Analyzing circuits with tapped inductors

M

M. Hamed

Jan 1, 1970
0
I am currently looking at the Hartley oscillator (yes I'm getting serious about that stuff), and I currently find it hard to analyze circuits with tapped inductors. I thought about treating them as two separate inductors, butthen there is mutual inductance which complicates matters.

Does anyone know of a good text that does this kind of analysis, possibly with worked examples and preferably in the context of semiconductor electronics ?

Thanks a million.

P

Phil Allison

Jan 1, 1970
0
"M. Hamed"

I am currently looking at the Hartley oscillator (yes I'm getting serious
about that stuff), and I currently find it hard to analyze circuits with
tapped inductors.

** The Hartley oscillator will operate with independent inductors.

I thought about treating them as two separate inductors, but then there is
mutual inductance which complicates matters.

** With independent inductors in series, the impedance ratio determines the
voltage ratio at a given frequency.

With a tapped inductor, it is the turns ratio that does it.

.... Phil

M

M. Hamed

Jan 1, 1970
0
** With independent inductors in series, the impedance ratio determines the

voltage ratio at a given frequency.

With a tapped inductor, it is the turns ratio that does it.

Phil, to analyze as series inductors, doesn't that assume the current going into the tap is negligible? Is it always a safe assumption?

For example in this circuit, http://www.mathworks.com/matlabcentral/fx_files/20391/1/Hartley.JPG I couldn't really figure out why I can assume there is small current going from ground into the tap. Thanks.

P

Phil Allison

Jan 1, 1970
0
"M. Hamed"
Phil, to analyze as series inductors,

** Not possible if the inductor **IS* in fact TAPPED.

For example in this circuit,
http://www.mathworks.com/matlabcentral/fx_files/20391/1/Hartley.JPG I
couldn't really figure out why I can assume there is small current going
from ground into the tap.

** Because I posted a reply to your question does NOT mean you can turn
around and ask me anything you like.

ALL questions MUST be posted to the group, unless you only want
clarification of what some poster has said.

Usenet 101 - pal.

..... Phil

J

Jamie

Jan 1, 1970
0
M. Hamed said:
Phil, to analyze as series inductors, doesn't that assume the current going into the tap is negligible? Is it always a safe assumption?

For example in this circuit, http://www.mathworks.com/matlabcentral/fx_files/20391/1/Hartley.JPG I couldn't really figure out why I can assume there is small current going from ground into the tap. Thanks.

I don't know what spice program you're using, if you are using one at
all, but with LTspice the L value also reflects the turn ratio.

There is another L model in Ltspice you can use which will allow you
to better specify the coil properties. Just look at the help file for that.

If there is a way to do this with a directive using the default L
model to specify turn ratio, I don't know what it is. Maybe a LtSpice or
spice in general expert can answer that question. I had to use the other
model one day to do what I needed. (John Chan) model, that allows you to
specify the turns with out changing my L value.

Jamie

F

Fred Abse

Jan 1, 1970
0
I am currently looking at the Hartley oscillator (yes I'm getting serious
about that stuff), and I currently find it hard to analyze circuits with
tapped inductors. I thought about treating them as two separate inductors,
but then there is mutual inductance which complicates matters.

Two inductors, which are magnetically coupled go as:

Ltotal = L1+L2+2*M, where M is the mutual inductance, one to the other.

Where coupling is unity (like two inductors wound on a common core which
has very low magnetic leakage, toroid or pot core for example),
M=sqrt(L1*L2).

Hence, for tightly coupled inductors, Ltotal = L1+L2+2*sqrt(L1*L2).

This brings us to the concept of coupling coefficient, K, which is
commonly used in SPICE analyses. It is simply M/sqrt(L1*L2), which is
always <= unity.

If you are interested in voltage ratios, it is easier to work with turns
ratio. V1/V2= N1/(N1+N2), for a tapped inductor.

Bear in mind that inductance is proportional to turns squared.

F

Fred Abse

Jan 1, 1970
0
I don't know what spice program you're using, if you are using one at
all, but with LTspice the L value also reflects the turn ratio.

More accurately the turns ratio squared. L1/L2=(N1/N2)^2

There is another L model in Ltspice you can use which will allow you
to better specify the coil properties. Just look at the help file for
that.

If there is a way to do this with a directive using the default L
model to specify turn ratio, I don't know what it is. Maybe a LtSpice or
spice in general expert can answer that question. I had to use the other
model one day to do what I needed. (John Chan) model, that allows you to
specify the turns with out changing my L value.

The Chan model requires a knowledge of the core hysteretic properties:

The parameters Bs, Br, and Hc define the permeability of the core, with
respect to the magnetizing force, H. L is then derived from that, the
number of turns, the magnetic path length, the effective area, and the air
gap.

For an air cored coil, Bs is infinite, Br=0, Hc=0, Lg=Lm. Hence the Chan
model is not suitable.

The inductance of an air cored solenoidal coil depends not only on turns
squared, but on its diameter, length, and winding pitch.

P

Phil Allison

Jan 1, 1970
0
"Fred Abse"
Where coupling is unity (like two inductors wound on a common core which
has very low magnetic leakage, toroid or pot core for example),
M=sqrt(L1*L2).

Hence, for tightly coupled inductors, Ltotal = L1+L2+2*sqrt(L1*L2).

This brings us to the concept of coupling coefficient, K, which is
commonly used in SPICE analyses. It is simply M/sqrt(L1*L2), which is
always <= unity.

If you are interested in voltage ratios, it is easier to work with turns
ratio. V1/V2= N1/(N1+N2), for a tapped inductor.

Bear in mind that inductance is proportional to turns squared.

** Put far more simply, if you measure the overall L figure and you have the
turns ratio at each tapping - the L value of each tap goes as the square of
the turns ratio.

Egs:

For a centre tap L = 1/4 of the total L.

For a 10% tap, L = 1/100 of the total L.

Too simple.

.... Phil

F

Fred Abse

Jan 1, 1970
0
** Put far more simply, if you measure the overall L figure and you have
the turns ratio at each tapping - the L value of each tap goes as the
square of the turns ratio.

Egs:

For a centre tap L = 1/4 of the total L.

For a 10% tap, L = 1/100 of the total L.

Too simple.

Quite so.

M

M. Hamed

Jan 1, 1970
0
** Because I posted a reply to your question does NOT mean you can turn
around and ask me anything you like.
ALL questions MUST be posted to the group, unless you only want
clarification of what some poster has said.
Usenet 101 - pal.

Noted. That wasn't my intention though.

I'm not sure why you're assuming a small current going from ground into
the tap -- I don't see it in that schamatic.

That was just a misunderstanding of what Phil said about series vs. tapped inductors.

M

M. Hamed

Jan 1, 1970
0
** Because I posted a reply to your question does NOT mean you can turn
around and ask me anything you like.
ALL questions MUST be posted to the group, unless you only want
clarification of what some poster has said.
Usenet 101 - pal.

Noted. That wasn't my intention though.

I'm not sure why you're assuming a small current going from ground into
the tap -- I don't see it in that schamatic.

That was just a misunderstanding of what Phil said about series vs. tapped inductors.

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