You get 1/(37.59*10^3 ) = cross section m². You are simply reverse engineering how the author came to the equation: He has put in the datasheet Ae value of 2.66 mm².

The other question is, how you calculate the core flux (and check it for possible saturation) under different operation conditions. The "application circuit" in the pulse data sheet and the respective equation have some problems.
- you would never connect a single wave rectifier to a current transformer secondary without some additional parallel load or freewheeling means
- the equation doesn't consider the diode voltage drop and is only correct for rectangular current waveforms and discontinuous mode