# Impedance Of Coil

1. Nov 1, 2007

### tonytuong

hi,

I have tested impedance of transformer coil and I found out that if I would like to have smaller input impedance then I have to reduce the diameter of magnetic wire.

I understand that If i reduce the cross setion of the wire then I will reduce the DC resistance.

I dont understand why impedance increase.

Thanks for any help in advance.

Regards,

2. Nov 1, 2007

### ranger

If you want to decrease the resistance of your wire, just use a larger diameter wire. The dc resistance of the wire will vary inversely with the cross sectional area of the wire. You should not confuse yourself about thinking of the ac resistance of the wire. This would never be constant. It varies with frequency. At low frequencies, the wire will use its entire cross section as a transport medium for charge, At higher frequencies, a increasing magnetic field at the center will deflect charges away from the center and increase the charge density around the perimeter of the conductor. This will increase the ac resistance of the wire.

Just use a wire gauge with a larger diameter to reduce the resistance of the coils' wire.

Last edited: Nov 1, 2007
3. Nov 1, 2007

### subtech

The impedance of a transformer(you didn't specify which type) is a function of several things. Listed below are but a few things that affect transformer impedance.

Size and physical arrangement of the primary and secondary conductors
Type, size, and material chosen for core material(Core material varies widely)
Number of windings per volt on either side of the transformer

It goes on and on, but you get the idea.
The main thing to remember about impedance is that anything that would change the
inductance value of a coil (or coils) will change the reactance and hence the impedance.
Hope this helps.

4. Nov 2, 2007

### tonytuong

Thanks ranger,

Subtech,

I am doing a project call Rotational variable difrential tranducer.

I am using 38 HML magnetic wire and have a problem to fit them in to the stator.

I reduce to 40 HML for both Pri and secondaries. it fit ok but impedance goes down.
Il likes Ranger mention AC impedance should be propotional to cross setion and is oposite to DC resistance.

Assume fix frequency( for my case 3500Hz) fix material for wire, stator, rotor. Would you know any theory or fomular to back up that reducing cross setion will decrese impedance?

Regards,

5. Nov 3, 2007

### lrsabug

i would like to share some ideas.

i agree with ranger. increasing the cross-sectional area of the WIRE will decrease the DC resistance of the coil.

R = ρ*L/A
ρ = resistivity of the material used for the wire
L = length of the wire
A = cross-sectional area of the wire

and then the inductance of the coil will decrease when the cross-sectional area of the COIL is decreased or if the number of turns is decreased. this is in virtue of the equation

L = μ(N^2)A/s
A = cross-sectional area
s = length of the coil
N = number of turns
μ = permeability of free space

because the magnetic field would not be uniform if N is decreased, it is safer if you make the cross-sectional area of the COIL smaller. @tonytuong: maybe this may help for your problem.

the inductance will affect the AC resistance <the impedance> because the impedance of the ideal transformer <ignoring DC resistance of the wire> is jωL. this is the case for LONE COILS.

j = sqrt(-1)
ω= angular freq
L = inductance

we got this in class. the proving part for here is too long :(

the internal resistance of the wires used in the inductor can be modeled as a small resistance in series with an ideal inductor

actually the input impedance reflected at the primary is dependent on the LOAD connected to the secondary side.

using the equation (1) V2 = (n2/n1)*V1
V2 = voltage in secondary || n2 = turns at the secondary side
V1 = voltage in primary || n1 = turns at the primary

equation (2) I2 = (n1/n2)*I1
I2 = current through the secondary || n2 = turns at the secondary side
I1 = current through the primary || n1 = turns at the primary side

equation (2) should be agreeing with (1) because POWER IN = POWER OUT.
try to multiply these two equations to get V1*I1 = V2*I2!

and then try to DIVIDE equation (1) with (2) and you shall get:
V2/I2 = (n2/n1)^2*V1/I1 ===========> V1/I1 = (n1/n2)^2*V2/I2
V1/I1 = impedance when viewed from the primary side (Z1)
V2/I2 = impedance when viewed from the secondary side (Z2)

so from the above equations we can conclude that the input impedance of the TRANSFORMER when VIEWING FROM THE PRIMARY SIDE is DEPENDENT on the impedance on the secondary side, which is (n1/n2)^2*Z2!

