Hey all, I ran into a wall while trying to design a common emitter amplifier. I was wondering if someone here would be able to point me in the right direction:
Specifications:
Common Emitter BJT amplifier with an Emitter Resistance and two coupling capacitors (assumed infinite for now)
Transistor 2N2222 (assume \(\beta\)=100 and \(v_{be}=0.7V\))
\(V_{CC}=15V\)
\(R_{sig}=50\Omega\)
\(R_L=47k\Omega\)
\(R_C=10k\Omega\)
Design Goals:
Voltage Gain of -20 V/V
Output Voltage Swing ΔV=10V peak-to-peak
Input Resistance greater than 10kΩ
Attempt at Solution:
I thought to set \(V_{C}=7.5V\) to best bias the circuit. Thus \( I_C=0.75mA\). This would give a transconductance of \(g_m=30mS\).
Since the signal generator impedance is low:
\(G_v=\frac{v_{out}}{v_{sig}} \simeq A_v=\frac{v_{out}}{v_{in}}\)
The gain equation is \(\frac{-(R_c||R_L)}{\frac{1}{g_m}+R_E}\). We can solve for \(R_E=379\Omega\). Which makes \( V_E=0.379V\), \(V_B=1.079V\), and \(V_C=7.5V\).
I'm using a rule of thumb and making the current through the voltage divider equal to one tenth of the emitter current. Thus \( \frac{V_{CC}}{0.075mA}=200k\Omega\). Since the sum of the two resistances needs to equal \(200k\Omega\) and the voltage divider has to make \(V_B=1.08V\), then \(R_2=14.4k\Omega\) and \(R_1=186k\Omega\).
\(R_{in}=\frac{R_1||R_2}{r_{\pi}+\beta R_E}=10k\Omega\)
Finally, the collector current will be at a maximum when the amplifier is in saturation:
\(i_{c,max}=\frac{V_{CC}-V_{BE,sat}}{R_C+R_E}=\frac{15V-0.3V}{10k+0.380k}=1.42mA\)
\(v_{c,min}=V_{CC}-i_{c,max}R_C=15V-(1.42mA)(10k\Omega)=0.848V\)
\(v_{c,max}=V_{CC}=15V\)
\(\Delta V=(2)(V_{CC}-V_{c,min})=2(7.5V-0.848V)=13.3Vpp\)
So by my math, the output signal should be linear up to 13Vpp. However, when I simulate this in SPICE (0.5V signal at 1kHz), my output signal is clipped above at 6.5V. I am sure that there is small mistake somewhere, but I have been unable to pick it out. I can provide more detail on how I derived my equations if necessary. Any suggestions on where to go from here?
P.S.: Sorry for the poor formatting, this is my first post!
Specifications:
Common Emitter BJT amplifier with an Emitter Resistance and two coupling capacitors (assumed infinite for now)
Transistor 2N2222 (assume \(\beta\)=100 and \(v_{be}=0.7V\))
\(V_{CC}=15V\)
\(R_{sig}=50\Omega\)
\(R_L=47k\Omega\)
\(R_C=10k\Omega\)
Design Goals:
Voltage Gain of -20 V/V
Output Voltage Swing ΔV=10V peak-to-peak
Input Resistance greater than 10kΩ
Attempt at Solution:
I thought to set \(V_{C}=7.5V\) to best bias the circuit. Thus \( I_C=0.75mA\). This would give a transconductance of \(g_m=30mS\).
Since the signal generator impedance is low:
\(G_v=\frac{v_{out}}{v_{sig}} \simeq A_v=\frac{v_{out}}{v_{in}}\)
The gain equation is \(\frac{-(R_c||R_L)}{\frac{1}{g_m}+R_E}\). We can solve for \(R_E=379\Omega\). Which makes \( V_E=0.379V\), \(V_B=1.079V\), and \(V_C=7.5V\).
I'm using a rule of thumb and making the current through the voltage divider equal to one tenth of the emitter current. Thus \( \frac{V_{CC}}{0.075mA}=200k\Omega\). Since the sum of the two resistances needs to equal \(200k\Omega\) and the voltage divider has to make \(V_B=1.08V\), then \(R_2=14.4k\Omega\) and \(R_1=186k\Omega\).
\(R_{in}=\frac{R_1||R_2}{r_{\pi}+\beta R_E}=10k\Omega\)
Finally, the collector current will be at a maximum when the amplifier is in saturation:
\(i_{c,max}=\frac{V_{CC}-V_{BE,sat}}{R_C+R_E}=\frac{15V-0.3V}{10k+0.380k}=1.42mA\)
\(v_{c,min}=V_{CC}-i_{c,max}R_C=15V-(1.42mA)(10k\Omega)=0.848V\)
\(v_{c,max}=V_{CC}=15V\)
\(\Delta V=(2)(V_{CC}-V_{c,min})=2(7.5V-0.848V)=13.3Vpp\)
So by my math, the output signal should be linear up to 13Vpp. However, when I simulate this in SPICE (0.5V signal at 1kHz), my output signal is clipped above at 6.5V. I am sure that there is small mistake somewhere, but I have been unable to pick it out. I can provide more detail on how I derived my equations if necessary. Any suggestions on where to go from here?
P.S.: Sorry for the poor formatting, this is my first post!
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