Bipolar Junction Transistor
Draw the CB, CE & CC connection of a transistor
Fig: Common Base Fig: Common Emitter Fig: Common Collector
Determine the current IE and IB & the voltage VCE and VCB
4−0.7
IE= 1.2
= 2.75mA
IE=IB+IC= 61IB
2.75
⇒ I B= 61
= 0.045mA
IC≅IE
VEE+VCC= ICRC+VCE+ ICRE
⇒ VCE= 4+10 -2.75*2.4 – 2.75*1.2= 4.1V
Also, VCC= ICRC+VCB ⇒ VCB= 10 – 2.75*2.4= 3.4V
Determine IC &VCE
20= 430IB+0.7+51*1IB
⇒ IB= 0.040mA
IC=𝛽IB= 0.040*50= 2mA
IC≅IE
VCE= 20 -2(2+1)= 14V
Bipolar Junction Transistor
Determine IC and VCE
𝛽RE> 10R2 150KΩ> 39KΩ Condition Satisfied
Rth= 39║3.9= 3.55KΩ
3.9∗22
Eth= = 2V
3.9+39
Eth= IBRTh +0.7 + 101*1.5IB
2−0.7
⇒ IB= 151.5+3.55= 8.38𝜇A
IC= 𝛽IB= 100*8.38= 838𝜇A= 0.838mA
VCE= 22 – 0.838(10+1.5) = 12.363V
Determine IC & VCE
VCC= (IB+IC) RC+ IBRB+VBE+ IERE
⇒ VCC= (IB+𝛽IB) RC+ IBRB+VBE+ (𝛽 + 1)IBRE
10−0.7 9.3
⇒ IB= 91∗4.7+250+91∗1.2= 427.7+250+109.2= 0.01181 mA
IC= 𝛽IB= 1.07mA
VCE= 10 -1.07(4.7+1.2) = 3.687V
Determine IE & VCE
IBRB +0.7 +IERE= 20
⇒ IBRB +0.7 + (𝛽+1) IBRE= 20
20−0.7 19.3
⇒ IB= 240+91∗2= 422 = 0.0457mA
IC= 𝛽IB= 4.11mA
IE=IC= 4.11mA
VCE= 20 – 4.11*2= 11.78 Volt
Bipolar Junction Transistor
Determine Vc & VB
9−0.7
I B= 100
= 0.083mA
IC= 𝛽IB= 3.73mA
VC= - IcRc= - 4.47V
VBE= VB –VE= VB +9 =0.7
⇒ VB= - 8.3V
Determine VC & VB
Rth= 8.2ǁ2.2= 1.73KΩ
𝐸𝑡ℎ−20 𝐸𝑡ℎ+20
+ =0
8.2 2.2
⇒ 0.576Eth = -6.65V
⇒ Eth= -11.55V
Eth+ IBRth+ VBE+ IERE=20
20−11.55−0.7 7.75
⇒ IB= 1.73+121∗1.8 = 1.73+217.8= 0.0353mA
IC= 𝛽IB= 4.23mA
VC= 20 – 4.23*2.7= 8.57V
VE= -20 + 4.23*1.8= -12.38V
VBE= VB –VE=0.7
⇒ VB= VE+0.7= -12.38+0.7= -11.68V
Bipolar Junction Transistor
Given Ic=2mA and VCE=10V determine the value of R1 and RC
VE= IERE=ICRE= 2.4V
VBE= VB –VE=0.7
⇒ VB=2.4+0.7= 3.1V
18∗18
VB= 18+𝑅1= 3.1
⇒ 18+R1= 104.51
⇒ R1= 86.51KΩ
VCE=VC –VE= 10
⇒ VC= 12.4V
18−12.4
So, RC= 2
= 2.8KΩ
Given the device characteristics determine Vcc, RB & RC
VCE= Vcc - IcRc
At cut off Ic=0, VCE= Vcc=20V
20
At saturation VCE=0 Rc= 8
= 2.5KΩ
Now Vcc= IBRB+ 0.7
20−0.7
⇒ RB = = 482.5KΩ
40
Bipolar Junction Transistor
Calculate the constant current I
5.1∗(−20)
VB= 5.1+5.1
= -10V
VBE=0.7=VB -VE
⇒ VE= -10 -0.7 = - 10.7V
20−10.7
I=IE= = 4.65mA
2
Calculate the constant current I
6.2−0.7
I=IE= = 3.06KΩ
1.8
Determine VCE
𝛽RE>10R2 ⇒ 132KΩ>100KΩ condition satisfied
−18∗10
VB= = -3.15V
10+47
VBE=VB -VE = -0.7
⇒ VE= VB+0.7= -3.15+0.7= -2.45V
𝑉𝐸
IE= 𝑅𝐸 = 2.22mA
VCE= -18 + 2.22(2.4+1.1) = -10.23V
Bipolar Junction Transistor
Determine whether the transistor is in saturation or not
VCE=VCC-ICRC
At saturation VCE=0
𝑉𝑐𝑐
IC-Sat= 𝑅𝑐
= 6.097mA
5−0.7
I B= 68
= 63.23𝜇A
𝐼𝑐(𝑆𝑎𝑡) 6.097
𝛽
= 125 = 48.77𝜇A
𝐼𝑐(𝑠𝑎𝑡)
Since IB> ; so transistor is in saturation mode
𝛽
Determine RB & RC for the transistor network if Ic-sat= 10mA
𝑉𝑐𝑐
Ic-sat= 𝑅𝑐
10
⇒ Rc= 10=1 kΩ
𝐼𝑐−𝑠𝑎𝑡 10
𝛽
= 250= 0.04mA
Choose IB= 0.06mA to ensure saturation
10−0.7
