BJT - Small Signal Analysis

Introduction:

The different circuit values used are the ones which is used in Exercise 2 – BJT DC Biasing, for easier understanding since the DC values are computed and simulated hence the student have at least a small foundation on the circuit. The added instruments are function generator, oscilloscope and bode plotter. The function generator is easier to used compared to AC supply since it will already generate a peak value thus serving as an AC supply for the circuits. Oscilloscope is used to observe the output waveforms of the circuit. For this case, to verify if the output is 180 degrees out of phase to input( since it is a common emitter configuration). The bode plotter is used for the preparation for the oscillator exercise, and to find out the cut-off frequency of the circuit. An additional AC voltmeter is also used to measure the AC output voltage.

re model is used to represent the small signal values of the circuit.

The convert the circuit into AC equivalent circui, the capacitors should be shorted and also the DC supply. Here is the equivalent AC circuit ( re model)…

The B in BIb and Bre, represent b, this is because b character cannot be encoded in the EWB software.

Input and output impedance(Zi and Zo):

Zi = Rb//bre Zo = Rc

Voltage Gain(Av):

Vo = -bIbRc

since Ib = Vi/bre

Vo = -bViRc/bre

Vo/Vi = -bRc/bre

Av = -bRc/bre

Av = -Rc/re

Current Gain(Ai):

By current divisor theorem:

Io = bIb/Rc Ib = RbIi/(Rb + bre)

Io/Ib = b/Rc Ib/Ii = Rb/(Rb + bre)

Ai = Io/Ii = (Io/Ib)(Ib/Ii) = {bRb/Rc(Rb+bre)}

Ai = {bRb/Rc(Rb+bre)}

Simulated Circuit:

Computed Values:

re = 26mV/2.492mA bre = 150*10.45Ω

re = 10.45Ω bre = 1.57kΩ

Zi = Rb//bre Zo = Rc

Zi = 495kΩ//1.57kΩ Zo = 1.8kΩ

Zi = 1.57kΩ

Av = -Rc/re Ai = -Avbre//Rc

Av = -1.8kΩ/1.57kΩ Ai = -(-171.97*1.57kΩ)/(1.8kΩ)

Av = -171.97 è 44.70dB Ai = 150

Input Capacitor: Output Capacitor:

Ci = 1/fRi Co = 1/fRo

Ci = 1/(10kHz*1.57kΩ) Co = 1/(10kHz*1.8kΩ)

Ci = 63.69nF Co = 55.56nF

Voltage Ouput:

Av = -Vo/Vi

Vo = -AvVi

Vo = -171.97*(10mV/2^1/2)

Vo = -1.22V

Note: the 10mv is not used since it in peak value, while the voltage measured by the voltmeter are in terms of Vrms.

The computed output closely matched to its simulated output. Only the measure output voltage does not have a negative sign, this may be because this is in AC, by simple means of understanding, AC have no polarities since they have positive and negative values and thus this software assumed it is in such.

Phase Analysis:

The input and output capacitors are important in determining the proper phase of the output. For this case, the computed value of the capacitor is correct since the output of the circuit is 180 degrees out of phase to its input. It is so since common emitter configurations have a 180 degrees out of phase to its input.

Legend: Ouput

Input

The boxes above are the measurements of the output and input waveforms. The input box which is the blue box waveform, shows the maximum and minimum voltages which is 10.0167mV and -10.1566mV. It also shows the input frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

The ouput box which is the red waveform, shows the minimum and maximum voltages which are -1.8460V and 1.5622V. It also shows the output frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

We can observed the peak values of the waveforms are not equal, this may due to some inconsistencies, but at least in our computation we should assume a fixed value. Also, we should also be reminded no matter how basic this is that the oscilloscope measures the peak value of voltages while voltmeters measures the rms voltages which will be used later.

Frequency Response:

The bode plotter is used to determine the frequency response of the circuit with respect to its gain in decibels.

The box shows the cutoff frequency x1 which is 1.5114k which already has a gain io 41.3636 db (y1). The ideal gain of the circuit is 44.7 db (y2) which is the response of the circuit at 53.4212. Since this circuit allows high frequency to pass, this is called as a high pass filter.

Av(db) = 20log(Vo/Vi)

Av(db) = 20log(1.22V/7.07mV)

Av(db) = 44.74dB , the gain of the circuit at 10kHz…

The convert the circuit into AC equivalent circui, the capacitors should be shorted and also the DC supply. Here is the equivalent AC circuit ( re model)…

The B in BIb and Bre, represent b, this is because b character cannot be encoded in the EWB software.

