EXPERIMENT 13

Circuit Components and Magnetic Fields - Inductors, Transformers, and …

Partly as preparation for the next project and partly to help develop a more complete picture of voltage sources, we will return to considering inductors. The extension we are primarily concerned with is the mutual inductor or transformer. The transformer has three uses: stepping up or down voltages, stepping up or down currents, and transforming impedances. Like other devices we have considered, the transformer does not work in an idea manner for all circumstances. We will first consider the basic ideas behind the transformer.

In the circuit above, the voltage source (V1) and the 50 ohm resistor (R2) represent a sinusoidal voltage source like the function generator we use in the studio. R1 is the load resistor and TX1 is the transformer. We make transformers by winding coils of wire around some kind of a core material. Sometimes the core material is just air when we wind the wire around a paper tube, for example, Sometimes the core material is iron or some other magnetic material. Inductors have larger values when the core material is magnetic. The value of the inductance will also depend on the geometry of the core material. Just for simplicity, we will address only one geometry, the cylindrical core. This produces the kind of inductors we have used in previous experiments. If you do not recall what they look like, get one out and look at it. If the core cylinder has a radius equal to a and we wind a coil N times around the cylinder to cover a length l, the inductor will have an inductance equal to:

L = (_o N² a²)/l Henries

where _o = 4 x 10^-7 Henries/meter. If the core is not air, but rather iron or some other magnetic material, replace _o with , which is usually many times larger than _o. By many times we can mean as much as 10⁵ times larger. You should know that this formula only works well when the length l is much larger than the radius a. For other applications, it is only useful to find a ballpark number.

When one makes an inductor, it is also obvious that the wires used can have a large variety of cross sectional areas. You probably noticed that there are some inductors in the studio made with very thick wires, while others are made with very thin wires. Thin wires permit one to wind many more turns of wire around a core and thus increase the inductance. However, since all wires have a resistance given by the expression

R = l/(A)

where l is the length of the wire, A is the cross sectional area of the wire (thickness), and is the conductivity of the wire material. For copper, the conductivity is about 6 x 10⁷ Siemens/meter. For sea water the conductivity is 5 Siemens/meter. There are many handbooks like the CRC Handbook of Chemistry and Physics that have the resistance per mile or meter or whatever of wires, given the thickness of the wire. It is also quite easy to calculate the resistance of a piece of wire using the formula above.

Now that we have seen how the inductors we have used are made, we should also notice that they look a lot like electromagnets we have seen back in science project days in grade school. If you have forgotten about this, you can check the website

http://schoolnet2.carleton.ca/english/math_sci/phys/electric-club/pag12.html

Inductors work by creating a magnetic field. If one inductor is placed near another inductor, then the magnetic fields of the two inductors will interact with one another. If you have ever built an electromagnet, you will know that the magnets will even attract or repel one another if the current in the coil is large enough. Even when we cannot sense that they are interacting, a current in one coil will induce a current in a nearby coil. To maximize this interaction, we usually wind the coils onto the same core, making them look as similar as possible. When we build a device to maximize this interaction, we call it a transformer. Transformers only work for time varying currents and voltages.

To analyze just how a transformer works, we have to add an additional kind of inductance, called mutual inductance. If a coil of wire of inductance L₁ is very near another coil of inductance L₂, there will be a mutual inductance M between the two coils where M² = k²L₁L₂. The constant k is the coupling coefficient. If the coils are perfectly coupled, k = 1. Usually k is a little less than 1 in a good transformer. The two loop or mesh equations that apply to the two current loops in the figure above are

V1 = i₁(R2 + jL1)- i₂ (jM)

0 = -i₁ (jM) + i₂ (R1 + jL2)

where L₁ is the inductance of the primary coil and L₂ is the inductance of the secondary coil. The primary connects to the source. The secondary connects to the load. The L and M terms have opposite signs because the loop currents go in opposite directions. We can solve for the input impedance of the transformer by taking the ratio of the source voltage V1 to the primary current i₁

Z_in = V1/i₁ = R2 + jL1 +(²L1L2)/(R1+jL2)

For an ideal transformer, the coupling constant k = 1 and then M² = L1L2 and

Z_in = R2 + jL1 + (²L1L2)/(R1+jL2)

Now, let both L1 and L2 become very large (tend to infinity), but their ratio should remain the same.

L2 = a² L1

then,

Z_in = R2 + R1/a²

Thus, the transformer transforms the load resistance R1 by the square of the turns ratio, a = N₂/N₁ where N₁ is the number of turns in the primary coil and N₂ is the number of turns in the secondary coil. This is the first, and most stringent, relationship for an ideal transformer. Remember that Z_in is the ratio of the input voltage to the input current. In any circuit that we build or simulate, we can determine Z_in by finding this ratio.

The relationship between the primary and secondary currents in the transformer can be found from the second loop equation. Solving again for the case where the inductances become very large, we find that

I₂/I₁ = 1/a, a² = L2/L1, or N₁I₁ = N₂I₂

The voltages across the primary and secondary coil connections are given by

V₁N₂ = V₂N₁

Note that both the voltage and current relationships shown actually contain no sign information. Depending on how the transformer is wired, it is possible for minus signs to appear in these expressions. Thus, you should consider that they hold only for magnitudes. As with Z_in, we can determine these ratios by measuring the voltages and currents separately and then taking their ratios.

