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Old 03-16-2012, 02:47 AM
DrWatts DrWatts is offline
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Default Output Signal Swing of CE amplifier

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:

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)

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:




\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!

Last edited by DrWatts; 03-16-2012 at 02:53 AM.
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Old 03-16-2012, 09:29 PM
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Ron H Ron H is offline
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I don't believe you included the capacitor and the 47k load resistor in your calculations.
Keep in mind that the cap always has 7.5V across it. When the transistor cuts off, you are left with a series circuit between +15V and ground consisting of 10k, 7.5V, and 47k. This means that the peak current into the load is (15-7.5)/(10k+47k). Multiply this current by the 47k load resistance, and you get the peak output voltage.
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