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I'm designing a simple amplifier for educational purposes. Please ignore that this amplifier is very sensitive to changes in the transistor's DC current gain. I calculated some component values, then simulated the circuit and realized my calculations were way off. Could you help me identify where I went wrong? Why is the gain not even close to what I calculated? How can I fix it?
Consider the given circuit, along with its constraints and requirements:
Constraints:
Requirements:
My train of thought:
First, I read in the transistor's datasheet that the DC current gain is between 100 and 300 for the following values: \( I_C \) = 150mA and \( V_{CE} \) = 10V. Since I have the supply voltage \( U_b \) = 12 V and it is recommended to set the collector voltage at the middle of the supply voltage rail, I decided to choose \( V_{CE} \) = 6 V. Further I did assumtion that \( I_E \approx I_C \).
This allows me to make the following equation:
\( U_b = I_C R_C + U_{CE} + I_C R_E \)
Next, I assumed that the voltage drop across the base-emitter is \( U_{BE} \) = 0.7 V. Further I now that \( I_C = \beta I_B \).
This allows me to make the following equation:
\( U_b = I_B R_B + U_{BE} + I_C R_E \)
Finally, I aimed for a gain of 20. Since the circuit is a common-emitter amplifier, the following applies:
\( A = -S r_{out}\), where
A = -20
\( S = \frac{I_C}{U_T} (1 + \frac{U_{CE}}{|U_A|}) \)
\( r_{out} = (R_C || r_{CE} || R_L)\) with \( r_{CE} = \frac{|U_A| + U_{CE}}{I_C}\) to model the Early effect.
My calculations:
First, calculate \( R_C \):
\( S = \frac{I_C}{U_T} (1 + \frac{U_{CE}}{|U_A|}) = \frac{150 mA}{26 mV} (1 + \frac{6 V}{100 V}) \approx 6.115 \frac{1}{\Omega}\)
\( A = -S r_{out} \implies \frac{A}{-S} = r_{out} \implies \frac{-20}{-6.115} = r_{out} \approx 3.271 \Omega \)
\( r_{CE} = \frac{|U_A| + U_{CE}}{I_C} = \frac{100 V + 6 V \cdot150 mA}{150 mA} \approx 672.67 \Omega \)
\( \frac{1}{r_{out}} = \frac{1}{R_C} + \frac{1}{r_{CE}} + \frac{1}{R_L} \implies \frac{1}{R_C} = \frac{1}{r_{out}} - \frac{1}{r_{CE}} - \frac{1}{R_L} \implies R_C \approx 3.29 \Omega \)
Second, calculate \( R_E \):
\( U_b = I_C R_C + U_{CE} + I_C R_E \implies R_E = \frac{U_b - I_C R_C - U_{CE}}{I_C} = \frac{12 V - 150 mA \cdot 3.29 \Omega - 6 V}{150 mA} \approx 36.71 \Omega \)
Finally, calculate \( R_B \):
\( I_C = \beta I_B \implies I_B = \frac{I_C}{\beta} = \frac{150 mA}{200} = 750 \mu A\)
\( U_b = I_B R_B + U_{BE} + I_C R_E \implies R_B = \frac{U_b - U_{BE} - I_C R_E}{I_B} = \frac{12 V - 0.7 V - 150 mA \cdot 36.71 \Omega}{750 \mu A} \approx 7.725 k\Omega\)
Simulation:
As you can see, the amplified signal (blue) is even weaker than the input signal (green).
Consider the given circuit, along with its constraints and requirements:
Constraints:
- Input \( u_e \) is a sine wave signal with the 50 mV peak-to-peak voltage and 1 kHz frequency.
- Transistor is 2N2222 with the SPICE model
.model 2N2222 NPN(IS=1E-14 VAF=100 BF=200 IKF=0.3 XTB=1.5 BR=3 CJC=8E-12 CJE=25E-12 TR=100E-9 TF=400E-12 ITF=1 VTF=2 XTF=3 RB=10 RC=.3 RE=.2 Vceo=30 Icrating=800m mfg=NXP) - Load resistance \( R_L \) = 2 kΩ.
- Supply voltage \( U_b\) = 12 V.
- Two coupling capacitors \( C_1 \) and \( C_2 \) with the capacitance of 10 μF.
- Temperatur voltage \( U_T \) = 26 mV.
Requirements:
- Gain factor should be -20.
My train of thought:
First, I read in the transistor's datasheet that the DC current gain is between 100 and 300 for the following values: \( I_C \) = 150mA and \( V_{CE} \) = 10V. Since I have the supply voltage \( U_b \) = 12 V and it is recommended to set the collector voltage at the middle of the supply voltage rail, I decided to choose \( V_{CE} \) = 6 V. Further I did assumtion that \( I_E \approx I_C \).
This allows me to make the following equation:
\( U_b = I_C R_C + U_{CE} + I_C R_E \)
Next, I assumed that the voltage drop across the base-emitter is \( U_{BE} \) = 0.7 V. Further I now that \( I_C = \beta I_B \).
This allows me to make the following equation:
\( U_b = I_B R_B + U_{BE} + I_C R_E \)
Finally, I aimed for a gain of 20. Since the circuit is a common-emitter amplifier, the following applies:
\( A = -S r_{out}\), where
A = -20
\( S = \frac{I_C}{U_T} (1 + \frac{U_{CE}}{|U_A|}) \)
\( r_{out} = (R_C || r_{CE} || R_L)\) with \( r_{CE} = \frac{|U_A| + U_{CE}}{I_C}\) to model the Early effect.
My calculations:
First, calculate \( R_C \):
\( S = \frac{I_C}{U_T} (1 + \frac{U_{CE}}{|U_A|}) = \frac{150 mA}{26 mV} (1 + \frac{6 V}{100 V}) \approx 6.115 \frac{1}{\Omega}\)
\( A = -S r_{out} \implies \frac{A}{-S} = r_{out} \implies \frac{-20}{-6.115} = r_{out} \approx 3.271 \Omega \)
\( r_{CE} = \frac{|U_A| + U_{CE}}{I_C} = \frac{100 V + 6 V \cdot150 mA}{150 mA} \approx 672.67 \Omega \)
\( \frac{1}{r_{out}} = \frac{1}{R_C} + \frac{1}{r_{CE}} + \frac{1}{R_L} \implies \frac{1}{R_C} = \frac{1}{r_{out}} - \frac{1}{r_{CE}} - \frac{1}{R_L} \implies R_C \approx 3.29 \Omega \)
Second, calculate \( R_E \):
\( U_b = I_C R_C + U_{CE} + I_C R_E \implies R_E = \frac{U_b - I_C R_C - U_{CE}}{I_C} = \frac{12 V - 150 mA \cdot 3.29 \Omega - 6 V}{150 mA} \approx 36.71 \Omega \)
Finally, calculate \( R_B \):
\( I_C = \beta I_B \implies I_B = \frac{I_C}{\beta} = \frac{150 mA}{200} = 750 \mu A\)
\( U_b = I_B R_B + U_{BE} + I_C R_E \implies R_B = \frac{U_b - U_{BE} - I_C R_E}{I_B} = \frac{12 V - 0.7 V - 150 mA \cdot 36.71 \Omega}{750 \mu A} \approx 7.725 k\Omega\)
Simulation:
As you can see, the amplified signal (blue) is even weaker than the input signal (green).
Last edited:
