Basic common emitter amplifier design
Posted 28th August 2010 at 08:43 PM by wakibaki
The amplifier is probably the single most important composite circuit element in electronics. Certainly such familiar devices as radios and music systems would be impossible without amplifiers. The very large majority of amplifiers these days are solid state, that is, they use transistors. While there are many types of transistors, the first mass-produced transistors were BJTs, or bipolar junction transistors, and understanding transistor amplifiers, for most people, begins with these.
The transistor has three terminals and can be arranged in three basic amplifier configurations, the common emitter, common base (or grounded base) and common collector or emitter follower amplifiers.
Common base amplifiers are not commonly employed at lower frequencies as, amongst other reasons, they have a low input impedance, although they can be found in amplifiers for e.g. moving coil microphones. They are sometimes employed as current buffers, having a current gain of 1, and are most commonly seen in audio applications in the guise of a [I]cascode[/I].
Emitter follower amplifiers provide no voltage gain. They provide current gain and hence power gain. While power gain can always be parlayed into voltage gain, an amplifier providing voltage gain directly is often required.
The common emitter amplifier is the basic circuit commonly employed to provide voltage gain at baseband (audio) frequencies. The gain of common emitter amplifiers is affected by the bias current (and by temperature) so the example shown employs negative feedback in the form of emitter degeneration (i.e. it has a resistor between the emitter and ground) to make the resultant amplifier dependant on the stable and predictable values of the surrounding components (resistors) rather than the transistor’s unpredictable intrinsic characteristics. It is widely known that transistor current gain, denoted as h[SIZE="1"]fe[/SIZE], or beta, is highly variable, even within a manufactured batch. The prototype for a common emitter amplifier with emitter degeneration can be seen here.
[ATTACH]238[/ATTACH]
This tutorial gives no consideration to the determination of supply voltage or output impedance, save to say that output impedance should be no higher than the input impedance of the sink that the amplifier is driving. Supply voltage will often be a given, determined by the supply required by the final output stage to deliver the power demanded into the load impedance. For the purposes of this design exercise a supply of +20V will be used. The calculations performed are engineering approximations and neglect some of the less significant factors.
====================================
When looked at in isolation the base-emitter terminals of a BJT are a diode. In fact the transistor resembles two back-to-back diodes, the base-emitter diode and the base-collector diode but this is irrelevant in this context. We are going to consider an NPN silicon BJT since these are the most common and most easily understood insofar as they operate from a positive supply. This amplifier has a voltage gain equal to (minus) the value of the collector resistor divided by the value of the emitter resistor, to a close approximation i.e. –R[SIZE="1"]C[/SIZE]/R[SIZE="1"]E[/SIZE]. The derivation of this equation is not given here for brevity, but can easily be found elsewhere. Note that the minus sign shows that the amplifier is an INVERTING amplifier, i.e. when the input voltage rises, the output voltage falls, and [I]vice versa[/I]
We will set out to design an amplifier with a voltage gain of (-)10. This requires that the ratio of the values of the collector resistor and emitter resistor to be 10, as stated above.
When the emitter is connected to ground and the base to a positive voltage, the transistor conducts current from positive to ground, just as a (silicon) diode (in the correct orientation) does. In the process of conduction, a small voltage is lost across the diode or base-emitter junction. Other than when the current flowing is very small, the voltage drop, or [I]forward[/I] voltage drop is about 0.6 volts. This voltage drop varies slightly with temperature, which is a complicating factor when high precision is necessary, but for our initial purposes it is sufficient to treat it as a constant 0.6V.
If a voltage of 0.6V or greater is applied to the base when the emitter is connected to ground in the absence of any other resistance to limit the current, the current flow will be very large, and the transistor will be destroyed. In the case illustrated, there is a resistor interposed between the emitter terminal and ground. Provided the resistor is sufficiently large to limit the current to a level which the transistor can tolerate, the voltage at the emitter-resistor junction will settle at a voltage equal to the base voltage minus 0.6V.
In order that the available voltage swing should be maximised, the voltage at the output (collector) should be approximately half the supply voltage, permitting the output to swing from rail to rail. While this will not be achieved in practise, it illustrates the basic intent.
