Impedance is the opposition to the flow of current in an AC circuit. Reactance is opposition to the flow of alternating current caused by capacitance or inductance. Reactance in an inductor or capacitor causes opposition to the flow of alternating current. The unit used to measure reactance or impedance is the Ohm.
As the frequency of the AC applied to a coil increases, the reactance increases. As the frequency of the AC applied to a capacitor increases, the reactance decreases.
When the impedance of an electrical load is equal to the internal impedance of the power source, the source can deliver maximum power to the load. Impedance matching is important so the source can deliver maximum power to the load. One reason to use an impedance matching transformer is to maximize the transfer of power.
One method of impedance matching between two AC circuits is to insert an LC network between the two circuits. Core saturation of a conventional impedance matching transformer should be avoided because harmonics and distortion could result. ALL of the following devices can be used for impedance matching at radio frequencies -- A transformer, A Pi-network and A length of transmission line.
A two-times increase or decrease in power results in a change of 3 dB [103/10≅2.0
]. The percentage of power loss that would result from a transmission line loss of 1 dB is 20.5% [10-1/10=0.796; 100%-79.6%=20.6%].The total current relate to the individual currents in each branch of a parallel circuit equals the sum of the currents through each branch.
If 400 VDC is supplied to an 800-ohm load, 200 watts of electrical power is used [P=IE and I=E ÷ R, so P=E2 ÷ R; P=4002 ÷ 800=200]. A 12-VDC light bulb that draws 0.2 amperes uses 2.4 watts of electrical power [P=IE, so P=12×0.2=2.4]. The voltage across a 50-ohm dummy load dissipating 1200 watts would be 245 volts [P=IE and I=E ÷ R, so P=E2 ÷ R, or E2=P × 50 = 60,000, and E=244.9 volts.]. When a current of 7.0 milliamperes flows through 1.25 kilohms, approximately 61 milliwatts are dissipated [P=IE and E=IR, so P=I2R; P=7.02×1.25=61.25].
Peak, peak-to-peak, and RMS voltages | |
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The ratio of peak envelope power to average power for an unmodulated carrier is 1.00 [The graph of an unmodulated carrier is a flat line; the peak and the average are the same]. The RMS value of an AC signal is equivalent to a DC voltage of the same value [A continuous, but lower, DC voltage is equivalent to a higher peak AC voltage.]. The peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts is 339.4 volts [If the RMS voltage is 120, the peak is 120 ÷ .707 = 169.4; so the peak-to-peak voltage is 2 × 169.7, or 339.4 volts.]. The RMS voltage of sine wave with a 17 volts peak is 12 volts [If the peak voltage is 17 volts, the equivalent RMS voltage is 17 × .707 or 12.0 volts].
The output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm resistor connected to the transmitter output is 625 watts. If an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output, the output PEP from a transmitter is 100 watts. If an average reading wattmeter connected to the transmitter output indicates 1060 watts, the output PEP of an unmodulated carrier 1060 watts.
Mutual inductance causes a voltage to appear across the secondary winding of a transformer when an
AC voltage source is connected across its primary winding.
The source of energy is normally connected in a transformer to the primary winding.
When no load is attached to the secondary of a transformer, there is still a
current in the primary winding of a transformer. It is called the magnetizing current.
Transformer Ratios | ||||||||||||||||
Voltages - V
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Impedances - Ω
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If the 2250-turn primary of a transformer is connected to 120 VAC the voltage across a 500-turn secondary winding is 26.7 volts [Vsecondary=120×500 ÷ 2250=26.66]. The turns ratio of a transformer used to match an audio amplifier having a 600-ohm output impedance to a speaker having a 4-ohm impedance is 12.2 to 1 [Ratio=√ 600 ÷ 4 = 12.2].
How electronic components combine | |
Values add |
Reciprocals add |
The component that should be added to an existing resistor in a circuit to increase circuit resistance is a resistor in series. If three equal value resistors in series produce 450 ohms of resistance, each resister is 150 ohms [X+X+X=450; 3X=450; X=150].
The total resistance of three 100-ohm resistors in parallel is 33.3 ohms [1 ÷ 100+1 ÷ 100+1 ÷ 100=.03; 1 ÷ .03=33.3]. The total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel is 5.9 ohms [1 ÷ 10+1 ÷ 20+1 ÷ 50=.17; 1 ÷ .17=5.88]. If three equal value resistors in parallel produce 50 ohms of resistance, each resister is 150 ohms [1 ÷ X+1 ÷ X+1 ÷ X=1 ÷ 50; 3 ÷ X=.02; X=150].
The component that should be added to a capacitor in a circuit to increase the circuit capacitance is a capacitor in parallel. The equivalent capacitance of two 5000 picofarad capacitors and one 750 picofarad capacitor connected in parallel is 10750 picofarads [5000+5000+750=10750].
The capacitance of three 100 microfarad capacitors connected in series is 33.3 microfarads [1 ÷ 100+1 ÷ 100+1 ÷ 100=.03; 1 ÷ .03=33.3]. The capacitance of a 20 microfarad capacitor in series with a 50 microfarad capacitor is 14.3 microfarads [1 ÷ 20+1 ÷ 50=.07; 1 ÷ .07=14.3].
The inductance of three 10 millihenry inductors connected in parallel is 3.3 millihenrys [1 ÷ 10+1 ÷ 10+1 ÷ 10=.3; 1 ÷ .3=3.3].
The component that should be added to an inductor in a circuit to increase the circuit inductance is an inductor in series. The inductance of a 20 millihenry inductor in series with a 50 millihenry inductor is 70 millihenrys [20+50=70].