高频电子线路实验指导书（英文版）

1st Lesson: Single-tuner resonant amplifier circuit

A. Experiment aim

1．Get familiar with the composing of Single-tuner resonant amplifier circuit and usage. 2．Test its performance parameters that take place the resonance.

3．Mastering the principle of high-frequency low signal tuner amplifier and the methods on designing high-frequency signal amplifiers.

B. Apparatus

1、 2、 3、

Oscillograph one Digital multimeter one

High-frequency experimentation case one

C. Experimental circuit principle

This experiment consult circuit diagram 1-1, using emitter to a total of Law. Load of collector of transistor is parallel resonance LC which can do the work of amplifying and frequency choosing at the same time. In the circuit, the standard point of the transistor depends on resistance of RB1, RB2 & RE.

The characteristics of tuner amplifier is the load of amplifier being not pure resistance, but the parallel resonance circuit build up by L & C. Because of the changing of impedance of LC, at the point of f0=

12?LC, the impedance get the apogee as a resistance performance with

the highest magnification. In this case, we can choose to amplify the very frequency signals. So, it is widely used as selected frequency amplifier in high and middle frequency environment.

Pic 1-1 Single-tuner resonant amplifier circuit

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D. Process

1、Get the modular G2 connected and the small high-frequency signal tuner amplifier parts JA1. Connect the +12V power supplier. 2、Transform K1 to right to make the LED on. 3、Adjust the standard point of the transistor.

Test to read the voltage of ejector of QA1 with the INA1 connected to ground. Adjust WA1 to make UEQ=2.25V. And then, read and notate the standard point. metrical VBQ VEQ VCQ VCEQ Whether Q1 working as amplifier depend on Vce Yes No Warrant: when Q1 is working as amplifier：VBEQ=VBQ-VEQ≈0.6V~0.7V，VCEQ equals to VCQ-VEQ

4、The frequency of circuit resonant f0

INA1 connects to signal of sine wave Ui with fs?10.700MHz and amplitude

V1P~P?20mV. Connect INB1 to oscillograph and fine-tuning CCA2 to make the graph on the oscillograph with the most amplitude. Draw down the wave graph. 5、Voltage magnification AV0

Based on process 4, the Ui is for input signal. Read out the quantity of Ui and the output signal Uo. So, the magnification can be calculated by Uo/Ui. 6、Band Δf0.7, Rectangular coefficient Kr0.1 and Q

Based on resonant state, change the input frequency with keeping the amplitude static. Read and note down the output voltage into the form below without the distortion ones.

Form 1-1 fS（MHz） Uo f0.1L f0.3L f0.5L f0.7L f0 f0.7H f0.5H f0.3H f0.1H Characteristics of amplitude-frequency Band： Δf0.7 = fH-fL

Q?

f0?f0.7；Rectangular coefficient：Kr0.1=

?f0.1 ?f0.7- 2 -

Pic. 1-2 Characteristics of amplitude-frequency

E. Requirements for the Report of Experiment 1. Finishing the experimental data;

2. Calculate DC points compare with experiment result;

3. Calculate Voltage gain when f0, bandwidth, Rectangular coefficient and Q according to form 1-2

4. Talk about the characteristics of single-tuner circuit resonant

F. Thinking

1.Talk about the characteristics of Single-tuner resonant amplifier circuit and RC-coupled amplifier

2.What effect does damping resistance R make to the gain?

3.What is the reason leading to unstable of high-frequency low signal tuner amplifier? How to make deal with self-incentive?

G．Words

单调谐回路谐振放大器: Single-tuner resonant amplifier

高频小信号调谐放大器: high-frequency low signal tuner amplifier 示波器: Oscillograph

选频放大器: selected frequency amplifier

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2nd Lesson: High-frequency resonant amplifier

A. Experiment aim

1、Understand principle of resonant power amplifier, effect of load impedance and voltage changing of encouraging signals.

2、Get mastery of characteristics of resonant power amplifier.

B. Apparatus

1、Oscilloscope one 2、Digital multimeter one 3、High-frequency experimentation case one

C. Experimental circuit principle

Resonant power amplifier uses selected frequency net as the load circuit. According to current amplifier angle range θ, we considered them as A, B,C, D 4 kinds. The smaller the angle, the more the larger efficiency η is. Kind A,θ=180,η no larger than 50%，when kind C with θ< 90o, η less or equal to 80%. A general as a power amplifier output power of the middle-class or smaller level at the end of the use of power amplifier, when general C amplifier is usually as a end-class to get greater power output and high efficiency. Picture 2-1 is one of the high-frequency amplifiers composed by two gear of amplifier.

