In a previous section (Switching Power Supplies), I touched on transformers. You should remember that a transformer has a primary winding and a secondary winding wrapped around a core. In car audio, the core is usually a donut shaped toroid. A transformer can be designed to step voltage up or down, and/or isolate. We discussed isolation in a previous section. This section will deal with stepping the voltage up or down.
Remember that the transformer windings are enamel coated magnet wire which is wrapped around the core (as seen below). The number of windings is determined by the number of times that a piece of wire makes a complete turn around the core. The primary winding is the winding which is driven (in car audio amplifiers, it's usually driven by transistors). The secondary winding is the output winding. The secondary is driven by the magnetic field that the primary induces in the core. A transformer with a ratio of 1:1 will not cause a voltage increase or decrease (disregarding small losses) from the primary to the secondary (as measured across each of the individual windings). If the ratio is 1:2 (primary:secondary), the voltage across the secondary will be twice the voltage across the primary. A ratio of 1:3 will result in a secondary voltage 3 times as high as the voltage on the primary. Of course all of this applies to a transformer which is very lightly or not loaded (minimal current flow). When current is drawn from the secondary winding, there may (will) be a voltage drop and therefore a primary to secondary voltage ratio which may not match the winding ratio exactly. This loss of voltage is primarily due to the less than 100% efficiency of the magnetic coupling of the primary and the secondary windings through the core and also some copper (resistance) losses. Remember that the primary and the secondary windings are not generally electrically connected together. This means that all of the power transfer between the primary and secondary is transferred (magnetically) through the core. The transformer below is similar to one that you would find in a small car audio amplifier. The winding ratio is 1:2. The different colors are each half of the center tapped primary and secondary. Notice that there are twice as many secondary windings as there are primary windings. The schematic symbol shows how the windings relate to each other. The center tap of the primary (red) is connected to the battery. The center tap of the secondary (black) is connected to ground.
As a side note:
The power driven into the primary will equal the power produced at the output of the secondary (if we ignore wire and core loss). If we have a 'step-up' transformer with a 1:2 ratio being driven with 24 volts A.C., the secondary output voltage (disregarding losses) will be 48 volts. If we load the secondary so that 5 amps of current is flowing through the secondary windings, the power output is P=I*E; P=5*48; P=240 watts. Since the power driven into the primary equals the power out of the secondary, we know that the power being driven into the primary is 240 watts. If we use the formula I=P/E, we see that the I=240/12; I=20 amps. If we were stepping the voltage down, the current flowing through the primary would be less than the current being drawn through the secondary windings.
When designing a transformer you have to calculate the number of primary windings so that the transformer will operate properly/efficiently. There are a few different variables that have to be taken into account.
Ac is the effective cross sectional core area. This number is supplied by the core manufacturer.
Flux density (B) is expressed in gauss. If the flux density is too high the core will saturate (effectively disappear from the magnetic circuit - very bad). Generally, in car audio amplifier switching power supplies operating at or below 35.000hz, the flux density is kept at or below 2000 gauss. Some cores will operate at higher flux density for frequencies below 35,000hz but 2000 gauss is a good conservative number. For higher frequencies, you have to design for lower flux density to prevent excess core heating. The following chart shows the approximate maximum flux density for a given frequency. For a more precise value for a given core material, refer to the core manufacturer.
The primary voltage for a push-pull system is double the primary input voltage. For car amplifier switching power supplies, the input voltage is 12VDC. This means that the total primary voltage is 24 volts. If we use 13.5 volts as the input voltage, the primary voltage would be 27 volts.
The operating (oscillation) frequency is simply the frequency at which the primary is driven. Generally between 25,000hz and 50,000hz in car audio amplifiers.
The number of primary turns returned by the calculator is the total number of turns on the primary side of the transformer. Of course, with a push pull system, the number of turns on each half of the primary must be equal. If the output says that you need 13 turns, you'd round up to 14 turns and each half of the primary would have 7 turns. From the previous diagram, you'd have 7 orange turns and 7 green turns on the core.
When wire (specifically, solid copper conductors as are used for transformers) is used to carry DC current, the entire cross sectional area of copper carries the current equally. When wire is used for AC current, the current is carried differently. At low frequencies, the current flow is not significantly affected by the skin effect. As you get into the higher frequencies (as those used to drive a transformer in a switch mode power supply), the current flow is carried disproportionately by the outer area of the copper wire (especially for large single solid conductors). This is called the skin effect. If, for example, you are using 14g wire at 100khz, the wire will not be able to carry the same amount of current as it could if it were passing DC. If your calculations told you that you needed to have ~4120 circular mils, you'd have a few choices. You could use 1 strand of 14g wire, 3 strands of 17g or 6 strands of 20g. All would have the same current carrying capacity if you were using it in a DC circuit but... If you were using it for AC, the 14g would only be suitable for frequencies below ~6000hz. Above that frequency, the voltage losses and power dissipation may be unacceptable (it would still work above 6000hz but not efficiently). The maximum frequency that you'd want to use with the 17g would be about 11,000hz. For 22,000hz the 6 strands of 20g would be a good choice.
As you can see in the following demo, the current flow is AC (because the direction of flow changes at regular intervals). You can also see that the electrons on the outside of the conductor change direction instantaneously. The electrons deeper in the conductor take a little longer to get going. When the frequency is high, the electrons deep in the conductor don't precisely follow the flow of the outer electrons.
The following shows how the area of current carrying copper (the copper colored area) diminishes as frequency increases. The up/dn buttons control frequency. If you have a large diameter solid round conductor, the conductor will appear to be increasingly more hollow at higher frequencies. At high frequencies, the wire will act more like copper tubing than a solid copper wire.
This is the maximum diameter solid wire that will have 100% current density (100% of the wire will have current flowing through it).
This is the maximum cross sectional area of round wire that will have 100% current density.
This is the maximum wire gauge that will assure ~100% current density at the frequency entered in the calculator above. If you're selecting the conductors based on copper area (in circular mils), you need to use enough strands of this gauge wire to add up to the desired number of circular mils.
When using large gauge wire for high frequencies, you have loss due to 'AC resistance'. When using a number of smaller strands to equal the desired/required number of circular mils, you reduce the loss. If each strand has a radius that's less than or equal to one skin depth, you will virtually eliminate the loss at high frequencies. In effect, the resistance to AC (at or below the frequency entered above) and DC will be the same.
Number of strands of calculated wire gauge (3) to equal the wire gauge that was entered in the input section of the calculator.
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