Common Emitter Amplifier
𝑪𝒊𝒏 is the input coupling capacitor
This capacitor is couple the ac generator voltage
to the base of the transistor
𝑪𝑬 is the bypass capacitor
This capacitor is used to neglect effect of
𝑅𝐸 at ac analysis
The common-emitter amplifier provides
• a high current gain,
• a high voltage gain, with
180𝑜 phase shift
• very high power gain.
IC mA
IC(sat)
Q point
0
VCE
V
1
Common Emitter Amplifier
Example 2-3:𝜷 = 𝟐𝟎𝟎
For the amplifier in the following figure, find the amplifier small signal
parameters, amplifier parameters, the over all voltage gain,
Solution 2-3
Step 1: DC analysis and find Q point
a. Replace any ac source with short circuit
Also replace all capacitor by open circuit
1
𝑋𝑐 = 2𝜋𝑓𝐶 & f=0Hz. (at dc analysis)
1
𝑋𝑐 = 0 = ∞ (Open circuit)
b. Try to find all DC currents
βR E = 200 240 = 48K
10𝑅2 = 10 3.6𝐾 = 36𝐾
As βR E > 10𝑅2  (stiff voltage divider)
(unload circuit) & 𝐼𝐵 ≅ 0
𝑅2
3.6
𝑉𝐵 =
𝑉𝑐𝑐 =
15 = 2.5𝑉
𝑅2 + 𝑅1
3.6 + 18
2
Common Emitter Amplifier
Assume tr. At active mode  𝑉𝐵𝐸 = 0.7𝑉
𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 2.5 − 0.7 = 1.8𝑉
𝑉𝐸
1.8
𝐼𝐸 =
=
= 7.5𝑚𝐴
𝑅𝐸 240
𝐼𝐶 ≅ 𝐼𝐸 = 7.5𝑚𝐴
𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 = 15 − 7.5𝑚 ∗ 1𝐾 = 7.5𝑉
c. Find Q point (VCE , IC )
=2.5V
𝑉𝐶𝐸 = 𝑉𝐶 − 𝑉𝐸 = 7.5 − (1.8) = 5.7𝑉
𝐼𝐶 = 7.5𝑚𝐴
Q point (𝟓. 𝟕𝑽, 𝟕. 𝟓𝒎𝑨)
As NPN Tr. & 𝑉𝐶 > 𝑉𝐵 or 𝑉𝐶𝐸 > 0.2𝑉 transistor will operate at active mode
Step 2: determined the small signal model parameter
𝑉𝑇 25𝑚𝑉
𝑟𝑒 =
=
= 3.3
𝐼𝐸 7.5𝑚𝐴
𝐼𝐶 7.5𝑚𝐴
𝑔𝑚 =
=
= 0.3−1
𝑉𝑇 25𝑚𝑉
𝛽
200
𝑟𝜋 =
=
= 666.7
𝑔𝑚
0.3
3
Common Emitter Amplifier
Step 3: in amplifier circuit :
replace all DC voltage source by short circuit
also replace all capacitors with short circuit
Step 4: replace the transistor with  model
Step 5: analysis the small signal circuit to find voltage gain
4
Common Emitter Amplifier
Step 5: analysis the small signal circuit to find voltage gain
𝒁𝒊𝒏
𝒁𝒊𝒏𝒃𝒂𝒔𝒆
𝒗𝒃
𝑖
𝑖
𝒁𝒊𝒏(𝒃𝒂𝒔𝒆) =
= 𝒓𝝅 = 𝟔𝟔𝟕
𝒊𝒃
+
+
𝑖𝑏
𝒗𝒊
𝒁𝒊𝒏 = = 𝑹𝟏 𝑹𝟐 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒊𝒊
𝑣𝑖
𝒁𝒊𝒏 = 𝟏𝟖𝒌𝟑. 𝟔𝒌667=545.6
18K
667
3.6K
𝑣𝑏
𝒁𝒐𝒖𝒕 = 𝑹𝒄 = 𝟏𝒌
_
𝒗
𝒗
𝑨𝒗 𝒕𝒓. = 𝒗𝒐𝒕𝒓. = 𝒗𝒄 =𝑨𝒗 𝒕𝒓. 𝑹 =∞
𝑳
𝒊𝒕𝒓.