Pic 2-1 high-frequency amplifier

The transistor VT1 composes general A amplifier, working the statue of lineal amplifier.

RB1 and RB2 are bias resistance at base with DC negative feedback RE1 and AC negative feedback RF1, to steady the point. The transistor VT2 composes general C amplifier, with θ=70o, base bias uses emitter current IEO that generates the voltageVBB on resistance of Re, with DC feedback 30Ω, AC feedback 10Ω, collector resonant circuit capacitance 82pF and load of 50Ω. The output uses transformer coupled output with middle tap for impedance matching.

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D. Process

Connect sending modular G1, according to high-frequency resonant amplifier in sending modular and connect the +12V power.

1、Switch K2 left and change WE1, to make voltage on emitter QE1 VE=2.2V. 2、Connect JE3 and JE6.

3、At INE1, input 10.700MHz carrier signal with VP-P=250mV. Change CCE1 and CCE2 to make the output wave max at ANTE1 from the Oscilloscope.

Details:

①．QE1 on collector, inching CCE1 to make the output max without distortion ②．QE2 on collector, inching CCE1 to make the output max without distortion

③．QE2 on ejector, the correct waveform is spires cosine pulse. You should decrease the rate if it concave. Note down the waveform and UA。 ④．Note down the waveform of ANTE1 4、Characteristics of load

After the 3rd process, change the load to make the resistance from 25Ω through 51Ω to 100Ω, with the circuit from owe-voltage through critical to pass-voltage on TTE1. In the circuit JE3，JE4 and JE5, the following resistances are 51Ω, 51Ω & 100Ω, so, R=25Ω（51Ω||51Ω）, that is, connect JE3，JE4 the same time. Note down the test digital below. Warrant： Po= （Uo* Uo）/8R fs=10.700MHz ，VP-P=UA R= 25Ω 51Ω 100Ω 5、Amplifier Characteristics：

①．After the 4th process, connect the JE3 & JE6 and note down the ejector voltage waveform on QE2.

②．Increase the input signal rate till the ejector on QE2 turn out to be concave, saying VT2 is in pass-voltage state. Note down the waveform.

③．Increase the input signal rate more; you can see the depth of the concave increasing.

Uo(Test) Po(Cal) E. Requirements for the Report of Experiment

1、Paint the pictures of current of three kinds of amplifier working 2、Paint the characteristics of load 3、Analyze the experience result

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F. words

高频谐振功率放大电路: High-frequency resonant amplifier 谐振功率放大器: resonant power amplifier

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3rd Lesson: Sine oscillator

A. Experiment aim

1、Get master of the transistor working state and the effect made by feedback to waveform 2、Get master of principal of advanced capacitance three-point sine oscillator and the way to test the effect

3、Research the effect made by environment

4、Compare the stability of IC oscillator and the crystal oscillator

B. Apparatus

1、Oscilloscope one 2、Digital multimeter one 3、High-frequency experimentation case one

C. Experimental circuit principle

1．Principal of LC sine oscillator

Sine oscillator is a widely used circuit. It can output nearly ideal sine waveform. There are 3 kinds of circuits of RC, LC and crystal. Now, the goal of this experience is to research the LC 3-points oscillator and crystal ones. The circuit is below:

Pic 3-1 Three-point Oscillator AC equivalent circuit

There are X1, X2 and X3, 3 reactor components in the circuit. According to the phase balance qualification, X1 and X2 must be the same reactor characteristics, when X3 must be the opposite, with the relationship:

X3??(X1?X2)

If X1 and X2 are capacitive reactance components and X3 is inductance reactance components, it is capacitive three-point oscillator. The opposite is inductance three-point oscillator.

