Learn about efficiency and selecting a heat sink for Class B amplifiers.

A generic circuit diagram of a Class B amplifier using FET control devices is shown in **Figure 1**.

**Figure 1 **A** **Class B amplifier using FET control devices

Many years ago in engineering school, we learned that the maximum efficiency of this Class B amplifier is 78.5%. A detailed derivation reveals that the general formula for efficiency of a Class B amplifier is:

Where π = 3.14159…

Vp = Peak voltage of the output sine wave (Vp sin(wt)).

Vdc = DC supply voltage (See Figure 1)

Vp cannot physically exceed Vdc because the amplifier can’t produce output voltages greater than its own supply voltage. So maximum efficiency occurs when Vp = Vdc:

Maximum Efficiency = π/4 = 0.785 = 78.5%

Equation 1 degenerates properly for maximum efficiency when Vp = Vdc, and what we learned in school many years ago is verified. In this condition, minimal voltage appears across the pass device, and efficiency is maximum. When Vp < Vdc, increased voltage appears across the pass transistor, more heat is generated, and the circuit becomes less efficient.

This equation assumes that a constant amplitude sinusoidal output voltage is delivered to the load. An example of an amplifier running at a constant amplitude sinusoidal output voltage occurs in Industrial Test Equipment's frequency converter product line. These units convert a local power line voltage and frequency to a different voltage and/or frequency. The Model 500C can convert 60Hz to 50Hz, or vice versa, 50Hz to 60Hz and in each case provides a constant 115VAC to the load.

However, having Vp = Vdc operates the amplifier too close to the edge. A slight change in Vdc or Vp might result in the top of the waveform being clipped. In other words, the amplifier cannot be driven to the point where Vp might go above Vdc. Clipping would be the result and is generally unacceptable. Vp = Vdc is not a practical operating point.

Allowing a 10% margin, maximum Vp could be selected as 0.9Vdc so that efficiency becomes:

Efficiency = π/4 x 0.9Vdc/Vdc = 70.7%

This is a more practical level of efficiency of a Class B amplifier running at a constant output voltage.

Efficiency can be improved using a switching (Class D) amplifier instead of a Class B amplifier. However, this introduces switching noise and more complicated circuitry. Class B circuits have little noise, simpler circuitry, and pure sinusoidal waveforms.

**Selecting a heat sink**

Consider next the general equation for efficiency:

Efficiency = n = Pload/(Pdevice +Pload)

Where Pload = Power (Watts) delivered to the load

Pdevice = Power that needs to be dissipated from the FETs

This equation can be rearranged so that:

Pdevice = Pload (1-n)/n (Equation 2)

This allows calculation of the required heat sinking for the system. For example, assuming n = 70.7%, and Pload =100W, the required heat sinking for a Model 500C is:

Pdevice = 500 (1-0.707)/0.707 = 207 Watts

This dissipation occurs when the system is delivering full output power (ie., 500W). At lower output power, dissipation is proportionately lower. For example at 50 Watts of output power, 20.7 Watts is dissipated.

All of our Class B power amplifiers come with a thermal switch mounted on the heat sink. The switch opens and turns the amplifier off when heat sink temperature exceeds 70 degrees C. When the heat sink cools to 50 degrees C, the switch closes and normal operation resumes automatically.

**Derivation of Equation 1**

The derivation is divided into two main parts: the basic equations and the results after integration. The step-by-step path between these parts is omitted because I believe most readers will probably prefer to avoid the tedious details. Enough information is provided so those interested can verify the results.

The integrations are necessary because the equations being integrated represent instantaneous power. All the instantaneous powers must be added up (integrated) and then divided by the period (π) to get the average power over the half cycle. We actually integrate over only ½ cycle of the sine wave because the positive and negative halves of the sine wave output are symmetrical and each half delivers identical power to the load.

**A derating curve**

From the above example, we have learned that with Vp/Vdc = 0.9, the Model 500S can deliver 500 Watts to the load, and 207 Watts are dissipated as heat (Pload = 500W and Pdevice = 207 Watts). How much power can be delivered when operating at a lower voltage (ie., Vp/Vdc < 0.9)?

This can be answered by rearranging Equation 2 and substituting for n from Equation 1. The result is:

Continuing with the example, we let Vp/Vdc vary from 1 to 0 in 0.1 increments, and let Pdevice = 207 Watts. This amount of heat sink dissipation is necessary to achieve full rated power output. Since it is available, we use it for all levels of output power.

The results are shown in Table 1, constructed using the above equation and an Excel spreadsheet.

Maximum load power (Pload) as a function of Vp/Vdc.

Vp/Vdc = 0.9 is taken as the best operating point (see previous discussion). Operating in the region 0.9 < (Vp/Vdc) < 1 is more efficient but operates too close to the point of clipping.

**Table 1**

The table shows that for Vp/Vdc = 0.9, and with 207 Watts being dissipated, Pload = 500W, as previously calculated. As Vp/Vdc decreases, the maximum power from the Model 500S also decreases. For example, reading from the table, at Vp/Vdc = 0.7, only about 250 Watts are available, and efficiency falls to 55%.

For questions or comments, please email Powertron@aol.com, Attn: Dana.

*Dana Geiger is an electronics engineer presently working part time for Industrial Test Equipment, Inc. in Port Washington, NY.*