Part A Simulation of Transformer

Setup the circuit shown above in PSpice. For the voltage source V1, use VSIN and assume an amplitude of 1 volt, a frequency of 1k, an AC voltage of 1 volt, and all other terms equal to zero. For the transformer(XFMR_LINEAR), you need to assume a coupling (choose 1), and primary and secondary inductances, L1 and L2. For your first simulation, assume L1 and L2 are both 1mH. Perform an AC sweep (use the decade option) from 1 Hz to 10 MHz (remember that the latter must be specified as 10meg). For what range of frequencies, if any, does this transformer work like an ideal transformer. Remember that all the above relationships must be satisfied. Also, the ratio of N₁ to N₂ is determined from the ratio of L1 to L2 since inductance L is proportional to N².

Trying a variety of values for L1 and L2, specify a transformer for which the voltage across the secondary is 10 times the voltage across the primary (not the source voltage) for frequencies of 10kHz and up. Use the smallest possible inductances to make this happen.. Write down the values for L1 and L2 that you have selected and print your PROBE plot that demonstrates that the transformer works as specified. Using the equations above, confirm that the transformer is working correctly. You will need to do this in class or you will not know if you have the correct values for L1 and L2. Also, you will need to be sure that you plot the three ratios above involving currents and voltages.

Now repeat the task of the last paragraph for frequencies of 100Hz and up. What are the new values of L1 and L2? What design advantages exist for circuits that are to work at higher frequencies rather than lower frequencies?

Part B Making an Inductor and a Transformer

First, we need something to wind our coil on. Take a sheet of paper and roll it into a tight cylinder. You can start by wrapping it around a pencil and then tighten the roll up. Tape the roll so it does not unwind. Next, take about 2 feet of enameled wire and wrap it tightly in a single layer around your roll of paper. Try to keep track of the number of times you wind the wire around the paper roll. This is the number of turns of your inductor. Leave about 1 inch at each end of the coil so you will have something to connect to. Tape the wire in place. Remove some of the enamel from the ends of the wire. The enamel is the insulation for this wire, so you cannot make electrical contact unless it is removed. There should be some sandpaper for this purpose or you can use a knife.

Now you have made an inductor. Estimate its resistance by first estimating the wire dimensions and then use the formula given on page 2 above. Next, measure the resistance using the multimeter, if you can. When you measure small resistances, it is important to first measure the resistance of the wires you are using to connect the coil to the meter. Then, you add the coil and see if there is a difference. Then, measure the resistance of just the wires again to be sure you did it correctly. Do your numbers agree at least roughly?

Estimate the inductance of your coil using the formula above on page 2. Check with other students in the studio to see if your estimate is about the same.

Based on work you have done on some of the early experiments in this course, develop a procedure for measuring the inductance of your coil using the function generator, other capacitors, inductors, resistors, a scope, etc. In Part C of this experiment, we will do such a measurement based on your plan. You can also discuss your plan with other students in the class.

Part C Measurement of Inductance

In the circuit above, V1 and R2 represent the function generator, R1 is a 50ohm load connected to the function generator by a series combination of an inductor and a capacitor. The inductor will have a positive imaginary impedance given by jL1 while the capacitor will have a negative imaginary impedance given by 1/(jC1). At the frequency =(L1*C1)^-1/2 , the voltage difference across the series combination of inductor and capacitor will go to zero. Using a known capacitance with an unknown inductance in this configuration will permit us to determine the inductance. We can use the approximate expression for the inductance to figure out the range of frequencies in which to look for this effect. You should simulate the response of this circuit using PSpice to be sure you know what to look for. This gives us at least one method to determine the inductance. There are many other methods that we can also use. You can try this method or any other reasonable method.

Even though we have an approximate expression for the inductance, the ideal model of the cylindrical inductor will generally over-estimate the inductance by quite a bit. We can get a larger inductance and also a more predictable inductance by winding the coil on a piece of iron rather than on a paper tube. The permeability of iron is at least about 1000 times larger than that of air. There are several iron rings available for this purpose. Remove your coil from the paper tube and wrap it tightly around the iron ring. Recalculate the value of the inductance including the additional factor of 1000. Try to measure this inductance value. Does it agree at all with the predicted value?

Part D Transformer

Since you now have one coil wrapped around the iron ring, the addition of a second coil will result in a configuration that can be used as a transformer. Using the function generator, find a frequency for which the transformer works more-or-less as expected. Justify your choice of frequency.

Part E Magnetic Pickup Coil

Now remove your coil from the iron ring and wind it into a small coil with a radius of about one inch. You may only get a few turns with this coil, since its radius is so much larger. Connect the coil to the oscilloscope. Obtain a small magnet from an instructor or a TA. Drop the magnet through the coil and observe the voltage signal across the coil. This is the same principle by which some speedometer designs work. Print a plot of the signal using the HPBenchlink software. Moving a magnet near a coil is basically the same as having a time varying current in one coil creating a magnetic field that is picked up by another coil. In the case of the transformer there is nothing that can be observed to move, but the principle is more-or-less the same.

Last Updated on 31 March 1998