𝒃
_
𝒁𝒐𝒖𝒕
𝑖𝑐
+
𝑣𝐶
_
𝒗𝒄 = −𝒊𝒄 ∗ 𝑹𝒄 = −(𝒈𝒎 ∗ 𝒗𝒃𝒆 ) ∗ 𝑹𝑪
𝒗𝒃 = 𝒊𝒃 ∗ 𝒓𝝅 = 𝒗𝒃𝒆
𝑨𝒗
𝒕𝒓.
=
−𝒊𝒄 𝑹𝑪
𝒊𝒃 𝒓𝝅
=−
𝜷𝒊𝒃 𝑹𝑪
𝒊𝒃 𝒓𝝅
=−
𝜷𝑹𝑪
𝒓𝝅
=−
𝟐𝟎𝟎∗𝟏𝑲
𝟔𝟔𝟕
= −𝟐𝟗𝟗. 𝟖𝟓
Or
𝑨𝒗
𝑰
𝒈𝒎=𝑽𝑪 =
−(𝒈𝒎 𝒗𝒃𝒆 )𝑹𝑪
=
= −(𝒈𝒎 )𝑹𝑪 =−𝟑𝟎𝟎
𝒕𝒓.
𝒗𝒃𝒆
𝐯
𝐯
𝐀𝐯 𝐨𝐯𝐞𝐫𝐚𝐥𝐥 = 𝐆𝐯 = 𝐯 𝐨. = 𝐯𝐜 =-300
𝐢𝐧.
𝐛
𝑨𝒊
𝒕𝒓
=
𝒊𝒄 𝜷𝒊𝒃
=
= 𝜷 = 𝟐𝟎𝟎
𝒊𝒃
𝒊𝒃
𝒗𝒐 = 𝒗𝒊𝒏𝒑𝒑 𝑨𝒗
𝑻
𝜷𝑰𝑩
𝑽𝑻
𝜷
=𝒓
𝝅
5
𝒐𝒗𝒆𝒓𝒂𝒍𝒍
= 𝟏𝟎𝒎𝑽 ∗ −𝟑𝟎𝟎 = −𝟑𝑽𝒑𝒑
Common Emitter Amplifier with Load Resistance
Example 2-4
Analysis the circuit shown to find voltage gain and amplifier parameters and
output voltage 𝒗𝒐 Note =200
Solution 2-4
Step 1: DC analysis and find Q point
a. Replace any ac source with short circuit
Also replace all capacitor by open circuit
b. Try to find all DC currents
𝐈𝐁
𝐈𝐄
𝐈𝐂
𝐕𝐁
𝐕𝐄
𝐕𝐂
0
7.5mA
7.5mA
2.5V
1.8V
7.5
c. Find Q point (VCE , IC )
Q point (𝟓. 𝟕𝑽, 𝟕. 𝟓𝒎𝑨)
Step 2: determined the small signal model parameter
𝒓𝝅
𝟔𝟔𝟔. 𝟕
𝒈𝒎
𝒓𝒆
𝟎. 𝟑 −𝟏 𝟑. 𝟑
6
Common Emitter Amplifier with Load Resistance
Step 3: in amplifier circuit :
replace all DC voltage source by short circuit
also replace all capacitors with short circuit
Step 4: replace the transistor with  model
Step 5: analysis the small signal circuit to find voltage gain
7
Common Emitter Amplifier with Load Resistance
Step 5: analysis the small signal circuit to find voltage gain
𝒁𝒊𝒏
𝒁𝒊𝒏𝒃𝒂𝒔𝒆
𝒗𝒃
𝑖
𝑖
𝒁𝒊𝒏(𝒃𝒂𝒔𝒆) =
= 𝒓𝝅 = 𝟔𝟔𝟕
𝒊𝒃
+
+
𝑖𝑏
𝒗𝒊
𝒁𝒊𝒏 = = 𝑹𝟏 𝑹𝟐 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒊𝒊
𝑣𝑖
𝒁𝒊𝒏 = 𝟏𝟖𝒌𝟑. 𝟔𝒌667=545.6
18K
667
3.6K
𝑣𝑏
𝒁
= 𝑹 = 𝟏𝒌 Exclude 𝑅
𝒐𝒖𝒕
𝒄
+
𝑣𝐶
_
_
_
𝑨𝒗
𝑖𝑐
𝒁𝑳
𝐿
𝒁𝑳 = 𝑹𝒄 𝑹𝑳 = 𝟏𝒌 1.5K=0.6K
𝑨𝒗𝒐 . = 𝑨𝒗
𝒁𝒐𝒖𝒕
𝒗
𝒕𝒓. 𝑹 =∞
𝑳
𝒗
= 𝒗𝒄
𝒃
𝒕𝒓.
