what is k factor if coupled inductor?

[treez]

Hello,

What is the K factor of the DRQ127-151 coupled inductor.?.........What is the equation for K factor?

DRQ series coupled inductors
**broken link removed**

 

[FvM]

What is the equation for K factor?

Formula with which input quantities? There's no parameter specified that allows to calculate K.

You need to measure it yourself.

 

[treez]

To measure it myself, i would need to have an expensive LCR meter, i would need to short one coil and measure the inductance of the other.

The series'd inductance given in the datasheet is exactly four times the paralleled inductance, which suggests coupling = K= 1.

K = M/sqrt(L1L2)

You have to assume L1 = L2

 

[FvM]

The series'd inductance given in the datasheet is exactly four times the paralleled inductance, which suggests coupling = K= 1.

This doesn't tell anything about K. You can determine K if you e.g. measure the inductance of a single winding and compare with parallel inductance. For K<1, the parallel inductance will be lower than single coil inductance.

Lparallel = (K+1)/2 * Lsingle

 

[treez]

thanks, i searched google al over to find such an equation but could nto find it.....i cannot understand coiltronics...why they havent given some indication of the k value for these inductors.

 

[albbg]

The theory of mutual coupling says that:

V1=S*L1*I1+S*M*I2
V2=S*M*I1+S*L2*I2

where V1 and V2 are the voltages on the coils and I1 and I2 the currents entering from the dots (if the current is in output of the dot then a sign "-" is required).
Applying the usual circuits rules it's easy to see that, for the connection shown in the data sheet (and assuming L1=L2=L)

0.5*(L+M) = Lparallel
2*(L+M) = Lseries

in your case

0.5*(150+M) = 144.9
2*(150+M) = 579.6

both of them gave the same result (that means the measurements are correct): M=139.8

Since k=M/L, then k=0.932

 

[treez]

sorry i dont see how you got

0.5*(L+M) = Lparallel
2*(L+M) = Lseries

i did some substitutions and get MI1L1 = L1L2I1 + MI2L2 - L1L2I2

...but i dont see how you got the above equations

 

[FvM]

0.5*(150+M) = 144.9
2*(150+M) = 579.6

both of them gave the same result (that means the measurements are correct): M=139.8

Since k=M/L, then k=0.932

Assumes that "rated inductance" in the datasheet equals the single coil inductance "L". This might be the case, but can't be derived from the datasheet.

 

[treez]

Thanks,

did this...

0.5*(L+M) = Lparallel
2*(L+M) = Lseries

come from this....

V1=S*L1*I1+S*M*I2
V2=S*M*I1+S*L2*I2

?
(where s = jw)

 

[chuckey]

If you take two isolated inductors (no coupling between them), then their inductance is .5 or 2 x when in parallel or series, here it seems to be .99 or 3.96. so the excess inductance must be due to the magnetic coupling between them. i.e. .5 + .49 and 2 + 1.96 (using 1 micro henry figures), so coupling (k) is about .49/.5 ~ .98 or 1.96/2 ~ .98. This is what you would expect from pulse transformers.
Frank

 

[treez]

thanks Chuckey, but your analysis assumes that 150uH is the inductance of one coil with the other open circuit........how can you make this assumption when it isn't stated in the datasheet?

I know that scenario would make sense, but you cannot presume that datasheet writer's have acted logically

 

[albbg]

Derivation of formulas is quite easy. Let's assume L1=L2=L

In parallel you will have same voltage across the two windings and half of total current entering the mutual coupled inductors, this means:

V=S*L*I/2+S*M*I/2
V=S*M*I/2+S*L*I/2

both gives the same result V=S*(L+M)*I/2

In series you will have same current but half voltage, that is:

V/2=S*L*I+S*M*I
V/2=S*M*I+S*L*I

both gives the same result V=2*S*(L+M)*I

Since, in general V=S*L*I ==> L=V/(S*I) where "S" is the Laplace operator that in time domain means "d/dt". Due to this:

in parallel the equivalent inductance will be 0.5*(L+M)
in series the equivalent inductance will be 2*(L+M)

FvM, usually "rated inductance" I think means the inductance of a single coil. Otherwise this information has no meaning and will be useless. It's clear that all the values: rated inductance, parallel inductance and series inductance have to be intended as "typical"

chuckey, I don't understand your analysis

 

[FvM]

Rated inductance is primarly "designed" inductance in contrast to measured inductance. Please notice the +/-20% tolerance for the measured value.

If the manufacturer wants to specify a coupling factor, he would explicitely, as others do.

The calculated k of 0.93 is nevertheless in a reasonable range.

 

[albbg]

Not a proof, but I've found an application notes of coiltronics in which the value of an inductor is specified as "rated inductance"

**broken link removed**

Of course in your case, as FvM stated the 20% tolerance apply.

 

[chuckey]

thanks Chuckey, but your analysis assumes that 150uH is the inductance of one coil with the other open circuit........how can you make this assumption when it isn't stated in the datasheet?

I know that scenario would make sense, but you cannot presume that datasheet writer's have acted logically

I said one micro henry - from the link :- **broken link removed**
Frank