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The transistor is a current-control device. For example, control the collector-emitter current by changing the base current. In a general voltage amplification occasion, this amplification effect comes from the use of resistors to convert current into voltage. In the small-signal model, the source of the base current is the ratio of the input voltage to the base-emitter dynamic resistance rbe, which is usually kΩ. So the base current is very small, and may only be a few tenths of mA. Through the amplification of the transistor, the base current is generated between the collector and the emitter by β times. This article will introduce how transistor works in the common-emitter amplifier circuit.
Transistor Amplifiers Circuit Introduction
| Ⅱ Common-emitter Amplifier Circuit Design |
| Ⅲ Common-emitter Amplifier Circuit Expansion 3.2 Low-voltage and Low-loss Circuit |
Here, take the common emitter amplifier circuit as an example:

Figure 1. Transistor Common-emitter Amplifier Circuit
△Vo=VCC-△ieRc=VCC-β△ibRc=VCC-△Vi·Rc/rbe
△Vi/rbe=△ib
Thus, the collector generates a current of β times ib:
△ie=β△ib
Furthermore, the output voltage can be obtained by the relative positive power supply potential:
△Vo=VCC-△ieRc=VCC-β△ibRc=VCC-△Vi·Rc/rbe
Thus, we can get an inverted amplified voltage signal by AC coupling and controlling the collector resistance Re. But generally the emitter will have a resistance to control the gain, so the above formula is not practical. When designing a circuit in non-extreme situations, we often hope that the circuit can work with most general-purpose transistors, avoiding the parameter that depends on component parameters such as rbe. At the same time, it is very cumbersome to consider the base current in the specific calculation. Therefore, in the general design process, the existence of the base current is ignored in an approximate calculation (In some circuits, although the base current is ignored, it is still necessary to give the base a certain current drive to make the circuit working normally). In addition, the calculation of gain is the external circuit resistance not the rbe.
Among them, the base-emitter tube voltage drop VBE is also a very important parameter, which is generally equal to 0.6V (silicon tube). The parameters of the transistor circuit can all be obtained according to VBE=0.6V and Ohm's law.
The cumbersome part of the transistor circuit lies in the setting of the static operating point. Usually, careless design will cause clipping and distortion of the output waveform. Therefore, the selected values of some experimental values can be used for reference. The overall design idea is: quantitatively determine the voltage and current to calculate the resistance.
The common-emitter amplifier circuit is a typical inverting amplifier, which has a wide range of applications and stable effects. First show the overall design ideas, and then explain the purpose and principles of the design in steps.
1) Determine the supply voltage VCC, and determine the static emitter current IE according to the frequency curve/noise curve/others.
2) Determine VE, where selects 1~2V to absorb temperature drift.
3) According to VE and IE, calculate the emitter static resistance RE ( IE≈IC).
4) Determine the magnification Av, and apply the relationship Av=RC/RE to calculate the static collector resistance RC. At this point, the static working point has been established.
5) Check whether the static operating point meets the requirements: positive output swing limit=VCC-IE·RC, negative output swing limit=IE·RC-VE. It is necessary to ensure that the amplified output voltage does not exceed the swing limit (usually the swing limit is larger). If RC is too large, there will be a downside clipping, so is the small RC. In addition, determine whether the power exceeds the limit: PC=VCE·IC.
6) Determine the base bias voltage as follows: According to VBE=0.6V, it is easy to get VB=VE+0.6 (divide the voltage from the power supply through the resistor). Since ib is considered to be small and negligible, the current IB0 flowing through the base voltage divider resistors (R1, R2 in the above figure) should be much larger than ib. ib is approximately calculated as IC/β, and IB0 is about an order of magnitude larger than ib, so R2=VB/IB0, R1=(VCC-VR2)/IB0.
7) Finally, determine the AC coupling capacitor value and the power supply decoupling capacitor value.
Let's first use a designed common-emitter amplifier circuit to intuitively understand the waveforms of the next parts:

Figure 2. Transistor Common Emitter Amplifier Circuit Design
As shown in the figure, the circuit uses 2SC2240 tube, 15V power supply, and the input and output are AC coupled. The output signals are as following:

Figure 3. 4-channel Signal Waves
The pale blue waveform is the input signal, selecting the sine wave of 1kHz, 1Vpp.
The green is the output signal, amplified by about 5 times, and it is inverted.
The blue is the base signal, which can be seen because the DC level is raised due to the influence of the base bias resistance.
The red is the emitter signal, which is only a fixed value away from the base signal.
First, perform a DC analysis, that is, determine the static operating point. In the initial design process, the design and verification of static operating points are also the first to proceed. The static potential of the base can be easily calculated according to the base bias resistance, and the static potential of the emitter can be determined according to the voltage drop of the base-emitter tube as a constant. Therefore, according to the magnitude of the emitter resistance, the magnitude of the collector-emitter current can be obtained, and then the collector static potential can be obtained from the power supply voltage.
Why is the static operating point important? Take the NPN transistor as an example, which is equivalent to two back-to-back diodes. If requiring the diode work, you must give it a proper bias to make it reasonably conductive. In the circuit, the base-collector diode prevents internal feedback, and the base-emitter diode is the key to achieving amplification. In other words, it is enough to design an external circuit so that the current flows normally in the base-emitter diode. This idea will be mentioned in the analysis of the carrying capacity of the emitter follower.
Find the AC voltage gain. When the input voltage changes △vi, it will cause the emitter current to produce an AC change △ie. Since the base emitter voltage drop is constant, it does not contribute to the AC change, so △ie=vi/RE. Therefore, the emitter AC output voltage can be determined as vo=△ieRC=vi·RC/RE, and the AC gain is Av=RC/RE. This conclusion can quickly analyze the magnification of the common-emitter circuit.
The output power rails are VCC and VE respectively, which are determined by the current characteristics of the transistor during operation, and there is generally no rail-to-rail output. According to the output power rail and the AC amplification factor, the circuit can be used.
When the input and output are not AC coupled, the input (especially for DC) will cause the output waveform to be distorted.
After understanding the circuit characteristics, you can design the common emitter circuit according to the design steps at the beginning of this section. The static operating point and magnification have been determined during the analysis, and the other parts are designed below.
Supply voltage: According to the swing of the output voltage, we can determine the size of the voltage. Usually the power supply voltage is larger than the output peak-to-peak value.
Transistor: Select the appropriate transistor according to the operating frequency, required power, noise level and β, etc.
Emitter current: Determine the size of the emitter current according to the frequency characteristics by consulting the device manual.
RC and RE: Determined by the emitter voltage and current, and the magnification, pay attention to review the upper and lower limits of the swing and the rated power.
Base bias resistance: VB is determined according to VE, thereby determining the voltage divider resistance of the power supply. Note that the current flowing through the voltage divider resistor should be one to two orders of magnitude higher than the base current. The base current is calculated by dividing the collector-emitter current by β.
Coupling capacitor: The AC coupling capacitor is generally 10uF. Note that the coupling capacitor of the output stage and the input impedance of the next stage will form a high-pass filter. The cutoff frequency of the filter should be handled carefully.
Through the method of AC analysis, we can obtain some characteristic parameters of the designed circuit, such as input and output impedance, magnification and so on.
Input impedance: According to AC analysis, the input impedance is the parallel value of the base bias resistance. In small signal analysis, the base emitter dynamic resistance rbe should also be connected in parallel.
Output impedance: The method to determine the output impedance is to add a load to the circuit. When the peak-to-peak output value drops to half of the no-load, the load impedance is the output value. Generally, the output impedance of the common-emitter amplifier circuit is the collector resistance RC.
Magnification: Due to the influence of the base current, the actual magnification is about 10% lower than the design value. So the design formula is more practical.
By improving the general common-emitter amplifier circuit, various application circuits with other characteristics can be obtained. This section introduces the means to increase the magnification, the low-voltage power supply circuit, the differential output circuit, and the tuning amplifier circuit.
According to the introduction of the design circuit, the voltage gain is mainly determined by the ratio of the collector resistance RC to the emitter resistance RE. So it is common to change the ratio of the resistance to change the gain. However, the problem arises: these two resistors are responsible for determining the working current at the same time. Because the DC operating point is changed arbitrarily, the circuit is likely to be distorted or even not work.
From another perspective, voltage gain belongs to the category of "AC Analysis", and the static operating point belongs to "DC Analysis". So add some reactive components to the circuit to change the ratio under the AC perspective, the resistance value during DC analysis does not change.
This can be achieved by connecting the emitter resistor in parallel, or making the resistor in parallel with the capacitor, that is, modifying the circuit in the first section:

Figure 4. Common-emitter Amplifier Circuit
Pay attention to the emitter in the above figure. In the AC analysis, the resistor R4 is short-circuited by the capacitor. At this time, it is equivalently considered that the emitter resistor is only R7 (330Ω). From the signal source and the oscilloscope, the signal has been amplified nearly 50 times at this time. It is much larger than the original design value (10k/2k=5), thus realizing the expansion of voltage gain. If the original emitter resistance is not split, but the entire capacitor is connected in parallel, the maximum gain βRC/rbe will be obtained at this time.
How to choose the capacitance value? It should be noted that after the capacitors are connected in parallel, the entire circuit will have high-pass characteristics, and the cut-off frequency is f=1/2πRC. If this high-pass characteristic is not required, the C capacitance value can be selected to a larger value between 47uF~100uF.
In addition, the capacitor C6 has the function of temperature compensation.
If the op amp circuit is powered by a dry battery (1.5V), it is not realistic, but the transistor circuit can be done. The key is to use the conduction voltage drop of the external diode to offset the base-emitter voltage and have small small. The circuit in the figure below can still amplify small signals as designed even under 1.5V power supply:

Figure 5. Common-emitter Amplifier Circuit
But the disadvantage is that the maximum voltage of the system is always below the supply voltage. Because of the small circuit loss, it is suitable for low power consumption.
Fully differential op amps can provide dual-mode output, and many transmission lines also require differential transmission. Transistor circuits can also perform differential output. In addition to the principle of a common emitter amplifier circuit, the principle of an emitter follower is also used. The following figure shows the circuit connection of the differential output.

Figure 6. Common-emitter Amplifier Circuit
It can be seen that two differential signals with the same shape and opposite phase are output. The collector signal is in phase with the input signal, and the emitter output signal is in phase with the input signal. However, the output impedance of the two signals is different due to the different lead-out positions. The output impedance of the inverted output is higher (RC), and the output impedance of the non-inverted output is lower, which is suitable for driving the load. The inverted output is generally connected to the emitter follower before driving.
In addition, the static potential of the base should be set between VCC and GND as much as possible to expand the undistorted output range.
The introduction of reactive components in the circuit will cause the properties of the circuit to change with the frequency. We can use this property to design LPF, HPF, and tuning amplifier commonly used in high-frequency circuits. Actually, it uses the characteristic that the impedance of the reactance element changes with the frequency, and then changes the voltage gain at the current frequency. The impedance at the resonance frequency is often purely resistive and has an extreme value to achieve frequency selective amplification. The following show low-pass, high-pass and frequency selective amplifiers at specific frequencies:
① LPF

Figure 7. Common-emitter Amplifier Circuit
As shown in the figure, a low-pass filter is constructed (the input of the bode tester is placed at the base instead of the output of the signal generator, because the input coupling capacitor will form a high-pass filter with the input resistor, which affects the observation effect), and its cut-off frequency is about 1.06kHz, calculated by f=1/2πRcC.
From the sinusoidal steady-state analysis, the impedance of the RC parallel loop is R/√(1+(wRC)^2). As the frequency increases, the impedance decreases, so the voltage gain decreases, forming a low-pass characteristic.
② HPF

Figure 8. Common-emitter Amplifier Circuit
As shown in the figure, a high-pass filter is constructed, and the calculation of its cut-off frequency is similar to that of LPF.
At the gain peak point, the voltage gain reaches 50dB, which is close to the β value of the transistor. Then the gain is attenuated due to the deterioration of the transistor's frequency characteristics.
③ 10.7MHz

Figure 9. Common-emitter Amplifier Circuit
By replacing RC with an LC network with a resonance frequency of 10.7MHz, a frequency selective amplifier can be obtained. As shown in the figure, the amplification factor is 35dB at 10.7M, while the amplification factor when detuning 1MHz is only 12.6dB. The disadvantage is that the pass-band is slightly wider, the rectangular coefficient is not good enough, and the equivalent quality factor of the loop is about 65.2, which is relatively large. In addition, the high-frequency decoupling capacitor has been changed to 1uF.
Resonant Amplifier Circuit Example:

Figure 10. Resonant Amplifier Circuit Example
Transistor amplifier circuit is the basis of an operational amplifier circuit, and common-emitter configuration is the most commonly used form. Drawing lessons from the feature that the amplifier's magnification can be easily determined by the ratio of two resistors, and the gain of the common emitter amplifier can also be approximated by the ratio of the two resistors.
1. What are transistor amplifiers used for?
Amplifiers are derived from the transistors because they are capable of operating under three regions active, cut-off and saturation. For the purpose of amplification, the focus will be on the active region. The main purpose of these amplifiers is to enhance the strength of the applied input signal without alteration.
2. How does a transistor amplify current?
Transistors are normally used as amplifiers. ... The small current travels from the voltage source into the base of the transistor. A current at the base turns on the transistor. The current is then amplified and travels from the emitter of the transistor to the collector.
3. What is a common emitter transistor amplifier?
The common emitter amplifier is a three basic single-stage bipolar junction transistor and is used as a voltage amplifier. The input of this amplifier is taken from the base terminal, the output is collected from the collector terminal and the emitter terminal is common for both the terminals.
4. Why common emitter is used in amplifier?
Common emitter (CE) configuration. ... Common emitter transistors are used most widely, because a common emitter transistor amplifier provides high current gain, high voltage gain and high power gain. This type of transistor gives for a small change in input there is small change in output.
5. What is the use of CE amplifier?
In electronics, a common-emitter amplifier is one of three basic single-stage bipolar-junction-transistor (BJT) amplifier topologies, typically used as a voltage amplifier. It offers high current gain (typically 200), medium input resistance and a high output resistance.
6. How does transistor work as amplifier?
A transistor acts as an amplifier by raising the strength of a weak signal. The DC bias voltage applied to the emitter base junction, makes it remain in forward biased condition. ... Thus a small input voltage results in a large output voltage, which shows that the transistor works as an amplifier.
7. What is common emitter amplifier circuit?
The Common Emitter Amplifier circuit has a resistor in its Collector circuit. The current flowing through this resistor produces the voltage output of the amplifier. ... The Base of the transistor used in a common emitter amplifier is biased using two resistors as a potential divider network.
8. What are the main parts of a transistor amplifier circuit?
A Single stage transistor amplifier has one transistor, bias circuit and other auxiliary components. The following circuit diagram shows how a single stage transistor amplifier looks like. When a weak input signal is given to the base of the transistor as shown in the figure, a small amount of base current flows.
9. What is the phase difference in common emitter amplifier?
The phase difference between the input and output voltage of CE amplifier circuit is. The phase difference of 1800 between the signal voltage and output voltage in a common emitter amplifier is known as phase reversal.
10. When an NPN transistor is used as an amplifier?
For a npn transistor to be used as an amplifier, forward bias has to be applied on the transistor. Thus, when an npn transistor is used as an amplifier, holes move from base to emitter. So, the correct answer is option D i.e. holes move from base to emitter.
11. When an NPN junction transistor is used as an amplifier in CE mode?
A transistor is used in the common emitter mode as an amplifier then: (A) the base emitter junction is forward baised. (B) the base emitter junction is reverse baised. (C) the input signal is connected in series with the voltage applied to bias the base emitter junction.
12. How is an NPN transistor used as an amplifier show with its circuit diagram?
The circuit of a common-emitter amplifier using an n-p-n transistor is shown below : In a common emitter amplifier circuit, the input signal voltage and output collector voltage are in opposite phase. i.e 180° out of phase. Thus the phase difference between the input signal and output voltage is 180°.
13. How does a common emitter amplifier work?
Operation of Common Emitter Amplifier
When a signal is applied across the emitter-base junction, the forward bias across this junction increases during the upper half cycle. This leads to an increase in the flow of electrons from the emitter to a collector through the base, hence increases the collector current.
14. What is β for a CE configuration?
Base Current Amplification Factor (β)
The base current amplification factor is defined as the ratio of the output and input current in a common emitter configuration. In common emitter amplification, the output current is the collector current IC, and the input current is the base current IB.
15. What is current gain CE configuration?
The current gain of a transistor in CE configuration is defined as the ratio of output current or collector current (IC) to the input current or base current (IB). The current gain of a transistor in CE configuration is high. Therefore, the transistor in CE configuration is used for amplifying the current.
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