2．Common base of colpitts oscillator basing on transistor The basic circuit is below:

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Pic. 3-2 Common base of colpitts oscillator basing on transistor

From the circuit above, we can see the 2 components connected to the ejector are same reactance C1 and C2, and the 2 components connected to base are opposite reactance C2 and L. According to the principal of LC oscillator, this circuit can meet the conditions of phase, so only if we meet the conditions of amplitude, it will oscillate as sine, that is,

A0?F?1

A0 is the voltage gain of little signal when it begins. F is the feedback coefficient. Y equivalent circuit of Pic. 3-2 is below:

Pic. 3-3 Y equivalent circuit

According to 3-3, have:

??yfbV0? A0???YVi- 8 -

?V??f?Z2 F?Z1?jx1V0yfbZ2??A0?F???

YZ2?jx1??F?VZ2C1Cf ???''?V0Z1?jx1C1?C2C2??F? increasing. But, if F is over much, because of Absolutely, increasing of F can make A0??F? decreasing and lead to it fail to oscillator. the gib decreasing the gain, may make the A0??F?＞1, Y should be much enough. Hence, we always make the If F is little and keep Afb0feedback coefficient at range of

0.125~0.5. When the feedback coefficient F is sure, the

only needed is A0, so we can calculate the coefficient of the other components in the circuit.

E. performance effected by oscillator state

When the load resistance and the feedback coefficient F are sure, the static point have effect to oscillating state. Generally, if the point higher, the state can working as saturated and the output resistance decreasing would lead to the distortion of the oscillating waveform badly even stop oscillating. But, the opposite, if the point lower, it is not easy to start oscillating because of approaching to the deadline. Factual, we always use self-bias circuit for improve the efficient of the low power oscillator and stability. Before it starts, the DC self-bias depends on the current IEO and Re,

VBEQ?VB?IEQ?RE

The static point should be low to easily oscillating. Beginning, the oscillating amplitude skyrocket but when feedback voltage Uf is positive-half to the base bias, the instantaneous voltage Ube?Ubeq?Uf turn to be more positive with ic increasing, the current charge to

Ce. But when Uf is the negative-half, the bias voltage decrease, even the negative, the charge on Ce will discharge to Re with the coefficient τdis=Re·Ce（τdis>>τcha）. During one cycle of Uf, the charge accumulated more than released, lead to the amplitude of Re with self-bias voltage. After several cycles, it get dynamic equilibrium to found stability voltage IEO·Re, and the voltage of BE is:

VBEO=VB—IEO·Re

Because of IEO＞IEQ, UBEO＜UBEQ. We can see the bias of BE decrease and the point moves to deadline. That is, at the beginning, the amplitude is low and it working at the A state with little changing of self-bias. Later, with the positive feedback, the amplitude increase soon to get the non-lineal zone with the self-bias makes the UBE deadline. But the oscillator’s

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non-lineal working, limits the increasing of the amplitude. Hence, only if Re and Ce are appropriate, the self-bias will change with the amplitude timely till the equilibrium time to output the static amplitude sine wave.

E. frequency stability of oscillator

The stability stands for the frequency changing range at the very time range or the changing of the condition. The less changing volume, the higher stability. To get stability advanced is to decrease the oscillating frequency affected by temperature, load, power and etc. The oscillating circuit is the main component leading to the oscillating frequency. So, the main way to oscillating frequency stability advanced is increase the ability of keeping the frequency when the condition changed.

Three ways to get stability of oscillating advanced 1）use better stability and higher Q capacitance and ICP

2）use capacitance with negative coefficient to implementation temperature

3）use the part connect way to decrease unstable transistor capacity effect to oscillating frequency.

According to the circuit, the oscillating frequency is:

1f0?

2?LCWhen f0?10.7MHz，L?2.2?H, have

C?1?100pF

(2? f0)2LC is the total capacity in the circuit. D. Process

Connect the cycle module G4, and the power of＋12V.

1、Turn the switch down. Disconnect all the junctions of VT1, and make W1 to VE=2V. 2、（1）Connect J54 and J52, without any of the junctions. Change the capacitance CC2, and watch the waveform of oscillating to make the frequency to 10.700MHz. Then, change W2 to the max of the output. Note down the coefficient.

（2）Connect J53 and J55 without all the junctions. Change CC1 to make the frequency 10.245MHZ and note down the coefficient.

3、watch the relationship between oscillating state and the working state of transistor Connect J52 and J54 with all the junctions opening. Watch the waveform, change W1, and watch the waveform on TT1 state, again with the ejector voltage. Then calculate the IE。IE = VE1/R4 ）。Warrant：R4 =1K

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