=
𝒗𝒐𝒕𝒓.
𝒗𝒊𝒕𝒓.
= 𝒗𝒄 =
𝒕𝒓.
=
𝒗𝒐𝒕𝒓.
𝒗𝒊𝒕𝒓.
= 𝒗𝒄 =
𝒃
=
𝑹𝑳 =∞
−𝒊𝒄 ∗𝒁𝑳
𝒊𝒃 ∗𝒓𝝅
=−
−(𝒈𝒎 𝒗𝒃𝒆 )𝑹𝑪
𝒗𝒃𝒆
𝜷𝒊𝒃 𝒁𝑳
𝒊𝒃 𝒓𝝅
=−
= −(𝒈𝒎 )𝑹𝑪 =−𝟑𝟎𝟎
𝜷𝒁𝑳
𝒓𝝅
=−
𝟐𝟎𝟎∗𝟎.𝟔 𝑲
𝟔𝟔𝟕
= −𝟏𝟕𝟗. 𝟗𝟏
Or
𝑨𝒗
𝐀𝐯
𝒗
𝒃
𝐯
𝐨𝐯𝐞𝐫𝐚𝐥𝐥
−(𝒈𝒎 𝒗𝒃𝒆 )𝒁𝑳
𝒗𝒃𝒆
𝐯
= 𝐆𝐯 = 𝐯 𝐨. = 𝐯𝐜 =-180
𝐢𝐧.
𝐛
= −(𝒈𝒎 )𝒁𝑳 =−𝟏𝟖𝟎
8
Common Emitter Amplifier with Load Resistance
𝑨𝒊
𝒕𝒓
𝒊𝒄 𝜷𝒊𝒃
= =
= 𝜷 = 𝟐𝟎𝟎
𝒊𝒃
𝒊𝒃
𝒗𝒐 = 𝒗𝒊𝒏𝒑𝒑 𝑨𝒗
𝒐𝒗𝒆𝒓𝒂𝒍𝒍
= 𝟏𝟎𝒎𝑽 ∗ −𝟏𝟖𝟎
= −𝟏. 𝟖𝑽𝒑𝒑
Note:
1. Adding load resistance 𝑅𝐿 reduce amplifier voltage gain 𝑨𝒗 𝒕𝒓 and reduce output
voltage 𝒗𝒐
2. The output impedance,𝑍𝑜𝑢𝑡 , of CE amplifier equals the value of collector resistance,
𝑅𝐶 , but not include the load resistance, 𝑅𝐿 . This is because the load resistance, 𝑅𝐿 , is
actually being driven by the amplifier.
9
Common Emitter Amplifier with Emitter Resistance (swapping
resistance)
The swapping resistance is a
resistance connected in common
emitter configuration to emitter
terminal and appear in ac analysis
Why?
• The actual value of 𝑟𝑒 =
𝑉𝑇
𝐼𝐸
may vary with the
type of transistor used, a shift in circuit bias, or
fluctuations in temperature.
• With any change in 𝒓𝒆 , 𝑨𝒗 may vary drastically.
This is undesirable.
