# Class-A Small-Signal Amplifiers

A class-A small-signal amplifier is usually used to boost the voltage of a signal.

By definition, a class-A amplifier operates equally on all 360 degrees of a sine-wave input. There is a problem if we use a simple common-emitter amplifier like the examples above. The input swings both positive and negative during its cycle. When the signal goes negative, the voltage at the base will be lower than the voltage at the emitter. This reverse biases the base-to-emitter junction of the transistor. When the base voltage drops below approximately 0.7 volts, the transistor starts to shut down. Only the positive half of the signal is amplified.

Input    Output

 A simple common-emitter amplifier as above will only amplify the positive half of the signal. The emitter is tied to ground (0 volts). It acts as an amplifying half-wave rectifier.

In the illustration above, when the input signal is in the positive half of the cycle (the voltage is above 0 volts) the transistor operates normally. While operating normally, as the input voltage goes up, the output voltage goes down. Here, VCC is 10 volts, so when the input voltage is at 0 volts, the output voltage is 10 volts. As the input voltage rises above 0.7 volts, the output voltage goes down. In this case, the combination of peak input voltage, transistor hFE and the value of the collector resistor, bring the output voltage down to near 0 volts, just above the saturation voltage. When the input voltage drops below +0.7 volts, the transistor acts goes into cut-off. It acts as an open switch. In that state, there is no current flowing through the collector resistor, so there is no voltage drop across the resistor and the output voltage is equal to VCC. The amplifier is not operating for the entire negative half of the input signal and then some. The result is the output shown above.

A class-A amplifier solves this problem by mixing DC with the AC input signal. When DC is mixed with AC, the AC part of the signal is simply offset. For example, imagine a pure AC signal that is swinging between -1 volt and +1 volt (2 VP-P) with a mean voltage of 0 volts. If you mix this with +2 volts DC, this will offset the mean voltage by +2 volts. Now the voltage will swing between +1 volt and +3 volts.

 2 VP-P centered on 0 volts 2 VP-P centered on +2 volts

The voltage now never falls below the 0.7-volt threshold required to keep the transistor operating.

Adding the DC voltage is done with a voltage divider and a capacitor. The voltage divider determines the DC voltage. In the following example the resistors are equal, so the voltage between them is ½ of VCC. The capacitor blocks the DC voltage from reaching the AC source but allows the AC to pass into the DC part of the circuit. At any moment in time, the voltage at the junction between R1 and R2 becomes the voltage set by the voltage divider plus the voltage that comes from the input source. In the example below the voltage at the junction of the resistors is +2 volts. Now, let's say we freeze time and at that moment the input voltage is at its positive peak voltage of +1 volt. This will push the voltage at the junction of the resistors up by 1 volt, making the final voltage +3 volts. The voltage at the junction of the resistors becomes the AC signal centered on the quiescent DC voltage at the junction. In this example that is 2 VP-P centered on +2 VDC

 Input to the bias network. The biasing network(resistors are equal) Output of the bias network. To the left is the 2 VP-P signal centered on 0V. The circuit in the center has 0V DC on the left of the capacitor and +2V DC to the right of the capacitor (R1 and R2 are equal). DC cannot pass from one side of the capacitor to the other, so the nominal voltage on the left remains at 0 volts and the nominal voltage at the right remains at +2 volts DC. The AC signal passes through the capacitor from left to right, mixing with the +2V DC, resulting in a 2VP-P sine wave centered on +2V. Another way to look at it is, when the signal to the left is at 0 volts, the voltage to the right is at +2 volts. The capacitor isolates these voltages from each other. As the voltage on the left rises, it exerts a force through the capacitor that pushes the voltage on the right in the same direction. When the voltage to the left is at +1 volt, it is pushing the voltage to the right one volt higher than it's nominal voltage. That forces the voltage to the right up one volt above the nominal +2 volts to +3 volts. When the voltage to the left drops, it exerts a force on the voltage to the right that pulls it in the same direction. When the voltage to the left is at -1 volt, it is pulling the voltage to the right down by one volt. This pulls it down from the nominal +2 volts to +1 volt. As long as the voltage to the left keeps changing (making it alternating current), it will continue to exert these pushing and pulling forces on the voltage on the right.

