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| Common Emitter Amplifier Design Reports |
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| 发布时间：2015-03-18 点击： 次 |
The front:
The function of the amplifier circuit is using the effect of current control triode, or a field effect of transistor voltage control, thus, to change the weak signals without distortion amplification into the required numerical implementation, and achieve the energy of DC power supply into according to the input signal and output signal changes bigger energy. The essence of the amplification circuit, is a smaller energy to control the energy conversion device larger energy conversion.
The principle of amplification circuit must be a DC power supply, and the power supply should be set to ensure the transistor or FET amplifying condition in the linear; The arrangement of components must ensure that the signal transmission, to ensure that the signal can be input from the input end of the amplification circuit, and the amplification circuit output from the output end; The selection of element parameters to signal can be amplified to ensure without distortion, and satisfy the performance index of the amplification circuit.
A. Complete the design of the amplifier, clearly showing all the design steps
One, the purpose of design
1,master the concept of DC channel of the amplifier circuit and communication channel ; And also comprehend the resistance of the static working point and Calculation and the input, the output.
2 learn to draw the DC circuit, communication channel correctly and can master the area which is divided into two paths. This is helpful to understand the circuit analysis method, also can improve students' ability of reading electronic circuit.
3 continually asking questions by using the heuristic ways of teaching, and give students full of imagination, arouse interest, cultivate the ability to analyze problems, and solve the problem.
Two, The composition and action of element
1,Collector power UCC is the energy of amplifying circuit. To provide energy for the output signal, and to ensure that the emitter junction is forward biased, the collector is reverse biased, and the transistor in the enlarged area, the value of UCC is generally a few volts to tens of volts.
2, Transistor V is the core component of amplification circuit. Using the control role of transistor in the current amplification area, namely current IC = β IB amplification, make the weak signal amplification.
3,RL is the collector of the transistor collector resistance load resistance. It will change in collector current to voltage changes, and realize the voltage amplification circuit. RL is generally thousands to tens of thousands of Europe.
4, Base resistance RB is to ensure amplification in the state. Change RB to make transistor suitable quiescent point. RB usually takes tens of ohms to several hundred kilohms.
5, coupling capacitors C1, C2 plays the role of the exchange blocking circulation. In the signal frequency range, that the capacitance is approximately zero. So the analysis of the circuit, in the DC path capacitance can be regard as open circuit, and the AC pathway can be regard as short circuit capacitance. C1, C2 generally stays in 10 micro method to tens of micro method with electrolytic capacitor polarity.
THREE, Working principle
1, UI should be directly added to the triode V between the base and the emitter base current, caused by the iB for the corresponding changes.
2, Amplification by V current, collector current iC V will also change.
3, The changes in iC cause the changes between the collector and emitter and Voltage uCE V.
4, When the AC component uce in uCE smoothly transmits to load RL after C2, it becomes the output AC voltage uo, the voltage amplification.
Four, The design procedure and the results
Known number:
VCC=10V, VC=5V, RL=1kΩ, Zout= 500Ω, Zin=1kΩ, Gain (small signal)= 5
As above, the equivalent figure can get corresponding
B . Implement the circuit in Multisim and report the value of VCq, Zin and Zout obtained. Analyse the differences.
Because Vc=1000 ,Zout was obtained by Rl and Rc, in order to know Rl=1000. after calculation, Rc=1000 so, Z is composed of two parts, respectively is Rl=1000. and Rc=1000.
C. Choose Vin at 50kHz. Plot Vin(t), Vc(t) and Vo(t) to verify the gain. Set the amplitude of input voltage so that output is undistorted. Show the graph.
The power supply voltage of Vcc and a triode Uceq static pressure is determined by the Uom,
Uceq>=Uom+Uces，Icq*（Rc//Rl）>=Uom.
Because the output voltage distortion conditions are: Uceq>=Uom+Uces, Icq* (Rc//Rl) >=Uom.
After calculation, Uceq>=Uom+Uce, does not meet the conditions of the distortion, distortion.
D . How large can you make input voltage before Vc(t) and Vo(t) distorts? Show the graph. Plot about 4 or 5 cycles of the signals. Comment on your results.
The static working point Q
IBQ=Ucc/Rb=10/1000=0..001A
Icq=B*IBQ=0.005A
Rbe-300+ (1+B) 26/Icq=300+ (1+5) 26/0.1=2460
Input resistance Ri=Rb//Rbc=500/2.46=203
The output resistor Ro=Rc=1000
E.What change in output voltage would result when the load current varies over its full range?
Impact, because Vin changes, will cause the voltage and current changes.
F.Change Vin frequency to 50MHz and explain the changes in various outputs
The frequency of Vin is changed to 50 MHz, Ie and Ve should be larger, Ic will become smaller, so the Vc will become big.
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