The output impedance of this amplifier is equal to the value of the collector resistor R[SIZE="1"]C[/SIZE], since the current supplied to the load is pulled through this resistor.
If we require an output impedance of 10k, then a number of things are immediately obvious.
The value of the emitter resistor must be 1k.
In order for the output voltage to be 10V (half the supply of 20V) the current through the collector resistor must be 1mA.
Since very little current flows into the base of the transistor, the current through the emitter resistor is close to 1mA, being the sum of the collector and base currents. This is a result of the current gain of the transistor, h[SIZE="1"]fe[/SIZE], which is the ratio of the current into the collector and the current into the base, and which is rarely < 100 in small signal transistors.
In order for the current through the emitter resistor to be 1mA, the voltage across it must be 1 volt. Since the voltage drop from base to emitter is 0.6V, the voltage applied to the base must be 1V + 0.6V = 1.6V.
Now we can design the bias network, i.e. select the values for the resistors designated RB1 and RB2 in the prototype diagram.
In order for the voltage at the junction of these two resistors to be relatively stable, it is conventional to arrange that the current flowing into the upper resistor is 10* (ten times) the current flowing into the base.
If the current gain of the transistor is 100, then the current into the base is 1mA/100, or 10uA (microamps), and the current down through the bias network should be 100uA. Since the supply voltage is 20V this gives us a net resistance of 200k for RB1 + RB2, and for the voltage at the base to be 1.6V the individual values must be 184k and 16k for RB1 and RB2 respectively.
This makes it possible to calculate the input impedance. Voltage sources have a very low impedance, making the positive rail effectively shorted to ground when considering AC. This is a feature which beginners frequently find difficult to swallow, but it simply must be accepted. This makes the input impedance 16k||184k||input impedance of the transistor. The input impedance of the transistor is approximately R[SIZE="1"]E[/SIZE]*h[SIZE="1"]fe[/SIZE] = ~100k. This gives a net value of ~12k8.
In order that the preceding stage’s DC voltage offset, if any, should not disturb the biasing arrangements, the amplifier is decoupled from the preceding stage with a capacitor. The low frequency cutoff of the amplifier is usually taken as the frequency where the output is 3dB down (half power) from the mid-band output. This is the point where the reactance of the capacitor (X[SIZE="1"]c[/SIZE]) is equal to the input resistance of the amplifier. At 30Hz the value of this capacitor is ~420nF.
(X[SIZE="1"]c[/SIZE] = 1/[2*pi*f*C])
(12k8 = 1/[2*pi*30*C])
A little rearranging of this equation gives the value of C @ 30Hz or any other frequency desired. You can see the final design here.
[ATTACH]237[/ATTACH]
These calculations are a simplification of the true state of affairs, but are adequate to many circumstances, although they give no indication of e.g. the noise performance of the stage. The values shown for the resistors and capacitor are not necessarily available in the E-series and some compromises are usually made in practise.
If an output impedance of, say, 5k, is desired, then this can be achieved by changing R(sub c) to 5k and doubling the standing current in the transistor. This would require changing the bias network to deliver 2.6V at the base, with an obvious knock-on effect on the bias network and the other parameters of the stage. More than one iteration of the design process may be required to achieve a satisfactory result.
The final step in the design is to choose a suitable transistor which will:
1. Tolerate the standing current in the stage i.e. has an adequate I[SIZE="1"]CM[/SIZE]
2. Tolerate the supply voltage i.e. has an adequate V[SIZE="1"]CEO[/SIZE]
3. Provide adequate current gain h[SIZE="1"]fe[/SIZE]. Transistors are commonly sorted into gain groups.
4. NOT have excessive gain at high frequencies (high [SIZE="1"]fT[/SIZE]). Pick one advertised as an audio amplifier or general purpose to minimise the likelihood of HF oscillation, not an RF transistor.
5. Have an adequate total dissipation P[SIZE="1"]tot[/SIZE]. The dissipation in the stage can be calculated as P = (V[SIZE="1"]C[/SIZE] – V[SIZE="1"]E[/SIZE])* I[SIZE="1"]E[/SIZE]
I hope this brief-as-possible but necessarily long-winded explanation of the common emitter design process is accurate. Please contact me if you discover any errors.
