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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