Transistor low frequency amplifier

Transistor low-frequency amplifiers are mainly used to amplify low-frequency small-signal voltages, with frequencies ranging from tens of hertz to one hundred kilohertz
First, the bias circuit of the transistor
In order for the amplifier to achieve linear amplification, the transistor must not only have a suitable static operating point, but must also stabilize the operating point. Due to the influence of temperature on the tube parameters β, Icbo, Ube, they are finally reflected in the change of Ic. In order to eliminate this effect, we stabilize the static operating point through the negative feedback of the DC or voltage biased by the transistor. The two kinds of bias circuits and working point stability principle are as follows
Table I, Bias circuit of transistor amplifier
Working point stability principle
Calculation formula
Negative current feedback
Set temperature T ↑, DC negative feedback process
As a result, Ic remains unchanged
U = (1 / 3-1 / 5) Ec
Re = (1 / 3-1 / 5) Ec / Ic
Rb = Rb1 // Rb2
=== (2-5) Re
Ub = Rb1Ec / (Rb1 + Rb2)
Voltage negative feedback
Set temperature T, DC negative feedback process
As a result, Ic remains unchanged
Rb = β (Ec-Ube) / Ic-βRc
Ic = Ec / (Rc + Rb / β)
According to experience, usually take
Rb / Rc = (2-10)
Second, the three circuit forms of the amplifier
The amplifier is a three-terminal circuit, one of which must be the common “ground” terminal of the input and output. It is a common circuit, which is connected to the base and is called a common circuit. These three have different performances, see the table below
Three circuit forms and their performance comparison
Voltage magnification
Current magnification
Input resistance
Output resistance
Common firing circuit




Common circuit


Due to different loads, up to about 50MΩ

Common base circuit


About 10-500Ω



3. Graphic method
The so-called graphical method is to use the input and output characteristic curves of the transistor to analyze the performance of the amplifier by drawing. The graphical method can intuitively and comprehensively indicate the working process of the transistor amplification, and can calculate some performance indicators of the amplifier. Examples to illustrate the graphical process of graphical methods,
Example: Knowing the parameters and input voltage Ui = 15sinωt (millivolt) in the circuit shown in the figure below requires graphically determining the static operating point parameters Ibq, Icq, and Iceq of the circuit, and calculating the voltage and current amplification factors Ku and Kio.

Graphical steps
1. Determine the static working point of the base loop, and select the midpoint Q of the straight line segment from the input characteristic curve (Ubeq = 0.7 V at this point, Ibq = 40 μA) as the static working point of the base loop, by selecting the appropriate Eb or Rb (usually by adjusting Rb) to meet the requirements of the working point,
2. Make the DC load line. From the above figure, the load line equation is Uce = Ec-IcRc. Its trajectory is a straight line. If Ic = 0, Uce = Ec = 20 volts, and mark N on the horizontal axis. Point; and let Uce = 0, Ic = Ec / Rc = 20V / 6kohm = 3.3mA, mark M point on the vertical axis, connecting M and N is the DC load line. It intersects with the output characteristic curve of Ib = 40 μA at Q, and finds Icq = 1.8 mA, Uceq = 9 V from point Q, point Q is the static operating point of the collector circuit. For simplicity, the static current , The voltage is no longer added with subscript Q, Ic, Ie is Icq, Ieqo
3. Make a waveform, make a waveform Ut = 15sinωt (millivolt) on the input characteristics, and make the waveforms of ib, ic, and Uce according to the waveform of Ut. The following points can be obtained from the graphical method (1) The sinusoidality of the waveform can be judged Whether the selection of static operating point Q is appropriate.
(2) From the diagram, it is known that the input voltage Ui is opposite to the collector output voltage Uo, and the base current ib and the collector current Ic are in phase with the input voltage Ui.
(3) The above graphic method is carried out under no-load conditions. If the effect of the load resistance RL is considered, the AC load should be RL = RC // RL. Since the AC load line and the DC load line both intersect at Q, a straight line M'N 'with an inclination angle a' = (arctg) 1 / RL is made through the Q point, which is called an AC load line.

4. Equivalent circuit method and h parameter
1. The simplified "parameter change" of the equivalent circuit of the h parameter means that the Ib, Ube, Ic, and Uce of the transistor change only slightly in the vicinity of the static operating point Q. Ib and Ube are the input variables of the transistor, and Ic and Uce are the output variables. If you regard the transistor as a two-port network with a controlled source, you can use four h parameters to simulate the physical structure of the transistor, so that the equivalent circuit of the h parameter of the transistor is shown in Figure 7-1-4. The definition of h is as follows :
hie = â–³ Ube / â–³ Ib ------- â–³ Uce = 0, --hfe = â–³ Ic / â–³ Ib ----- â–³ Uce = 0
hre = â–³ Ube / â–³ Uce ------ â–³ Ib = 0, --hoe = â–³ Ic / â–³ Uce ---- â–³ Ib = O
Several parameters have their own physical meanings: hie is the input resistance when the output is short-circuited, that is, the reciprocal of the slope of the input characteristic curve; hfe is the current amplification factor of the short-circuit at the output, ie β (common emitter) or a (common base Pole); hre is the internal feedback coefficient of the open circuit at the input terminal, which indicates the degree of the influence of the output voltage on the input voltage; hoe is the output conductance when the input terminal is open circuit, that is, the slope of the output characteristic curve. When the transistor works at low frequency, hre and The two parameters of hoe are so small that they are negligible. Usually, two parameters of hie and hre are used to simulate the low-frequency transistor circuit. This is called a simplified equivalent circuit of h parameters, as shown in Figure 7-1-3. , Β is the above hie, hfe. The current amplification factor β (or hfe) can be obtained from the output characteristic curve or tested by the instrument, and the input resistance rbe is calculated by the following formula:
rbe = rb + (β + 1) 26 (millivolt) / Ie (milliampere)
In the formula: Rb is the base resistance, about a few hundred ohms, Ie is the static emitter current. The method to find the micro-variable equivalent circuit of the transistor amplifier is as follows: (1) The equivalent of the transistor shown in Figure 7-1-3 Analog type replacement; (2) All DC power supplies, DC blocking capacitors and bypass capacitors are regarded as short circuits; (3) Other components are drawn according to their original relative positions,
The equivalent circuit can be used to find the amplifier's amplification factor, input resistance, output resistance, and analyze the frequency characteristics of the amplifier.

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