10
Common Emitter Amplifier with Emitter Resistance (swapping
resistance)
Example 2-5
Analysis the circuit shown to find voltage gain and amplifier parameters and
output voltage 𝒗𝒐 Note =200
Solution 2-5
Step 1: DC analysis and find Q point
a. Replace any ac source with short circuit
Also replace all capacitor by open circuit
b. Try to find all DC currents
𝐈𝐁
𝐈𝐄
𝐈𝐂
𝐕𝐁
𝐕𝐄
𝐕𝐂
0
7.5mA
7.5mA
2.5V
1.8V
7.5
c. Find Q point (VCE , IC )
Q point (𝟓. 𝟕𝑽, 𝟕. 𝟓𝒎𝑨)
Step 2: determined the small signal model parameter
𝒓𝝅
𝟔𝟔𝟔. 𝟕
𝒈𝒎
𝒓𝒆
𝟎. 𝟑 −𝟏 𝟑. 𝟑
11
Common Emitter Amplifier with Emitter Resistance (swapping
resistance)
Step 3: in amplifier circuit :
replace all DC voltage source by short circuit
also replace all capacitors with short circuit
Step 4: replace the transistor with  model
Step 5: analysis the small signal circuit to find voltage gain
12
Common Emitter Amplifier with Emitter Resistance (swapping
resistance)
Step 5: analysis the small signal circuit to find voltage gain
𝒁𝒐𝒖𝒕
𝒁𝒊𝒏
𝒁
𝒊𝒏𝒃𝒂𝒔𝒆
𝒗𝒃 𝒓𝝅 𝒊𝒃 + 𝑹𝑬𝟐 𝒊𝒆
𝑖𝑖
𝒁𝒊𝒏 𝒃𝒂𝒔𝒆 =
=
𝒊𝒃
𝒊𝒃
+
+
+
𝑖𝑏
𝒓𝝅 𝒊𝒃 + 𝑹𝑬𝟐 (𝜷 + 𝟏)𝒊𝒃
=
𝑣𝑖
𝒊𝒃
18K 667
3.6K
(𝒓𝝅 + 𝑹𝑬𝟐 (𝜷 + 𝟏))𝒊𝒃
𝑣𝐶
=
𝑣𝑏
𝒊𝒃
_
𝑖𝑒
= 𝒓𝝅 + 𝜷 + 𝟏 𝑹𝑬𝟐
_
𝑅
=
60
_
= 𝟔𝟔𝟕 + 𝟐𝟎𝟏 ∗ 𝟔𝟎
𝐸2
= 𝟏𝟐. 𝟕𝟐𝑲
𝒗𝒊
𝒁𝒊𝒏 = = 𝑹𝟏 𝑹𝟐 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒊𝒊
𝒁𝒊𝒏 𝒃𝒂𝒔𝒆 = 𝒓𝝅 + 𝜷 + 𝟏 𝑹𝑬𝟐
𝒁𝒊𝒏 = 𝟏𝟖𝒌𝟑. 𝟔𝒌𝟏𝟐. 𝟕𝟐𝑲 =𝟐. 𝟒𝟐𝟓𝑲
𝒁𝒐𝒖𝒕 = 𝑹𝒄 = 𝟏𝒌
Exclude 𝑅𝐿
𝒁𝑳
𝑅𝐿
Resistance transferring
𝒁𝑳 = 𝑹𝒄 𝑹𝑳 = 𝟏𝒌 1.5K=0.6K
13
Common Emitter Amplifier with Emitter Resistance (swapping
resistance)
𝑖
𝑨𝒗𝒐 = 𝑨𝒗
=𝒊
𝒕𝒓.
𝑹𝑳 =∞
−𝒊𝒄 ∗𝑹𝑪
𝒃 ∗𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒗𝒄
=
𝒗𝒃
= −𝒊
𝑖𝑖
𝑐
+
𝑹𝑳 =∞
𝜷𝒊𝒃 𝑹𝒄
𝒃 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝜷𝑹𝒄
𝟐𝟎𝟎 ∗ 𝟏𝑲
=−
=−
𝒓𝝅 + 𝜷 + 𝟏 𝑹𝑬𝟐
𝟏𝟐. 𝟕𝟐𝑲
𝑣𝑖
3.6K
18K
+
𝑖𝑏
+
667
𝑅𝐿
𝑣𝐶
𝑣𝑏
_
𝑖𝑒
= −𝟏𝟓. 𝟕𝟐𝟑
−𝒊𝒄 𝒁𝑳
𝜷𝒊𝒃 𝑹𝒄 //𝑹𝑳
𝒗𝒐𝒕𝒓. 𝒗𝒄
=−
𝑨𝒗
=
=
=
𝒊𝒃 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒗𝒊𝒕𝒓. 𝒗𝒃 𝒊𝒃 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒕𝒓.
