Inductance question

[Carl WA1KPD]

I am working on a coil for a transmitter PI net
The inductances I need are:

31 mh
16 mh
7.9 mh

I have wired up the coil on a ceramic form checking the inductance with my Aade digital meter. It is one continuous coil with a tap. I get some very odd readings. The sum of the readings is  less then the total. I think a diagram would illustrate it best:
                                             
              I_______31MH________I
              I                               I
----------UUUUUUUUUUUUUUUUUU----------------
              I              I                I
              I___16MH_I_____9MH__I
              I              I                I
In other words the total coil reads 31mh and the two pieces either side of the tap read 16mh and 9mh

Is my meter flukey or my lack of kowledge showing?
Thanks
Carl / KPD

[N3DRB The Derb]

are the tpi different on one side? That would sure do it.

[Carl WA1KPD]

are the tpi different on one side? That would sure do it.

Derb

Both are tight wound enamel. I ran into this because I  wound enough for the 31 mh, looked for a tap from one side for the 16 mh and soldered it in place. Then, looking for the next tap things did not make sense. So I checked it 7 ways to Sunday.

When I got my license inductance in series was additive, so I thought maybe when they dropped the code requirement, that changed too.....

[WA1GFZ]

Try shorting the unused section of coil?

[Tom WA3KLR]

Hi Carl,

If the two sections had no coupling to each other, the total inductance would simply be the 9 mH + 16 mH.

But, I assume they are right next to each other on the same form, so there is a high degree of coupling between them.

For example, if they both measured 9 mH and are wound next to each other, this can also be viewed as a coil with a center tap. If the whole length of both coils couple to each other perfectly, the total inductance would be 36 mH, the total inductance being the result of the turns ratio squared (9 mH X 2 squared).

In your coil, if the two sections had perfect mutual coupling, I quickly calculate that the total inductance would be about 49 mH.  However there probably is very little coupling between the two ends of the coil.  The total result must be greater than 25 mH and less than 49 mH. 

The difference between your result of 31 mH and the 49 mH I calculate must be due to the less than total coupling between all of the turns of the two sections.  So your answer of what you measured seems reasonable.

[Rob K2CU]

Consider the classuic equation for the approximate inductance of a single layer solenoid:

L = N^2  x  A^2  /   (9A +10B), where,

N = number of turns
A = diameter in inches
B = length in inches
L in microhenries

Not knowing the values in this case, I guessed at some typical sizes. For example,

N        A       B        L

20       1       2         13.79
40       1       4         32.65

I figured twice as many turns meant twice as long.

I put this equation in an Excel spreadsheet so it was easy to manipulate the value to get results close to the original.

N        A       B        L

22       1       2.2     15.61
15       1       1.5      9.375
37       1       3.7      29.76


The difference beteen the sum of the individual coils and the total coil is due to the mutual inductance.

Classic equations have the voltage on a coil related to the current as follows:

V  =  L  dI/dt

and when there are two coupled coils, L1 and L2:

V1  = L1 dI1/dt  + M dI2/dt
V2  = L2 dI2/dt  + M dI1/dt

where M is the mutual inductance.

For our combined coil, Vtotal = V1 + V2 and I1 = I2:

Vtot = (L1 + L2 + 2M)  dI/dt

So then M in this example, is 1/2 the difference between the sum of the two coils and the total coil inductance, or 2.39 uH. This all also assumes that we are not near the self-resonance (capacitive effects) of the coils.

Another point is the often preference for shorting unused turns to prevent high voltage arcing (the Tesla effect). Here, if the smaller coil was unused for a higher frequency band, it would not pay to short thr coil. The induced voltage would be related by N2/N1 time V N2 or .6 N2....not a problem for switch arcing. But, the problem would be in the reduction of the inductance of the larger half of the coil. From the two equations above, the resulting inductance (resistive losses aside) would be L = L2 - M^2. The 16 uH half would be reduced to nearly 10 uH.

As Frank said, it would be interesting to see wht your meter reads with one part shorted.
 

[Carl WA1KPD]

Try shorting the unused section of coil?

Hi Frank,
Tried it, minimal difference.

Thanks Tom and Rob--
Looks like you nailed it

Just for my own knowledge I am going to put a known C across each combination and dip it with my GDO and calculate the L
Carl /KPD

[WA1GFZ]

I thought the short would parallel the coupled inductance with a low inductance to reduce the effects???

[AB2EZ]

Frank et. al.

Actually, a short on the unused portion of the coil has the same effect as placing a short on the secondary of a transformer.

See the attached schematic

When current flows in the "primary" (the part of the coil that is not shorted) it produces a magnetic field... which, in turn, induces a current in the opposite direction in the "secondary" (the shorted portion of the coil).

Likewise, the current that flows in the shorted section of the coil (the "secondary") produces a magnetic field that subtracts from the magnetic field being produced by the current in the "primary".

The above is represented by the equations that Rob posted earlier.

Since the coupling between the two parts of the coil is less than 1 ... i.e., only a portion of the total magnetic flux (the "B-field") produced by one part of the coil passes through the core (i.e., the area inside of each of the turns) of the other part of the coil...  the composite coil (i.e., both parts) acts like an ideal transformer in series with a significant "leakage inductance" . This leakage inductance is associated with the portion of the magnetic flux that "leaks" out of one part of the coil, without passing through the core of the other part of the coil. In addition, the equivalent circuit of the composite coil (both parts) includes a "magnetizing" inductance across the primary of the transformer. The values of the series leakage inductance and the parallel magnetizing inductance in the equivalent circuit of the composite coil are given by the formulas that Rob provided.

If you short out the secondary of the ideal transformer in the above equivalent circuit, then the inductance looking into the primary is equal to the series leakage inductance.

If you don't short out the secondary of the ideal transformer (i.e., if you leave the unused portion of the coil open circuited)... then the inductance looking into the primary is equal to the series leakage inductance + the parallel magnetizing inductance.

Stu   

[Rob K2CU]

What happens when you try doing too many things at once ad get interrupted mid stream?  Well, I dropped a term in the equation solving for shorting a term. Solving the two simultaneous equations when you set V2 = 0, yields,

V1 = (L1 - (M^2)/L2) dI1/dt

So, the 15.6 uH would be reduced by only .6 uH, not 5. The minimal reduction amount Carl observed.

For small coils the resistance is minimal, but has some effect, as does the distributed capacitance, etc.


I have always had a problem with the term "leakage inductance." Is spillage involved?  Do we need flux absorbers? Will older transformers required Depends?