All About Circuits Forum Output Signal Swing of CE amplifier
 User Name Remember Me? Password
 Blogs FAQ Members List Today's Posts Search Today's Posts Mark Forums Read

 Homework Help Stuck on a textbook question or coursework? Cramming for a test and need help understanding something? Post your questions and attempts here and let others help.

#1
03-16-2012, 02:47 AM
 DrWatts New Member Join Date: Mar 2012 Posts: 1
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:

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!

Last edited by DrWatts; 03-16-2012 at 02:53 AM.
#2
03-16-2012, 09:29 PM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

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.

 Tags amplifier, output, signal, swing

 Related Site Pages Section Title Worksheet Performance-based assessments for analog integrated circuit competencies Worksheet Performance-based assessments for semiconductor circuit competencies Worksheet Class A BJT amplifiers Worksheet Bipolar junction transistors in active mode Textbook BJT quirks : Bipolar Junction Transistors Textbook Feedback : Bipolar Junction Transistors Textbook Biasing calculations : Bipolar Junction Transistors Textbook The cascode amplifier : Bipolar Junction Transistors Textbook The common-collector amplifier : Bipolar Junction Transistors Textbook The common-emitter amplifier : Bipolar Junction Transistors

 Similar Threads Thread Thread Starter Forum Replies Last Post kaname08 The Projects Forum 27 03-09-2012 03:35 PM icydash The Projects Forum 25 07-05-2011 11:21 PM tzitzikas Radio and Communications 11 05-28-2010 10:15 PM sarang2502 General Electronics Chat 16 04-27-2010 10:16 PM JUAN DELA CRUZ General Electronics Chat 17 06-03-2008 02:33 AM

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Electronics Forums     General Electronics Chat     The Projects Forum     Homework Help     Electronics Resources Software, Microcomputing, and Communications Forums     Programmer's Corner     Embedded Systems and Microcontrollers     Computing and Networks     Radio and Communications Circuits and Projects     The Completed Projects Collection Abstract Forums     Math     Physics     General Science All About Circuits Commmunity Forums     Off-Topic     The Flea Market     Feedback and Suggestions

All times are GMT. The time now is 05:15 AM.