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Discussion:
Using the measurements made in the procedure, calculate the quiescent current IE, and the internal emitter resistance re.
The quiescent current IE and internal emitter resistance re can be calculated using the DC analysis carried out in Step#1 and Step#2 of the procedure. The DC voltage measured across RE is used to measure the biasing current across the emitter of BJT. The voltage is measured to be 793mV. Since RE = 1KΩ, so:
IE=VERE=0.7931000=0.000793A=0.793mA.
Now, we can calculate the internal resistance of emitter using the formula provided in the procedure. So:
re=0.026IE=0.0260.000793=32.786Ω
Using the values obtained in the experiment; draw the small-signal ac equivalent circuit for the amplifier.
In this part, we will draw the small scale AC equivalent mode of the circuit connected in Fig. 3 as shown below:
In order to find the small scale AC equivalent model of the circuit, we need to do some transformations. The DC sources don’t have any frequency, so they don’t pose any change in AC analysis. As a result, we assume DC sources to be shorted. Moreover, the capacitors also act as short circuits to AC signals. Therefore, we replace capacitors and DC sources in circuit above to make the small scale AC equivalent model as shown below:
Using the equivalent circuit, calculate the theoretical values for AV, VL/VS, rin and rO. Compare these with the measured values.
Av≅Rcre=560032.786=171
The practically measured value of open-loop voltage gain is 167 that is quite close to the theoretical value.
Now, we find the value of ratio of load voltage to source voltage VLVS.
IE=Vsre=20mVpp32.786=0.61mA
Ic=αIE≅IE=0.61mA
VL=0.61mA×RC×RLRC+RL=2.19Vpp
VLVS=2.1920m=109.5
The measured value of VLVS is 84 that is quite close to the theoretically calculated result. The deviation is due to neglecting the internal resistance of the source during calculations.
The internal resistance of the circuit is:
rinstage=RE×reRE+re=1000×32.7861000+32.786=31.745Ω
The measured value of internal resistance is 32Ω that is nearly equal to the calculated value using formula. The output resistance is:
rostage=RC=5.6KΩ
It is equal to the measured value.
Explain the distortion observed in the procedure. Why does the output waveform distort when the amplitude of the input is increased above a certain value?
The BJT amplifiers act as amplifiers only in a limited region called as the linear region. This region is dependent on the input DC bias, source amplitude and the quiescent current. Beyond this region, the output will not be amplified as desired. It will either be clipped or assume some irregular shape. The main reason for voltage saturation is the DC bias voltage. Beyond this bias voltage, the output voltage will be clipped. In order to operate BJT for amplification, the quiescent point should ideally be set near the middle of the load characteristic curve. If the operating current is not in the middle, the BJT cannot work as amplifier for its full cycle of input voltage. Apart from clipping, the distortion may result if the operating point is either too low or too high in the load curve.
Conclusion:
In this lab, we understood the performance of the common base BJT as an amplifier. The mathematical model is developed to find voltage gains, input and output resistances. We carried out the small scale AC analysis to compare calculated and measured values. The formulas to compute open-loop voltage gain is developed along-with Load to Source voltage gain. We saw how to transform the actual BJT circuit into small scale AC equivalent model for simplification of circuit calculations. The unknown parameters can easily be calculated using the developed model. At the end, the phenomenon of amplifier distortion is discussed in detail that is a characteristic of every BJT acting as an amplifier.
