Tutorial 15 - Power Amplifiers

Using a Single Transistor as a Power Amplifier

We can make a very simple amplifier using the circuit shown:

This circuit is called a follower because the output follows closely the input voltage.  The voltage gain is slightly less than 1, but there is a high current gain.  We can explain this further:

• The output voltage is slightly less than the input

• The output current is greater than input

• Gain = output current / input current (since the bipolar transistor is a current controlled device)

Here is another simple amplifier based on a single bipolar transistor:

This amplifier has a few disadvantages:

• Its impedance is high, while the load (a loudspeaker) has a low impedance, typically about 8 W.  This results in poor power transfer.  This can be overcome with a suitable output transformer to drive the loudspeaker.

• The transformer is costly, bulky, and can cause distortion, so the power of such an amplifier is limited.

• Transistor amplifiers have a linear region in their transfer characteristic between 0.5 and 0.7 volts, shown in the graph.

The transfer characteristic graph above tells us:

• If the base voltage falls below 0.5 volts, the transistor does not conduct at all

• However if the voltage goes above 0.7 V, the output is at Vcc (the supply voltage).

We can set the zero point of the transistor, i.e. when the signal is at zero, in the middle of the linear region, where VBE = 0.6 V.  This will give a voltage of VCE of Vcc/2.  This is not very efficient, and Class A amplifiers can get quite hot, especially if Vcc is in the order of tens of volts.

There is a DC voltage offset, which means that there could be a voltage of Vcc/2 between the terminals of the amplifier.  This could result in quite a heavy current flowing through the 8 ohm speaker coils, which would get hot.  The DC voltage offset (0.6 V) is blocked from the loudspeaker if there is a transformer.  This is because the transformer can only produce an output if the input signal is changing.  There is another snag, which is that the transformer is an inductive load, which means that it has a greater reactance to high frequencies.

MOSFET amplifiers have much higher impedance than the bipolar transistor amplifier.  Here is a simple MOSFET single stage amplifier.

The potentiometer is there to bias the MOSFET so that the voltage can move up and down a long way.  The gate voltage would be have to be at about 2.2 V to be biased.

The input impedance is about 20 MW, much higher than the input impedance possible with a bipolar transistor.  The circuit is a voltage-controlled device and will act as a voltage follower.

The amplifier here has a voltage gain of about 25, as MOSFETs are voltage controlled devices.

 What are two advantages of using a MOSFET instead of a bipolar transistor?

Push-Pull Amplifier

The main problem with single stage amplifiers is that the amplifier has to be biased so that the signal zero is in the middle of the linear region.

 Why is this a problem?

A way to overcome this is to use a push-pull circuit as shown in the diagram.  This is sometimes called a Class B amplifier.

Notice that the transistors are in complementary pairs.  This means that the npn transistor is matched with a pnp transistor.  There are three basic differences between the pnp and npn transistor:

1.      Conventional current flows into the emitter and out of the collector.

2.      The base voltage needs to be 0.7 V below the emitter to turn the transistor on fully.

3.      Current flows out of the base for current to flow out the collector.

When the input signal is positive the npn transistor has a positive voltage at the emitter compared with the 0 V on the far side of the load.  Conventional current flows through the load to the 0 V line.  The current is “pushed” through the load.

If the input is negative, the collector of the pnp transistor is negative compared to the zero point, and conventional current flows from the 0 V through the load to the pnp transistor.  The current is "pulled" through the load.

Current pushed through the npn transistor and the load:

Current pulled through the pnp transistor and the load:

In both diagrams, the transistors in faint lines are turned off.

The transfer characteristic of the push pull circuit is shown:

Notice that there is a flat region in the graph around the origin.  This can result in crossover-distortion.  The input compared with the output looks like this.

The picture below shows crossover distortion on a CRO:

The human ear is sensitive to electronic distortion, but it is overcome using an operational amplifier.  The push-pull circuit is placed in the feedback loop as shown:

The op-amp holds the base-emitter voltage of the npn transistor at 0.7 V.  When there is a negative signal, the op-amp changes over, dropping 1.4 V, so that it turns on the pnp transistor.  The voltage gain is 1, but the power gain is almost infinity.

 (Harder) Can you explain the role of the op-amp in reducing cross-over distortion?

MOSFET power amplifier

This push-pull follower uses MOSFETs:

• The n-channel MOSFET has a resistance of about 1 W when its gate is above 2.5 V

• The p-channel MOSFET has a resistance of 1 W when its gate is below – 2.5 V.

• Therefore the circuit cannot respond to signals with a peak to peak value of less than 5 volts.

• This can be changed by incorporating the op-amp into the circuit.  The op-amp forces the gate voltage to be held at whatever voltage is needed to ensure that the input voltage is the same as the load voltage.

• The gain is 1, but the power gain is very large.