RB = = 155kΩ
0.060
Choosing RB=150KΩ which is standard value
10−0.70
I B= 150
= 0.062mA
𝐼𝑐−𝑠𝑎𝑡
So, IB> 𝛽
condition is satisfied So, Choosing RB=150KΩ & Rc= 1 kΩ
Bipolar Junction Transistor
Calculate IE, IB, IC & VC where 𝜷=50 & voltage at the emitter is -0.7V
VBE= VB -VE =0.7
⇒ VB= VE+0.7= -0.7=0.7=0volt
VB+10=VBE+IERE
10−0.7
⇒ IE= 10
= 0.93mA
IE=IB+IC= (𝛽+1)IB
0.93
⇒ I B= 51
= 0.01823mA
IC= 𝛽IB=0.91mA
VC= 10 – 5*0.91= 5.45V
If VB=1V and VE=1.7V, then what are 𝜶 and 𝜷 & VC for the transistor
10−1.7
IE= = 1.66mA
5
1
I B= = 0.01mA
100
IC= 1.66 – 0.01= 1.65mA
𝐼𝑐
𝛼=𝐼𝑒= 0.99
𝐼𝑐
𝛽=𝐼𝑏= 165
VC +10= ICRC
⇒ VC= 1.65*5 -10 = -1.75V
Determine the all node voltage & current where 𝜷=100
4=0.7+101IBRE
4−0.7
⇒ IB= 101∗3,3=0.0099mA
IC=𝛽IB=0.99mA
IE= IB+IC=0.99mA
VB=4V
VC= 10 -4.7*0.99= 5.35V
VBE=0.7= VB -VE ⇒ VE= 4 - 0.7= 3.3V
Bipolar Junction Transistor
Determine the all node voltage and current where 𝜷=100
VBE=VB-VE= -0.7 ⇒VE=0.7V Where VB=0
10−0.7
IE= 2
= 4.65mA
𝛼
𝛽=1−𝛼=100
⇒ 100 -100𝛼=𝛼
100
⇒ 𝛼= 101= 0.99
IC=𝛼IE= 4.61mA
4.61
𝛽= ⇒ IB= 0.0461mA
𝐼𝑏
VC+10= 4.61
⇒ VC= -5.39V
Determine the all node voltage and current where 𝜷=30
VBE= VB -VE = -0.7
⇒ VE= VB+0.7
VC-VE= - 0.2= VCE
⇒VC= VE -0.2= VB+0.7 – 0.2= VB+0.5
5−𝑉𝐸
IE= = 5-VE=4.3 -VB mA
1
𝑉𝑏
IB=10 = 0.1VB mA
𝑉𝑐+5
IC= = 0.1VC+ 0.5= 0.1(VB+0.5) +0.5 = 0.1VB+ 0.55 mA
10
IE=IB+IC
⇒4.3 -VB= 0.1VB + 0.1VB+ 0.55
3.75
⇒ VB= 1.2
= 3.125V
Now place the value of VB
VE= 3.83V, VC=3.63V, IE= 1.17mA, IC=0.86mA, IB=0.31mA
Bipolar Junction Transistor
𝒗𝒐
Determine the voltage gain , assume 𝜷=100
𝒗𝒊
3−0.7
I B= = 0.023mA
100
IC= 𝛽IB= 2.3mA
IE=IB+IC= 2.323mA
26
re= 2.323= 11.20Ω
Zi= 100+1.12= 101.12KΩ
Zo= 3KΩ
𝑉𝑖
Vo= - 𝛽IB*Rc= -𝛽* 𝑍𝑖 *Rc
𝑉𝑜 3000
Voltage gain Av= = - 100* = -3 (Approx.)
𝑉𝑖 101.12∗1000
Determine whether the transistor is in saturation, cut off or linear region
VCE= VCC -ICRC
For a)
12−0.7
I B= = 0.15mA
75
At Sat VCE=0
𝑉𝑐𝑐 12
Ic-sat= 𝑅𝑐
= 1
=12mA
𝐼𝑐−𝑠𝑎𝑡
𝛽
= 0.12mA
𝐼𝑐−𝑠𝑎𝑡
As Ib> 𝛽
, So transistor is in saturation mode
For b)
12−0.7
I B= 150
= 0.0753mA
At Sat VCE=0
𝑉𝑐𝑐 12
Ic-sat= 𝑅𝑐
= 1
=12mA
𝐼𝑐−𝑠𝑎𝑡
𝛽
= 0.12mA
𝐼𝑐−𝑠𝑎𝑡
As Ib < 𝛽
, So the transistor is in linear region
Bipolar Junction Transistor
For c)
12−0.7
I B= 75
= 0.15mA
At Sat VCE=0
𝑉𝑐𝑐 12
Ic-sat= = =6mA
𝑅𝑐 2
𝐼𝑐−𝑠𝑎𝑡
𝛽
= 0.06mA
𝐼𝑐−𝑠𝑎𝑡
As Ib > 𝛽
, so transistor is in saturation mode.
Some Important Formulae
In the circuit diagram shown in figure draw the d.c load line, what will be the Q-point if zero
signal base current is 20𝝁A and 𝜷=50
VCE=VCC -ICRC
At cut off IC=0
VCE=VCC=12V
At saturation VCE=0
𝑉𝑐𝑐 12
Ic-sat= 𝑅𝑐 = 6
=2mA
Q-point:
IC= 𝛽IB= 1mA
VCE= 12 – 1*6= 6V
Bipolar Junction Transistor
Equivalent re model
Equivalent re model
Determine Ii, Zi, Vo, Io & Ai
12−4
Ii= 1
=8𝜇A
4
Zi= 8= 500Ω
Vo= -180*4=- 720mV
720
Io=0.51= 1.41mA
𝐼𝑜
Ai= = 176.25
𝐼𝑖
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