this means that the transformer <along with the load on the secondary side> can be modeled ON THE PRIMARY SIDE as an impedance with Z = (n1/n2)^2*Z2 just to simplify calculations.

remember:
DC resistance α cross-sectional area of the WIRE
AC resistance α cross-sectional area of the COIL WINDING

please correct me if something in my post is wrong. im just a student. sorry =)

6. Nov 4, 2007

### tonytuong

lrsabug,

I totally agree with you that

the inductance of the coil will decrease when the cross-sectional area of the COIL is decreased or if the number of turns is decreased.

But the thing that bother me is that when I decrease the cross section of the magnetic wire the input impedance decrease.

I can agrue that when i decrease the side of the magnetic wire given the same turn number and every thing else is the same, the cross sectional area of the coil will decrease and as the result the impedance will decrease eventhough the DC resistance of the coil will increase??? what is about the capacitance if the coil?

Regards

7. Nov 6, 2007

### lrsabug

1. There is no such thing as a capacitance of a coil. A coil only stores energy by means of the magnetic field (hence, the coil exhibits what is called inductance). A capacitor, in the other hand, stores energy by the potential difference between its plates, which is in turn caused by the electric field between them (the plates display capacitance in a circuit).

Remember:
Coil displays only inductance
Capacitor displays only capacitance

2. There is no such thing as a DC resistance of a coil. A coil can only display inductance in a circuit with changing input (sinusoid input is an example). A "DC resistance" shall arise along with the material that you coil up (the resistivity of the material).

3. AC resistance (impedance) is independent of the DC resistance. Impedance of a coil depends on the frequency of the signal and the inductance (look at the formulas in my last post). DC resistance depends on the material you actually coiled up (also refer to my last post).

I cannot explain DC resistance increased ALONG WITH the decrease in impedance when you decreased the cross-sectional area of the coil. How did you measure the "DC resistance" of the coil?

8. Nov 6, 2007

### tonytuong

lrsabug,

I use the Impedance Bridge to measure the Impedance and DC resistance. I can measure inductance and capacitance of the coil.

I was able to measure Inductance of the coil and small amount of capacitance was measured as well. There is some capacitance exist in the coil.

I know for sure that decreasing the cross section of the magnetic wire will decrease the input impedance of the coil but I dont have any thing to back up my experiments.

I understand the theories for Inductor and capacitor.

Again thanks so much.

Best Regards

9. Nov 6, 2007

### NoTime

I'm afraid that only applies to artificially constrained classroom discussions.
Real devices always will have both properties.

10. Nov 6, 2007

### Averagesupernova

I feel I need to jump in here and point a few things out. irsabug, have you never heard of parasitic and interwinding capacitance? Have you never heard of equivalent series inductance in a capacitor? It is usually just the inductance of the leads of the device. Do some RF design and you will learn this very quickly.
-
You also have some things mixed up concerning the DC resistance of a coil. Impedance is a ratio of voltage to current. It lumps together the DC resistance of the coil, which is real since no conductor is without SOME copper loss, with the REACTANCE which only applies to AC.

11. Nov 6, 2007

### tc_kid

eh?.... not only capacitors have capacitance you know. any conductor has capacitance as any conductor can store energy. now that capacitance may be minimal but it is still there. some capacitors are made with coils too (the coils being the plates on each side of the insulator), the idea is that the inductance of each's field draws in the current from the opposing poles with more "force" to pull more energy into the capacitor and obtain higher discharge peaks.
Although i dont quite get the point of it seeing as a coil would have more resistance than a plate.

12. Nov 6, 2007

### tc_kid

exactly...

13. Nov 9, 2007

### lrsabug

ok ok. i understand what tc_kid, Averagesupernova, and NoTime. thanks for the correction. as indicated in my first post, i am still a student.

14. Nov 9, 2007

### lrsabug

@Averagesupernova: i agree with you. i just wrongly interchanged impedance (R + jX ohm) with reactance (jX ohm). sorry.