Next:- how to improve the gain by bypassing the emitter resistor.
w
The transistor has three terminals and can be arranged in three basic amplifier configurations, the common emitter, common base (or grounded base) and common collector or emitter follower amplifiers.
Common base amplifiers are not commonly employed at lower frequencies as, amongst other reasons, they have a low input impedance, although they can be found in amplifiers for e.g. moving coil microphones. They are sometimes employed as current buffers, having a current gain of 1, and are most commonly seen in audio applications in the guise of a [I]cascode[/I].
Emitter follower amplifiers provide no voltage gain. They provide current gain and hence power gain. While power gain can always be parlayed into voltage gain, an amplifier providing voltage gain directly is often required.
The common emitter amplifier is the basic circuit commonly employed to provide voltage gain at baseband (audio) frequencies. The gain of common emitter amplifiers is affected by the bias current (and by temperature) so the example shown employs negative feedback in the form of emitter degeneration (i.e. it has a resistor between the emitter and ground) to make the resultant amplifier dependant on the stable and predictable values of the surrounding components (resistors) rather than the transistor’s unpredictable intrinsic characteristics. It is widely known that transistor current gain, denoted as h[SIZE="1"]fe[/SIZE], or beta, is highly variable, even within a manufactured batch. The prototype for a common emitter amplifier with emitter degeneration can be seen here.
[ATTACH]238[/ATTACH]
This tutorial gives no consideration to the determination of supply voltage or output impedance, save to say that output impedance should be no higher than the input impedance of the sink that the amplifier is driving. Supply voltage will often be a given, determined by the supply required by the final output stage to deliver the power demanded into the load impedance. For the purposes of this design exercise a supply of +20V will be used. The calculations performed are engineering approximations and neglect some of the less significant factors.
====================================
When looked at in isolation the base-emitter terminals of a BJT are a diode. In fact the transistor resembles two back-to-back diodes, the base-emitter diode and the base-collector diode but this is irrelevant in this context. We are going to consider an NPN silicon BJT since these are the most common and most easily understood insofar as they operate from a positive supply. This amplifier has a voltage gain equal to (minus) the value of the collector resistor divided by the value of the emitter resistor, to a close approximation i.e. –R[SIZE="1"]C[/SIZE]/R[SIZE="1"]E[/SIZE]. The derivation of this equation is not given here for brevity, but can easily be found elsewhere. Note that the minus sign shows that the amplifier is an INVERTING amplifier, i.e. when the input voltage rises, the output voltage falls, and [I]vice versa[/I]
We will set out to design an amplifier with a voltage gain of (-)10. This requires that the ratio of the values of the collector resistor and emitter resistor to be 10, as stated above.
When the emitter is connected to ground and the base to a positive voltage, the transistor conducts current from positive to ground, just as a (silicon) diode (in the correct orientation) does. In the process of conduction, a small voltage is lost across the diode or base-emitter junction. Other than when the current flowing is very small, the voltage drop, or [I]forward[/I] voltage drop is about 0.6 volts. This voltage drop varies slightly with temperature, which is a complicating factor when high precision is necessary, but for our initial purposes it is sufficient to treat it as a constant 0.6V.
If a voltage of 0.6V or greater is applied to the base when the emitter is connected to ground in the absence of any other resistance to limit the current, the current flow will be very large, and the transistor will be destroyed. In the case illustrated, there is a resistor interposed between the emitter terminal and ground. Provided the resistor is sufficiently large to limit the current to a level which the transistor can tolerate, the voltage at the emitter-resistor junction will settle at a voltage equal to the base voltage minus 0.6V.
In order that the available voltage swing should be maximised, the voltage at the output (collector) should be approximately half the supply voltage, permitting the output to swing from rail to rail. While this will not be achieved in practise, it illustrates the basic intent.
The output impedance of this amplifier is equal to the value of the collector resistor R[SIZE="1"]C[/SIZE], since the current supplied to the load is pulled through this resistor.