_
_
𝜷𝑹𝒄 //𝑹𝑳
𝟐𝟎𝟎 ∗ 𝟎. 𝟔𝑲
=−
=−
= −𝟗. 𝟒𝟑
𝒓𝝅 + 𝜷 + 𝟏 𝑹𝑬𝟐
𝟏𝟐. 𝟕𝟐𝑲
𝑨𝒗
𝒐𝒗𝒆𝒓 𝒂𝒍𝒍
=
𝒗𝒐 𝒗𝒄
𝒗𝒐 𝒗𝒃𝒆
=
=
∗
𝒗𝒊 𝒗𝒃 𝒗𝒃𝒆 𝒗𝒊
Chain rule
𝒗𝒐 = −𝒊𝒄 𝒁𝑳 = −(𝒈𝒎 𝒗𝒃𝒆 )𝑹𝒄 //𝑹𝑳 = −𝟎. 𝟑 ∗ 𝟎. 𝟔𝒌 ∗ 𝒗𝒃𝒆 = −𝟏𝟖𝟎𝒗𝒃𝒆
𝒗𝒃𝒆 =
𝒓𝝅
𝟔𝟔𝟕
𝒗𝒊 =
𝒗 = 𝟎. 𝟎𝟓𝟐𝟒𝒗𝒊
(𝜷 + 𝟏)𝑹𝑬𝟐 +𝒓𝝅
𝟐𝟎𝟏 ∗ 𝟔𝟎 + 𝟔𝟔𝟕 𝒊
14
Common Emitter Amplifier with Emitter Resistance (swapping
resistance)
𝑖
𝑨𝒊
𝒕𝒓
𝒊𝒄 𝜷𝒊𝒃
= =
= 𝜷 = 𝟐𝟎𝟎
𝒊𝒃
𝒊𝒃
𝒗𝒐 = 𝒗𝒊𝒑𝒑 𝑨𝒗
𝒐𝒗𝒆𝒓𝒂𝒍𝒍
𝑐
𝑖𝑖
𝑖𝑏
= 𝟏𝟎𝒎𝑽 ∗ −𝟗. 𝟒𝟑
3.6K
18K
667
𝑅𝐿
= −𝟗𝟒. 𝟑𝟑𝒎𝑽𝒑𝒑
𝑖𝑒
Swapping resistance
1.
2.
3.
4.
Increase value of 𝑍𝑖𝑛 and 𝑍𝑖𝑛(𝑏𝑎𝑠𝑒)
Reduce all voltage gain parameters
Increase stability of voltage gain
Reduce distortion
advantage
disadvantage
Advantage
Advantage
15
Common Emitter Amplifier With Generator Impedance
Example 2-6
Analysis the circuit shown to find voltage gain and amplifier parameters and
output voltage 𝒗𝒐 Note =200
Solution 2-6
Step 1: DC analysis and find Q point
a. Replace any ac source with short circuit
Also replace all capacitor by open circuit
b. Try to find all DC currents
𝐈𝐁
𝐈𝐄
𝐈𝐂
𝐕𝐁
𝐕𝐄
𝐕𝐂
0
7.5mA
7.5mA
2.5V
1.8V
7.5
c. Find Q point (VCE , IC )
Q point (𝟓. 𝟕𝑽, 𝟕. 𝟓𝒎𝑨)
Step 2: determined the small signal model parameter
𝒓𝝅
𝟔𝟔𝟔. 𝟕
𝒈𝒎
𝒓𝒆
𝟎. 𝟑 −𝟏 𝟑. 𝟑
16
Common Emitter Amplifier With Generator Impedance
Step 3: in amplifier circuit :
replace all DC voltage source by short circuit
also replace all capacitors with short circuit
Step 4: replace the transistor with  model
Step 5: analysis the small signal circuit to find voltage gain
17
Common Emitter Amplifier With Generator Impedance
Step 5: analysis the small signal circuit to find voltage gain
𝒁𝒊𝒏𝒔𝒐𝒖𝒓𝒄𝒆
𝒗𝒃
𝒁𝒊𝒏(𝒃𝒂𝒔𝒆) =
= 𝒓𝝅 = 𝟔𝟔𝟕
𝒁𝒊𝒏
𝒊
𝒃
𝒗𝒊
𝒁𝒊𝒏 = = 𝑹𝟏 𝑹𝟐 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝑖𝑖𝑛 +
𝒊𝒊
𝒁𝒊𝒏 = 𝟏𝟖𝒌𝟑. 𝟔𝒌667=545.6
𝒗𝒊𝒏
𝑣𝑖
𝒁𝒊𝒏𝒔𝒐𝒖𝒓𝒄𝒆 =
= 𝑹𝑮 + 𝒁𝒊𝒏
𝒊𝒊𝒏
𝒁𝒊𝒏𝒔𝒐𝒖𝒓𝒄𝒆 = 𝟔𝟎𝟎 + 545.6=1.1456K
_
𝒁𝒐𝒖𝒕 = 𝑹𝒄 = 𝟏𝒌
𝑨𝒗𝒐 = 𝑨𝒗
=
𝒗
𝒕𝒓. 𝑹 =∞
𝑳
𝒗𝒐𝒕𝒓.
