B² Spice

Case Study: Amplifier Max Power Testing in B² Spice

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Measuring Maximum Output Power
An amplifier’s maximum power output is the one specification that everyone, no matter how non-technical, want to know. Fortunately, this is one of the easier measurements to make in SPICE. In short, a voltage source is placed at the input of the amplifier and its AC value is incremented until the waveform distorts and then the output voltage is converted to an RMS value. RMS stands for the Root equals the Mean of sum all the instantaneous values within a period Squared. The Root refers to the equivalent DC value that would result in the same power dissipation. For example, a 1-ohm across a 5-volt DC power supply will generate 25W of heat, but requires a 7.07-volt AC sine wave or a 8.66-volt triangle wave to generate the same amount of heat. Why the difference?

The waveform stipulates the percentage of power that would be generated by a DC voltage and the percentage varies with the waveform. For example, a square wave produces the same amount of heat as the DC voltage, which makes perfect sense, as a square wave is really a DC voltage that periodically switches polarity. A 5-volt square wave starts at +5 volts and then instantly switches to –5 volts and then back again. So as a thought experiment, imagine a 1-ohm resistor placed across a 5 volt battery; the resistor will dissipate 25 watts as long it is connected to the battery. Now, if the resistor is flipped across the battery’s terminals, will the resistor dissipate more or less heat? Obviously, the dissipation will remain the same in both cases. Increasing the number switches in any given time period make no difference, assuming the switch time is instantaneous.

The triangle waveform, assuming equal rise and fall slopes, will only produce one third the heat that the DC voltage would, so its RMS value is equal to V/ 1.7332. If you guessed one half instead, the geometry probably misled you. Yes half the area is covered, but wattage is the result of the voltage being squared, not linearly multiplied against the resistance; half the voltage equals one forth the power.

The sine wave’s curves make for more complex math, but the quick answer is that the sine wave produces 1/v2 or 0.707107 times as much heat as a DC voltage that equals the waves peak value.

Thus, as Ohms law tells us that power equals voltage squared divided by resistance for DC voltages, when the waveform is a sine wave, we must divide the peak voltage by 1.414 or multiply by 0.707 first. This also applies to peak currents, so when taking a sine wave’s peak voltage and current, we divide both by the square root of 2, before multiplying each against each other to get the amount of power dissipated. Notice that since both values are being reduced by 0.707, 0.707 is being effectively squared as well, which results in 0.5, the inverse of 2. thus, we can multiply the peak voltage and current values and then divide by 2 to get the same result. Much easier.

In B2 A/D Spice, we can place a current meter in series with the load and a voltage meter in parallel with the load to measure these values. Then by defining a new plot, we can display power delivered into the load impedance in a graph.

Note that we are including the RMS transformation of the voltage and current into account in the plot’s expression. Now when we view the graph the peak of the sine wave will equal the power delivered into the load.