The following circuit is a basic class-A small-signal amplifier. It is a simple common-emitter amplifier with a voltage divider biasing circuit. This biasing circuit works as in the above illustration to set the base voltage such that the input signal never forces the base voltage below 0.7 volts.

 A Simple Class-A Small-signal Amplifier

Recall the simple common-emitter amplifier in the previous chapter as we examine this circuit. The small swing in the base voltage causes a large swing in the collector current (the swing of the collector current depends on the hFE of the transistor and the value of RC). This causes the collector voltage to vary correspondingly. Notice that as the base voltage increases that the collector voltage decreases and vice versa. The amplified output to be 180 degrees out-of-phase with the input.

 Input Signal Output Signal For a common-emitter amplifier, the output signal is 180 degrees out-of-phase with the input signal

The biasing network is fine-tuned so when the input voltage is 0 volts (the quiescent condition) the collector voltage is ½ of VCC. In the illustration above, VCC is +10 volts, so the quiescent collector voltage is set to +5 volts. This allows the output voltage to go down during the positive half of the input cycle and to go up during the negative half of the input cycle. The voltage at the base of the transistor never drops below +0.7 volts, so the transistor never goes into cut-off. With careful selection of the base bias resistors, matching them to the hFE of the transistor and the value of RC—and properly matching the gain of the amplifier to the amplitude of the input signal—the amplifier never goes into saturation or cut-off. The amplifier now operates on the entire input signal cycle, making it a Class-A amplifier.

Now that we have the amplifier operating normally, what happens if we increase the input voltage? When the input voltage reaches the positive peak voltage the amplifier goes into saturation; the output voltage bottoms out at 0 volts. When the input voltage reaches the negative peak the amplifier goes into cut-off; the output voltage tops out at VCC. This causes a flattening of the top and bottom of the output wave as viewed on an oscilloscope. This is called clipping or clipping distortion.

 A sine wave with clipping distortion. A sine wave with clipping distortion. This output signal is from an amplifier with a VCC of 12 volts. The output voltage is limited to 12 volts resulting in clipping when the output signal is too big.

Clipping is caused when the input signal has too much amplitude (goes to high or too low) or the amplifier gain is too high for the input signal. For example, the distortion you hear when someone speaks too close to the microphone of a public address system is clipping distortion. To prevent clipping distortion, you must either reduce the gain of the amplifier or reduce the input level. In the case of the PA system, the speaker can move farther from the microphone or the volume can be turned down.

Setting the quiescent collector voltage is accomplished by carefully selecting the voltage divider resistors and the collector resistor so that, with the hFE of the particular transistor, the voltage drop across the collector resistor is ½ of the supply voltage. This is a problem because, not only do different transistors have different hFE values, but two supposedly identical transistors are likely to have slightly different hFE values. Also, the hFE value for any transistor changes with base current, collector current and temperature.

This problem is solved by putting a resistor between the emitter and the ground (RE in the diagram below).

 A simple Class-A small signal amplifier with emitter resistor.

With the emitter resistor in place, as the collector current rises, so does the emitter current. The emitter current goes through RE. Therefore, when the collector current rises, so does the emitter voltage and vice versa. This is opposite to the voltage at the collector, which drops as the collector current rises. The emitter voltage is always going the opposite direction to the collector voltage. This works against changes the base current and works against changes in the collector voltage, decreasing the gain of the amplifier[1]. The gain of the amplifier as a whole becomes RC ÷ RE instead of being dependent on the hFE of the transistor and RC.

The voltage divider bias and the emitter resistor create an amplifier with a gain that is independent of the hFE of the transistor. This means the biasing is stable with temperature and with differences in hFE from transistor to transistor. This is great for the stability of the biasing, but it also reduces the AC gain of the circuit. If we can get the AC signal to ignore the emitter resistor, we can increase the signal gain without changing the DC gain. The amplifier would still have a stable bias but with increased signal gain. This can be achieved by bypassing the emitter resistor with a capacitor.