𝒗𝒊𝒕𝒓.
𝑨𝒗
𝒕𝒓.
𝐀𝐯
𝐨𝐯𝐞𝐫𝐚𝐥𝐥
= 𝒗𝒄
𝒃
𝒗
= 𝒗𝒄 =
𝒃
=
𝑹𝑳 =∞
−𝒊𝒄 ∗𝒁𝒐𝒖𝒕
𝒊𝒃 ∗𝒓𝝅
𝒗
𝒗
−(𝒈𝒎 𝒗𝒃𝒆 )𝑹𝑪
𝒗𝒃𝒆
=−
𝜷𝒊𝒃 𝒁𝒐𝒖𝒕
𝒊𝒃 𝒓𝝅
𝒗
𝒗
𝒁𝒊𝒏𝒃𝒂𝒔𝒆
+
𝑖𝑏
𝑖𝑐
𝒁𝒐𝒖𝒕
+
667
𝑣𝑏
_
𝑣𝐶
_
= −(𝒈𝒎 )𝑹𝑪 =−𝟑𝟎𝟎
=−
𝜷𝒁𝒐𝒖𝒕
𝒓𝝅
=−
𝟐𝟎𝟎∗𝟏 𝑲
𝟔𝟔𝟕
= −𝟑𝟎𝟎
= 𝑨𝒗𝒐
= 𝐆𝐯 = 𝒗 𝐨. = 𝒗 𝐜 = 𝒗 𝐜. ∗ 𝒗 𝒃
𝐢𝐧.
=𝑨𝒗
𝒊𝒏
𝐛.
𝒊𝒏
𝒗
𝒃
𝒕𝒓. * 𝒗
= −𝟑𝟎𝟎 ∗ 𝟎. 𝟒𝟕𝟔=142.877
𝒁𝒊𝒏
𝟓𝟒𝟓. 𝟔
𝒗𝒃 =
𝒗𝒊 =
𝒗 = 𝟎. 𝟒𝟕𝟔𝒗𝒊
𝑹𝑮 + 𝒁𝒊𝒏
𝟔𝟎𝟎 + 𝟓𝟒𝟓. 𝟔 𝒊
𝒊𝒏
18
Common Emitter Amplifier With Generator Impedance
𝑨𝒊
𝒕𝒓
𝒊𝒄 𝜷𝒊𝒃
= =
= 𝜷 = 𝟐𝟎𝟎
𝒊𝒃
𝒊𝒃
𝒗𝒐 = 𝒗𝒊𝒏𝒑𝒑 𝑨𝒗
𝒐𝒗𝒆𝒓𝒂𝒍𝒍
= 𝟏𝟎𝒎𝑽 ∗ −142.877
= −𝟏. 𝟒𝟐𝟗𝑽𝒑𝒑
Note:
1. Adding generator resistance 𝑅𝐺 reduce over all voltage gain 𝐴𝑣
output voltage 𝑣𝑜
𝑜𝑣𝑒𝑟𝑎𝑙𝑙
and reduce
19
Common Emitter Amplifier
• In CE amplifier the input applied to the
base and output signal is taken from the
collector.
• A CE amplifier has high voltage gain and
current gain and very high power gain
• In CE amplifier 𝑣𝑖𝑛 and𝑣𝑜𝑢𝑡 are 180𝑜 out
of phase
20
Common Emitter Amplifier cont.