Input and output impedance(Zi and Zo):

Zi = Rb//bre Zo = Rc

Voltage Gain(Av):

Vo = -bIbRc

since Ib = Vi/bre

Vo = -bViRc/bre

Vo/Vi = -bRc/bre

Av = -bRc/bre

Av = -Rc/re

Current Gain(Ai):

By current divisor theorem:

Io = bIb/Rc Ib = RbIi/(Rb + bre)

Io/Ib = b/Rc Ib/Ii = Rb/(Rb + bre)

Ai = Io/Ii = (Io/Ib)(Ib/Ii) = {bRb/Rc(Rb+bre)}

Ai = {bRb/Rc(Rb+bre)}

Ai = -AvZi/Rc

Simulated Circuit:

Computed Values:

re = 26mV/2.492mA bre = 150*10.45Ω

re = 10.23Ω bre = 1.53kΩ

Zi = Rb//bre Zo = Rc

Zi = 375kΩ//1.53kΩ Zo = 1kΩ

Zi = 1.53kΩ

Av = -Rc/re Ai = -AvZi/Rc

Av = -1.8kΩ/1.57kΩ Ai = -(-97.66*1.57kΩ)/(1kΩ)

Av = -97.66 è 39.79dB Ai = 149.42

Input Capacitor: Output Capacitor:

Ci = 1/fRi Co = 1/fRo

Ci = 1/(10kHz*1.53kΩ) Co = 1/(10kHz*1kΩ)

Ci = 65.36nF Co = 100nF

Voltage Ouput:

Av = -Vo/Vi

Vo = -AvVi

Vo = -171.97*(10mV/2^1/2)

Vo = -1.22V

Note: the 10mv is not used since it in peak value, while the voltage measured by the voltmeter are in terms of Vrms.

Phase Analysis:

Legend: Ouput

Input

The boxes above are the measurements of the output and input waveforms. The input box which is the blue box waveform, shows the maximum and minimum voltages which is 10.0178mV and -10.1612mV. It also shows the input frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

The ouput box which is the red waveform, shows the minimum and maximum voltages which are -1.0431V and 844.8063mV. It also shows the output frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

Frequency Response:

The bode plotter is used to determine the frequency response of the circuit with respect to its gain in decibels.

The box shows the cutoff frequency x1 which is 1.5114k which already has a gain io 36.2840 db (y1). The ideal gain of the circuit is 39.70 db (y2) which is the response of the circuit at 10kHz. Since this circuit allows high frequency to pass, this is called as a high pass filter.

Av(db) = 20log(Vo/Vi)

Av(db) = 20log(97.66)

Av(db) = 39.8dB , the gain of the circuit at 10kHz…

The convert the circuit into AC equivalent circui, the capacitors should be shorted and also the DC supply. Here is the equivalent AC circuit ( re model)…

The B in BIb and Bre, represent b, this is because b character cannot be encoded in the EWB software.

Input and output impedance(Zi and Zo):

Zi = Rp//bre ,where Rp = R1//R2 Zo = Rc

Voltage Gain(Av):

Vo = -bIbRc

since Ib = Vi/bre

Vo = -bViRc/bre

Vo/Vi = -bRc/bre

Av = -bRc/bre

Av = -Rc/re

Current Gain(Ai):

By current divisor theorem:

Io = bIb/Rc Ib = RpIi/(Rp + bre)

Io/Ib = b/Rc Ib/Ii = Rp/(Rp + bre)

Ai = Io/Ii = (Io/Ib)(Ib/Ii) = {bRp/Rc(Rp+bre)}

Ai = {bRb/Rc(Rp+bre)}

Ai = -AvZi/Rc

Simulated Circuit:

Computed Values:

re = 26mV/2.485mA bre = 150*10.45Ω

re = 10.48Ω bre = 1.57kΩ

Zi = Rp//bre ,where Rp = R1//R2 Zo = Rc

Zi = 490.83kΩ//1.57kΩ Zo = 1kΩ

Zi = 373.92Ω

Av = -Rc/re Ai = -AvZi/Rc

Av = -1kΩ/10.48Ω Ai = -(-958.42*373.92Ω)/(1kΩ)

Av = -95.42 è 39.59dB Ai = 35.68

Input Capacitor: Output Capacitor:

Ci = 1/fRi Co = 1/fRo

Ci = 1/(10kHz*1.57Ω) Co = 1/(10kHz*1kΩ)

Ci = 96.40µF Co = 100nF

Voltage Ouput:

Av = -Vo/Vi

Vo = -AvVi

Vo = -95.42*(10mV/2^1/2)

Vo = -674.7mV

Note: the 10mv is not used since it in peak value, while the voltage measured by the voltmeter are in terms of Vrms.