If we require an output impedance of 10k, then a number of things are immediately obvious.
The value of the emitter resistor must be 1k.
In order for the output voltage to be 10V (half the supply of 20V) the current through the collector resistor must be 1mA.
Since very little current flows into the base of the transistor, the current through the emitter resistor is close to 1mA, being the sum of the collector and base currents. This is a result of the current gain of the transistor, h[SIZE="1"]fe[/SIZE], which is the ratio of the current into the collector and the current into the base, and which is rarely < 100 in small signal transistors.
In order for the current through the emitter resistor to be 1mA, the voltage across it must be 1 volt. Since the voltage drop from base to emitter is 0.6V, the voltage applied to the base must be 1V + 0.6V = 1.6V.
Now we can design the bias network, i.e. select the values for the resistors designated RB1 and RB2 in the prototype diagram.
In order for the voltage at the junction of these two resistors to be relatively stable, it is conventional to arrange that the current flowing into the upper resistor is 10* (ten times) the current flowing into the base.
If the current gain of the transistor is 100, then the current into the base is 1mA/100, or 10uA (microamps), and the current down through the bias network should be 100uA. Since the supply voltage is 20V this gives us a net resistance of 200k for RB1 + RB2, and for the voltage at the base to be 1.6V the individual values must be 184k and 16k for RB1 and RB2 respectively.
This makes it possible to calculate the input impedance. Voltage sources have a very low impedance, making the positive rail effectively shorted to ground when considering AC. This is a feature which beginners frequently find difficult to swallow, but it simply must be accepted. This makes the input impedance 16k||184k||input impedance of the transistor. The input impedance of the transistor is approximately R[SIZE="1"]E[/SIZE]*h[SIZE="1"]fe[/SIZE] = ~100k. This gives a net value of ~12k8.
In order that the preceding stage’s DC voltage offset, if any, should not disturb the biasing arrangements, the amplifier is decoupled from the preceding stage with a capacitor. The low frequency cutoff of the amplifier is usually taken as the frequency where the output is 3dB down (half power) from the mid-band output. This is the point where the reactance of the capacitor (X[SIZE="1"]c[/SIZE]) is equal to the input resistance of the amplifier. At 30Hz the value of this capacitor is ~420nF.
(X[SIZE="1"]c[/SIZE] = 1/[2*pi*f*C])
(12k8 = 1/[2*pi*30*C])
A little rearranging of this equation gives the value of C @ 30Hz or any other frequency desired. You can see the final design here.
[ATTACH]237[/ATTACH]
These calculations are a simplification of the true state of affairs, but are adequate to many circumstances, although they give no indication of e.g. the noise performance of the stage. The values shown for the resistors and capacitor are not necessarily available in the E-series and some compromises are usually made in practise.
If an output impedance of, say, 5k, is desired, then this can be achieved by changing R(sub c) to 5k and doubling the standing current in the transistor. This would require changing the bias network to deliver 2.6V at the base, with an obvious knock-on effect on the bias network and the other parameters of the stage. More than one iteration of the design process may be required to achieve a satisfactory result.
The final step in the design is to choose a suitable transistor which will:
1. Tolerate the standing current in the stage i.e. has an adequate I[SIZE="1"]CM[/SIZE]
2. Tolerate the supply voltage i.e. has an adequate V[SIZE="1"]CEO[/SIZE]
3. Provide adequate current gain h[SIZE="1"]fe[/SIZE]. Transistors are commonly sorted into gain groups.
4. NOT have excessive gain at high frequencies (high [SIZE="1"]fT[/SIZE]). Pick one advertised as an audio amplifier or general purpose to minimise the likelihood of HF oscillation, not an RF transistor.
5. Have an adequate total dissipation P[SIZE="1"]tot[/SIZE]. The dissipation in the stage can be calculated as P = (V[SIZE="1"]C[/SIZE] – V[SIZE="1"]E[/SIZE])* I[SIZE="1"]E[/SIZE]
I hope this brief-as-possible but necessarily long-winded explanation of the common emitter design process is accurate. Please contact me if you discover any errors.
Next:- how to improve the gain by bypassing the emitter resistor.
w
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