Example 2-7
Analysis the circuit shown to find voltage gain and amplifier parameters and
output voltage 𝒗𝒐 Note =200
Solution 2-7
Step 1: DC analysis and find Q point
a. Replace any ac source with short circuit
Also replace all capacitor by open circuit
b. Try to find all DC currents
𝐈𝐁
𝐈𝐄
𝐈𝐂
𝐕𝐁
𝐕𝐄
𝐕𝐂
0
7.5mA
7.5mA
2.5V
1.8V
7.5
c. Find Q point (VCE , IC )
Q point (𝟓. 𝟕𝑽, 𝟕. 𝟓𝒎𝑨)
Step 2: determined the small signal model parameter
𝒓𝝅
𝟔𝟔𝟔. 𝟕
𝒈𝒎
𝒓𝒆
𝟎. 𝟑 −𝟏 𝟑. 𝟑
21
Common Emitter Amplifier cont.
Step 3: in amplifier circuit :
replace all DC voltage source by short circuit
also replace all capacitors with short circuit
Step 4: replace the transistor with  model
Step 5: analysis the small signal circuit to find voltage gain
22
Common Emitter Amplifier cont.
Step 5: analysis the small signal circuit to find voltage gain
𝒁𝒊𝒏
𝒁𝒊𝒏𝒃𝒂𝒔𝒆
𝒗𝒃
𝒁𝒊𝒏(𝒃𝒂𝒔𝒆) =
= 𝒓𝝅 + 𝜷 + 𝟏 ∗ 𝑹𝑬𝟐
𝒊𝒃
+
𝑖𝑖
+
𝑖𝑏
𝒗𝒊
𝒁𝒊𝒏 = = 𝑹𝟏 𝑹𝟐 𝒁𝒊𝒏(𝒃𝒂𝒔𝒆)
𝒊𝒊
𝑣𝑖
𝒁𝒐𝒖𝒕 = 𝑹𝒄
Exclude 𝑅𝐿
𝑣𝑏 667
𝒁𝑳 = 𝑹𝒄 𝑹𝑳
𝒗𝒄
−(𝒈𝒎 𝒗𝒃𝒆 )𝑹𝑪
_
𝑖𝑒
𝑨𝒗𝒐 =
=
_
𝒗𝒃
𝒗𝒃
𝑹𝑳 =∞
𝒁𝒐𝒖𝒕
𝑖𝑐
𝒁𝑳
+
𝑹𝑳
𝑣𝐶
_
𝒓𝝅
𝒗𝒃𝒆 =
∗ 𝒗𝒃
𝒓𝝅 + (𝜷 + 𝟏)𝑹𝑬𝟐
𝒓𝝅
𝑨𝒗𝒐 = −𝒈𝒎 𝑹𝑪
𝒓𝝅 + (𝜷 + 𝟏)𝑹𝑬𝟐
𝑨𝒗
𝒕𝒓.
=
𝒗𝒐𝒕𝒓.
𝒗𝒊𝒕𝒓.
𝒗
= 𝒗𝒄 = 𝒊
𝒃
𝐯
−𝒊𝒄 ∗𝒁𝑳
𝒃 ∗(𝒓𝝅 +(𝜷+𝟏)𝑹𝑬𝟐 )
𝐯
𝐯
= −𝒊
𝜷𝒊𝒃 𝒁𝑳
𝒃 (𝒓𝝅 +(𝜷+𝟏)𝑹𝑬𝟐 )
𝐯
= 𝐆𝐯 = 𝐯 𝐨. = 𝐯𝐜 ∗ 𝐯 𝒃 = 𝑨𝒗 𝒕𝒓. * 𝐯 𝒃
𝐢𝐧.
𝒊𝒏
𝒊𝒏
𝐛
𝒁𝒊𝒏
𝒗𝒃 =
∗ 𝒗𝒊𝒏
𝒁𝒊𝒏 + 𝑹𝑮
𝜷𝒁𝑳
𝒁𝒊𝒏
𝐀𝐯
= 𝐆𝐯 = −
∗
𝒓𝝅 + 𝜷 + 𝟏 𝑹𝑬𝟐 𝒁𝒊𝒏 + 𝑹𝑮
𝐨𝐯𝐞𝐫𝐚𝐥𝐥
𝐀𝐯
𝜷𝒁𝑳
𝝅 +(𝜷+𝟏)𝑹𝑬𝟐
= −𝒓
𝐨𝐯𝐞𝐫𝐚𝐥𝐥
23
𝑨𝒊
𝒕𝒓
=
𝒊𝒄 𝜷𝒊𝒃
=
= 𝜷 = 𝟐𝟎𝟎
𝒊𝒃
𝒊𝒃