 (Harder) Compare the MOSFET power amplifier with a power amplifier based on npn and  pnp bipolar transistors.

Power Considerations in a Power Amplifier

Let us go back to a transistor push pull circuit as before.  The circuit shows values added.

Let us assume that gain of the transistor is about 50.

• Since the voltage drop is 0.7 V, the voltage across the load = 7.0 – 0.7 = 6.3 V

• The current I = V/R = 6.3 ¸ 100 = 0.063 A = 63 mA.

• The base current Ib = Ic/hFE = 63 ¸ 50 = 1.26 mA

• Power input = VIb = 7 V ´ 0.00126 A = 0.00882 = 8.82 mW.

• Power output = VIL = 6.3 V ´ 0.063 A = 0.397 W = 397 mW.

• Power gain = 397 ¸ 8.82 = 45.

The extra power needed comes from the supply lines, but not all the power is delivered to the load.  Our voltage to the load is 6.3 V.  So 15 – 6.3 = 8.7 V is the voltage drop across the transformer.  Therefore the transistor must be dissipating that energy as heat.  Power lost in the transistor = 8.7 ´ 0.063 = 0.55 W or 550 mW.

More power is wasted in a transistor than the power applied to a load.  Without going into a specific heat calculation, it is easily seen that a transistor under that kind of load will start to get hot.

MOSFETs have a much lower resistance than bipolar transistors, typically about 1 W.  Suppose we connected the amplifier to a loudspeaker of resistance 8 ohms.  The positive supply rail is at a p.d. of 15 V.  We can work out the power dissipation in the MOSFET.

• Total resistance = 8 + 1 = 9 W

• Current = V/R = 15 V ¸ 9 W = 1.67 A.

• Power lost in the MOSFET = I2R = 1.672 ´ 1 = 2.79 W

• Power in the loudspeaker = 2.79 ´ 8 = 21.3 W.

• Only 1/8 (12.5 %) of the power is dissipated in the MOSFET, compared with 60 % with the bipolar transistor.

Heatsinks

Almost 3 W of power dissipated in a MOSFET can have quite a significant heating effect.  As semiconductors get hotter, their resistance falls, allowing a bigger current to flow.  This increases the heating effect yet further, so they got hotter.  Eventually they can burn out.

To prevent this from happening, we attach the component to a heatsink.  Heatsinks have the following features:

• They are made of metal which is a good conductor

• They have a large surface area, which allows heat to escape into the air by convection.

• They are often painted black to assist in passing heat by radiation.

There are many patterns of heatsink, and two of these are shown in the photograph.

 Use the picture to explain why heatsinks are  effective.

The thermal resistance for heatsinks are measured in Celsius per watt (oC/W), which means the temperature rise for each watt dissipated.  Suppose we had a heatsink rated at 1 oC/W, a 10 W power dissipation would result in a 10 oC temperature rise.  The kind of heatsink on the right of the picture has a thermal resistance of about 10 oC/W.

Where there are heavy currents there are massive heatsinks that are cooled by fans to aid convection.

The central processing unit (CPU in a computer has a heatsink cooled by a fan.

You will hear the fan running when you turn your computer on.  When the CPU is working hard, the chip can get quite hot.  The fan runs continually and the chip temperature is monitored by a thermistor.  This can give a warning when the temperature gets above about 60 oC, which is quite warm.

As chips have become faster, they get hotter.  An old 386 will run with just a passive heatsink.  A Pentium 4 (a 786) will fry within a few seconds without a fan, as would a modern dual or quad core.  The fastest gaming computers are either water cooled, or even refrigerated.

An excellent article on the use of heatsinks can be accessed at this web-address:

### H-bridge amplifiers

This is a motor control circuit called an H-bridge.  Here is one that controls a motor:

Image from Talking Electronics

It is called an H-bridge because of the H in the middle of the schematic.  In effect the MOSFETs are acting as four switches in this arrangement.

Image from Wikimedia

There are different combinations of the switches.

 1.      Complete the table.  Two have been done as an example. S1 S2 S3 S4 Result 1 0 0 1 Motor goes forward 0 1 1 0 0 0 0 0 0 1 0 1 1 0 1 0 Motor brakes 1 1 0 0 0 0 1 1 1 1 1 1

The advantage of using an electronic H-bridge circuit is that you can connect the inputs of the H-bridge to a range of electronic devices, including gates.

This arrangement is often called a Class D amplifier.

There are a number of advantages in using the Class D amplifier:

·        There is a lot less power dissipation, so the amplifiers do not get as hot.

·        Therefore it is more efficient.

·        The maximum possible power is greater.

However the Class D amplifier is a switching device.  It amplifies pulses; audio-signals are not pulses.  Therefore a modulator is needed to turn the audio input into pulses.  A low pass filter is needed to remove the pulse before the signal passes through the amplifier.

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