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Mia!

June 5th, 2008

Here’s some links on transformation geometry

http://regentsprep.org/Regents/Math/coreflec/reflect.htm

http://www.mathsisfun.com/geometry/reflection.html

Power amplifiers are typically used in the final output stages of circuits. In communications, they can be hooked up to an antenna or a transmitter, and in audio, to provide signal power to a speaker system. As the name implies, they dissipate large amounts of power, so heat sinks or cooling fans are important. They are physically larger than small signal transistors, and may have cooling fins built in. Power amplifiers deliver power to the load. Therefore the power gain (Ap) is the ratio of the power to the load (Pl), to the input power (Pin).

Ap = Pl/Pin

where:
Pl is the load power, calculated by

Pl = Vl^2/Rl

and Pin is the input power, calculated by

Pin = Vin^2/Rin

Rin being the input resistance.

This is usually expressed in RMS, which is .707 times the peak voltage. If you measure AC voltage with an RMS voltmeter, this is the way to calculate load power. More often you are looking at the AC output voltage with an oscilloscope. In this case use

Pl = Vpp^2/8Rl

When the voltage gain is known, another equation that can be used is

Ap = Av^2(Rin/Rl)

Assume a common collector amplifier has an input resistance of 10k ohms, and a load resistance of 100 ohms. Voltage gain is approximately one for common-collector, so the power gain is

Ap = 1(10k/100)
Ap = 100

The AC and DC load line

AC load line During the positive half cycle of ac source voltage, the collector voltage swings from the Q-point towards saturation. During the negative half cycle, the collector voltage swings from the Q-point towards cutoff.

Maximum output voltage can be achieved when the Q-point is at the center of the AC load line. This differs from the DC load line in amplifiers (common emitter for example) because the DC and AC collector resistances are not equal. The DC collector resistance is simply the collector resistance, where the AC collector resistance is the collector resistor in parallel with the load resistance.

In the image of the AC load line for a CE amplifier, point a is the AC saturation point, and is calculated by

Icq + (Vceq/Rc)

and point b is the AC cutoff, calculated by

Vceq + IcqRc

DC quiescient power

Pdq = (Icq)(Vceq)

This is saying the power dissipation of a transistor with no signal input will just be the product of q-point Ic and Vce. Class A power amplifiers must maintain a quiescient current that is at least as large as the peak current requirement for the load current. The output power is

Pout = Vl(rms)Il(rms)

This formula can be used to determine the output power maximum.

Pout(max) = .5(Vceq)(Icq)

The Efficiency of an amplifier is the ratio of the signal power to the load, to the power supllied from the DC source.

%Eff(max) = (Pout/Pdc)100

Classifications of Power Amplifiers

There are a few classifications of power amplifiers, and they are based on the percentage of the input cycle that the amplifier operated in the linear region. In the previous examples, reaching cutoff or saturation was undesirable and resulted in clipping and distortion.

Amplifiers operating solely in the linear region are known as Class A amplifiers. They are usually mid-point biased to maximize the available gain. Any distortion or clipping is undesired. They are usually constructed in a common-emitter or common-source configuration. The amplifier conducts for the full 360 degrees of the input cycle, always in the linear region, and the output wave is 180 degrees out of phase with the input. Class A efficiency is usually around 25%.

Class B amplifiers have the q-point at cutoff. For this reason, they operate for 180 degrees of the input signal, and since Icq = 0 and Vce = Vce(cutoff), the transistor is not conducting until an AC signal is applied. Two transistors are usually used in class B amplifiers to create a push-pull configuration. Each transistor conducts for 180 degrees of the input signal, and the full signal is sent to the load. Class B amplifiers have a 79% maximum efficiency.

Bipolar junction transistors have the .7 (silicon) or .3 (germanium) volt drops that must be overcome, or the signal becomes distorted as it flips between the two transistors. This is known as crossover distortion. Diode biasing can be used to overcome it. The diodes compensate for the base-emitter voltage drops and produce a undistorted signal.

Class AB is a modified form of Class B push-pull operation when biasing resistors are used to put the push-pull stages into slight conduction, even when there is no input signal applied.

Basic Class C amplifiers are biased so they conduct for even less than 180 degrees of the input cycle. More power can be obtained, but the output is very distorted, and so Class C is used more often in RF applications. They are biased way below cutoff, and therefore much less heat is generated from this momentary conduction. A negative voltage is applied from the base, and the transistor conducts only when Vin exceeds this negative voltage and the voltage from the base to the emitter.

Power dissipation is very low for a class C, and can be found through

Pd(avg) = (time on/T)(Vce(sat)Ic(sat))

Remembering that the voltage drop across a transistor is around .2 volts, this will usually be a pretty small amount.

In tuned operation, a tank circuit containing an inductor and a capacitor set for resonance is used. This tank circuit would normally start out at a full wave form, and then slowly discharge with one pulse from the input. These circuits are tuned so each pulse from the input keeps the oscillation of the tank circuit going. Efficiency for Class C operation can approach 100%!

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