 A complete class-A small-signal amplifier with emitter resistor and bypass capacitor. The DC gain is approximately RC/RE, but the signal gain (the AC gain) is dependent on the span style="font-family: Arial, Helvetica, sans-serif"> hFE of the transistor and the value of RC.

The finished circuit also has a capacitor on the output to block the DC voltage at the collector of the transistor from reaching the next circuit in the sequence.

## FET Small Signal Amplifiers

The common-emitter small signal amplifier design adapts easily to field-effect transistors. The name of the amplifier changes because the names of the terminals of the FET are different.

 A junction-FET

The source terminal of the FET is equivalent to the emitter terminal of the BJT. Therefore, the FET equivalent of a common-emitter amplifier is a common-source amplifier.

 A common-source amplifier.

The layout of the amplifier shown above is identical to the common-emitter except it has a JFET instead of a BJT. The biasing currents and voltages will be those appropriate for the JFET, but the circuit is otherwise the same. The most notable difference in the biasing is that the gate voltage must be lower than the source voltage (using an N-channel JFET as above). This will reverse bias the junction, being the normal biasing for a JFET.

## Variations

### Fixed Base Biasing

Fixed base biasing uses a single resistor between VCC and the base of the transistor.

 A small signal amplifier with fixed base biasing.

This biasing scheme is dependent on the hFE of the transistor. To get the correct quiescent bias point, RB has to be matched precisely to the hFE of the transistor. In practical application, it is quite fiddly and nearly impossible to get right. Then, a change in temperature or replacing the transistor messes everything up. This is why you see this circuit in textbooks but rarely in the real world.

### Self Biasing

 A self-biased JFET small signal amplifier.

Self-bias is the FET equivalent of fixed base biasing for the BJT. RG is tied to ground. There will be current flowing through RS, making the FET source voltage (the voltage at the top of RS) higher than ground. This reverse biases the junction as is normal for JFETs. No current flows through RG, so there is no voltage across it, putting the gate voltage at 0 volts. The gate voltage is not dependent on any condition other than that it is tied to ground through RG. This makes a self-biased FET amplifier much more stable than the fixed biased BJT circuit. This biasing scheme may be the most common biasing scheme for FET amplifiers.

## Review

1. A Class-A amplifier operates on how much of an AC signal?
• 100% or all 360 degrees
2. What happens if you put a sine wave, centered on 0 volts, to the input of a simple common-emitter amplifier?
• Only half of the signal is amplified. The other half is cut off when the input signal goes below zero volts.
3. How does the bias network solve the problem of cutting off half of the input signal?
• It adds some DC offset so that the input signal is no longer centered on zero volts.
4. What voltage should the input signal be offset to?
• The input signal must center on a voltage high enough that the input signal will not go below 0.7 volts.
5. What happens if the output voltage tries to go above the supply voltage or below 0 volts?
• The result will be clipping distortion.
6. What can be done to prevent clipping distortion?
• Reduce the input level or reduce the amplifier gain.
7. What if clipping is only on one half of the signal?
• Adjust the quiescent voltage to prevent the output from hitting the saturation voltage or cut-off voltage respectively.
8. What is likely to happen if you replace the transistor in a fixed base biased small signal amplifier?
• The bias voltages will change because the hFE of the new transistor will be different from that of the transistor that is being replaced.
9. How can the small signal amplifier be designed so that the biasing is independent of the hFE of the transistor?
• Add an emitter resistor to the amplifier.
10. What is the DC gain of the amplifier when there is an emitter resistor?
• The DC gain is the value of the collector resistor divided by the value of the emitter resistor?
11. What happens to the AC gain when an emitter resistor is added to the small signal amplifier?
• The AC gain is reduced.
12. How can the AC gain be increased without changing the DC gain and thus changing the biasing?
• Bypass the emitter resistor with a capacitor.
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