BJT Common Emitter Amplifier
David Baird
2002 Nov 4
1
Intro
I intend to build a two stage BJT common emitter amplifier. Here are my design constraints:
Accept a source resistance of 10kΩ and voltage of 40mVp−p
Drive a 2kΩ load
Have a gain near 200
2
Setup
+15V
IEE2 4mA
C4 100µF
RCC1
22k
4.7V
-4.7V
VBB1
Rs 10k Vi
VCC1 4V
ZIN
-7V
VCC2
22k
Vo
-5.4V
VEE1
5.6k
2x 2N3906
Ree1
1k
C2 10µF
82Ω
5.4V
2N3904
C1 .47µF 10k
Ree2
VEE2
IEE1 .4-.6mA
C5 10µF
2k
ZOUT
RCC2
2k
Re2=25mV/4.5mA = 5.56Ω
C3 10µF
-15V
Re1=25mV/.5mA = 50Ω
David Baird
xcircuit Drawing
Lab 9, Final Circuit
EE321
Figure 1: BJT amplifier with theoretical DC bias values
Figure 1 shows the setup I used. I used current sources to greatly simplify the biasing of the
transistors. The first BJT stage has a gain of about 20 and the second has a gain of about 10. The
first stage input is bootstrapped. The goal of this is to make the resistor divider bias disappear
at high frequency so that the signal source only sees the (β + 1)(re + Ree ), which should be much
larger than Rs = 10kΩ. The second BJT stage uses a darlington to unload the first stage. Most of
my calculations assume the input impedance to the transistor is negligible.
1
+15V
.51mA
100k
470Ω
4.7k
13.1V
-12.3V
12.4V
-13V
22k
3.9k
22k
4mA
-15V
Figure 2: Current sources used in the amplifier
All the capacitors form high pass filters with a 3dB point at 20Hz or less. The following table
shows the R and C values, and the 3dB frequency for each capacitor:
Capacitor
C1
C2
C3
C4
C5
R [Ω]
> 100k
3k
1k
82
4k (?)
C [µF ]
.47
10
10
100
10
f3dB [Hz]
< 3.4
5.3
16
19
4.0
The emitter bypassing required the largest capacitors, as you can see, because they had the
smallest source resistances.
The amplifier gains are determined approximately by Rcc /(Ree + Re ). This gives stages 1 and
2 gains of 22k/(1k + 50) = 21 and 1k/(82 + 5.6) = 11.4. The composite gain should then be
21 · 11.4 = 239. ...so, the gain is a little greater than 200.
3
3.1
Data
DC values
Honestly, the actual values were so close to my expected values that I didn’t really care to investigate the exact reasons for deviation.
2
Parameter
Vsup+ 15
Vsup− -15
VBB1
VCC1
VEE1
VCC2
VEE2
IEE1
IEE2
3.2
Expected
-4.7
4
-5.4
-7
5.4
.51
4
Comment
f=10kHz
f=10kHz
f=10kHz
f=10kHz
Vop−p /Vip−p
Vo /Vs
Expected
21
11.4
239
Actual
45.6
35.2
.840
8.24
23.9
9.81
234
181
Unit
mV
mV
V
V
-
AC gains
f [Hz]
100
1k
10k
3.4
Unit
V
V
V
V
V
V
V
mA
mA
AC values
Parameter
Vs p-p
Vi p-p
Vee1 p-p
Vo p-p
Av1
Av2
Av
AlternateAv
3.3
Actual
14.86
-14.82
-4.7
4.12
-5.25
-6.64
5.35
.50
4.0
Vs [mV]
40
40
40
Vi [mV]
36
36
36
Vo [V]
7.4
7.76
8.24
Av = Vo /Vi
206
216
229
Vo /Vs
185
194
206
AC Impedances
Figure 3 shows the setups I used to measure ZIN and ZOU T . To measure ZIN , I kept adding
resistance to the input until the output decreased by a factor of 2. At this point, the total input
resistance must be equal to the resistance looking into the amplifier. The process for finding Z OU T
is similar: decrease output resistance until the output voltage decreases by a factor of two. This
resistance must then be the resistance looking into the output of the amplifier.
ZIN is the only value that deviated much from what I had expected. I’m not totally sure why,
but I would guess that the bootstrap I made isn’t working as well as expected. I expect the AC
resistance looking into the base of the BJT should be over 100kΩ, which means the AC resistance
of the bootstrap must be around 150kΩ (if base=150kΩ) in order for ZIN = 76kΩ. Then again,
considering that the bias network goes from about 10k at DC to 150k (factor of 15) at AC makes
bootstrapping a pretty neat trick!
RL [kΩ]
inf
2.0
1.0
Vo [Vp−p ]
15.2
7.68
5.16
3
(a)
39.2mV p-p 17mV p-p
10k
56k
4.88V p-p
1kHz
(b)
10k
36mV p-p
RL
1kHz
Figure 3: Circuits used when (a) determining ZIN , (b) determining ZOU T
Parameter
ZIN
ZOU T
4
Comment
Expected
> 100
2
Actual
76
2.0
Unit
kΩ
kΩ
Conclusions
I built an amplifier with 2 common emitter BJT stages. I used bootstrapping for my first time and
found that it works quite nicely. It increased the input impedence of the bias network by a factor
of 15 for AC signals.
I also found that using current sources at the transistor emitters helps to simplify biasing
because I no longer had to consider emitter DC resistors.
I found the input impedance to be a little less than 100k and the output impedence to be equal
to the collector resistor (2k).
I had some blunders along the way. I learned that you cannot simply replace the collector
resistors with current sources in a common emitter amplifiers to achieve very high gain. A better
way to get high gain is to use a differential input stage with a current source on one side and
take negative feedback from the output. I was having trouble getting my experimental circuits
(differential + follower) to have enough open loop gain (I think) to work well for a closed loop gain
around 200 (they worked great for 20!).
4