Phase Analysis:

Legend: Ouput

Input

The boxes above are the measurements of the output and input waveforms. The input box which is the blue box waveform, shows the maximum and minimum voltages which is 9.9993mV and -9.9993mV. It also shows the input frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

The ouput box which is the red waveform, shows the minimum and maximum voltages which are -1.1504V and 55.8634mV. It also shows the output frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

Frequency Response:

The bode plotter is used to determine the frequency response of the circuit with respect to its gain in decibels.

The box shows the cutoff frequency x1 which is 10.33Hz which already has a gain of 33.4455db (y1). The ideal gain of the circuit is 39.5767 db (y2) which is the response of the circuit at 10kHz. Since this circuit allows high frequency to pass, this is called as a high pass filter.

Av(db) = 20log(Vo/Vi)

Av(db) = 20log(95.42)

Av(db) = 39.7dB , the gain of the circuit at 10kHz…

The convert the circuit into AC equivalent circui, the capacitors should be shorted and also the DC supply. Here is the equivalent AC circuit ( re model)…

The B in BIb and Bre, represent b, this is because b character cannot be encoded in the EWB software.

Input and output impedance(Zi and Zo):

Zi = re/{ (1/b) + (Rc/Rb)} Zo = Rc//Rb

Voltage Gain(Av):

Vo = -bIbRc

since Ib = Vi/bre

Vo = -bViRc/bre

Vo/Vi = -bRc/bre

Av = -bRc/bre

Av = -Rc/re

Current Gain(Ai):

By current divisor theorem:

Io = bIb/Rc Ib = RpIi/(Rp + bre)

Io/Ib = b/Rc Ib/Ii = Rp/(Rp + bre)

Ai = Io/Ii = (Io/Ib)(Ib/Ii) = {bRp/Rc(Rp+bre)}

Ai = {bRb/(Rb+bRc)}

Simulated Circuit:

Computed Values:

re = 26mV/4.45mA bre = 150*5.84Ω

re = 5.84Ω bre = 876Ω

Zi = re/ { (1/b) + (Rc/Rb)} Zo = Rc//Rb

Zi = 5.84Ω/{ (1/150) + (1kΩ/126.67kΩ)} Zo = 1kΩ//126.67kΩ

Zi = 401.07Ω Zo = 992.17kΩ

Av = -Rc/re Ai = {bRb/(Rb+bRc)}

Av = -1kΩ/5.84Ω Ai = { (150)(126.67kΩ) }/ { (126.67kΩ+(150)(1kΩ)}

Av = -171.23 è 44.67dB Ai = 68.68

Input Capacitor: Output Capacitor:

Ci = 1/fRi Co = 1/fRo

Ci = 1/(10kHz*131.24kΩ) Co = 1/(10kHz*1kΩ)

Ci = 249nF Co = 100nF

Voltage Ouput:

Av = -Vo/Vi

Vo = -AvVi

Vo = -171.23*(10mV/2^1/2)

Vo = -1.21V

Note: the 10mv is not used since it in peak value, while the voltage measured by the voltmeter are in terms of Vrms.

Phase Analysis:

Legend: Ouput

Input

The boxes above are the measurements of the output and input waveforms. The input box which is the blue box waveform, shows the maximum and minimum voltages which is 10mV and -10mV. It also shows the input frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

The ouput box which is the red waveform, shows the minimum and maximum voltages which are -1.8196V and 1.5270. It also shows the output frequency which is 10Khz (1/dx), this frequency is generated by the function generator.

We can observed the peak values of the waveforms are not equal, this may due to some inconsistencies at the start of the charging and discharging of the capacitors, but at least in our computation we should assume a fixed value. Also, we should also be reminded no matter how basic this is that the oscilloscope measures the peak value of voltages while voltmeters measures the rms voltages which will be used later.

Frequency Response:

The bode plotter is used to determine the frequency response of the circuit with respect to its gain in decibels.

The box shows the cutoff frequency x1 which is 1.5kHzHz which already has a gain of 41.2650db (y1). The ideal gain of the circuit is 44.48 db (y2) which is the response of the circuit at 10kHz. Since this circuit allows high frequency to pass, this is called as a high pass filter.

Av(db) = 20log(Vo/Vi)

Av(db) = 20log(171.23)

Av(db) = 44.67dB , the gain of the circuit at 10kHz

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