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Output Impedance

No source of power is perfect. A perfect source of voltage (from which power is derived) would not drop in output no matter how significant the load. The drop in output is generally due to internal resistance. A real-world amplifier acts as a perfect/ideal amplifier with a resistor connected in series with it's output.

Batteries:
The simplest example is the battery. You know that the voltage across the battery terminals will fall when current is drawn from it. The larger the current draw, the larger the voltage drop. When the current draw stops, unless you have significantly discharged the battery, its voltage returns to the original level (or very close to it). It is very similar to having a resistor in series with a 'perfect' 12 volt power source.

Power Supplies:
The same thing happens when you draw power from a regulated power supply (like a 12v D.C. power supply). For this example, let's say that we have a 12 volt regulated* power supply rated to deliver 30 amps of current. If it would be a "perfect" power supply, the output voltage would remain at EXACTLY 12 volts as long as the current draw is less than 30 amps. In the real world, this is generally not the case. Even though the power supply is regulated, the output voltage would fall slightly when current is drawn from it. The greater the draw, the greater the voltage drop. If we wanted to calculate the output resistance of the supply, we would first measure the output voltage with no load. Then we would draw current from the supply by connecting some type of load to it. We would then measure the current flowing through the load and the output voltage of the supply while it is loaded. By finding the difference between the 'no load' and the 'loaded' output voltages, we will have the internal voltage drop of the supply. If we use one of the Ohm's law formulas, we can find the the internal resistance. We will use the formula V=I*R >> R=V/I. Let's say that the current draw for the test was 20 amps and the voltage difference (loaded voltage minus no load voltage) was .25 volts. Plug in the numbers R=.25/20. The internal resistance of the supply is .0125 ohms or 12.5 milliohms. By using this information, you can predict the voltage drop from the output of the supply for any current draw less than 30 amps (the supply is only rated for 30 amps).


Amplifier Output Impedance

A.C. output impedance:
The output impedance of an amplifier is very similar to the internal resistance of the power supply above. The biggest difference is that the output of the amplifier is an A.C. signal. We know that impedance is the opposition to A.C. current. Another difference is that the output impedance of the amplifier will vary with frequency.

Standards:
If you are going to test more than one amplifier, you should adopt some standards. You should pick a single frequency or a defined set of frequencies. You should also pick a load with a commonly used impedance (such as 2 or 4 ohms). For this example let's say that the test frequency is 50 hertz and the test load's impedance is 4 ohms.

Testing:
For this test, you want to drive the amplifier to a sufficiently high level to produce large measurements (voltage, current...). The larger numbers will help increase the accuracy of the tests. You will drive the amplifier to a level which will be below clipping when loaded with the test load. You must be absolutely sure that the level of the input signal is not changed throughout the test procedure. Measure and make note of the output voltage of the amplifier with no load. Connect the 4 ohm load to the amplifier's output terminals and measure the output voltage again. Make note of it. Please note that the voltage measurements should be taken as close to the amplifier as possible. Since the resistor will change in value when it heats up, after reading the voltage quickly disconnect the resistor from the amplifier and measure its resistance, including the wire used to connect the resistor to the amp. Make note of the resistance.


Performing the Calculations

Method 1:
If the NO LOAD voltage was 20 volts and the LOADED voltage was 19.95 volts, the difference in loaded and no load voltages is .05 volts. If the resistor measured 4.15 ohms and we know that there was 19.95 volts across it we know that the current through the resistor was:


I=V/R
I=19.95/4.15
I=4.8072 amps.

If we use the formula Z=V/I (Z denotes impedance - we would use R if this were a DC circuit), we get:
Z=0.05/4.8072
Z=0.0104 ohms.

The output impedance of the amplifier is 0.0104 ohms.

Method 2:
Output impedance=(Load resistance*(Vunloaded minus Vloaded))/Vloaded
Z=(4.15*(20-19.95))/19.95
Z=0.0104 ohms

Note:
If you are using a digital multimeter, you should check the frequency response of the meter by measuring the output voltage of the tone generator at various frequencies. Some multimeters are designed to measure only a small range of A.C. frequencies. If the voltages change significantly (more than 1 or 2 percent), you must take that into consideration in the various measurements you make. If you have a 'true RMS' meter, you are likely going to get accurate readings over the entire audio spectrum.


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You should remember:
1.All voltages sources have internal resistance or output impedance.
2.When making tests, adopt standards which you will use in all future tests.


 

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