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BJT, Bipolar Junction Transisor
Base Current
Controls
Output current
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AGENDA
BJT transistorman
Transistor types
Bipolar Junction Transistor
BJT models
parameters
water model
NPN and PNP
operation modes
switch open
switch closed
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BJT
linear, controlled current source
active operation
characteristics
DC input characteristics
ac input characteristics
BJT DC biasing circuits
base bias
base bias + collector feedback
base bias + emitter feedback
voltage divider
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BJT, transistor man
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Transistor
Types
BJT =
Bipolar Junction Transistor
FET =
Field Effect Transistor
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Output current controlled
by input current
Output current controlled
by input voltage
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BJT, Bipolar Junction Transisor
Transistor = Transfer Resistor
BE Forward bias,
So low ohmic
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BC Reverse bias
high ohmic
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BJT, Bipolar Junction Transisor
Emitter = Sent electrons
Base
= Base
Collector = Get electrons
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BJT, Models
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BJT, parameters
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BJT, Water model
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BJT, Water model
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BJT, NPN and PNP
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BJT, Operation modes
Cut-off and saturation;
BJT is used as a switch
Active operation
Quiecent Point;
BJT is used as a
controlled
current source,
or analog amplifier
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BJT, Switch open
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BJT, Switch closed
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BJT, Lineair, controlled current source
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BJT, active operation
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BJT, characteristics
DC model; Vbe = 0V7
ac model; re = 26mV/Ie
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DC model ac model
Ube, Uce, Ic, Ib, Ie Capitals
ube, uce, ic, ib, ie Low cases
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BJT, DC input characteristics
Vbe = 0V7
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BJT, AC input characteristics
re = 26mV/Ic
The dynamic resistor can be
calculated by the DC current Ic
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BJT, characteristics
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BJT, DC biasing circuits
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A base bias
B base bias + emitter feedback
C base bias + collector feedback
D voltage divider
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BJT, base bias, introduction
Base current determined by Vcc, Rb and Vbe
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BJT, base bias
Vcc  U Rb  U be
Vcc  I b Rb  U be
Calculate Ib and then Ic
Ic    Ib
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Ic directly dependent on ß variation
So, for stability a “bad” circuit
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BJT, base bias load line
Q-point
= Quiecient point
= Working point
Load line is the loading of the transistor seen from Uce (>0V7)
Vcc and Rc determines the;
“open voltage” and the “short circuit current” 24
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BJT, base bias load line
Reliable circuit
= Q-point stability
Load line is the loading of the transistor seen from Uce (>0V7)
Vcc and Rc determines the;
“open voltage” and the “short circuit current” 25
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BJT, base bias load line
Vce always > 0V7
BC junction REVERSE
If Rc too big, transistor in saturation;
then;
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BJT, base bias load line
Vce always > 0V7
BC junction REVERSE
If Vcc too small, transistor in saturation;
then;
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BJT, base bias example
Calculate;
I b, I c
URc, Uc, Uce
Draw output caracteristic
Calculate now;
Ib = 47 uA, Ic = 2,35 mA,
URc = 5,17 V, Uc = 6,83 V, Uce = 6,83 V
Uce (for ß = 40) = 7,86 Ξ 15 %
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Uce if ß = 40
How many % did Uce Change
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BJT, base bias example
Ib = 33 uA, Ic = 2,9 mA
URc = 7,9 V, Uc = 8,1 V
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Rb = 282,5 kΩ, Ic = 3,2 mA,
Rc = 1,855 kΩ
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BJT, base bias example
ß = 200, VRc = 8,8 V
Vcc = 16 VRb = 765 kΩ
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BJT, base bias + emitter feedback
Base current determined by Vcc, Rb, Vbe and Ve
More stable for ß variations, than base bias.
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BJT, base bias + emitter feedback
Vcc  U Rb  U be  U Re
Vcc  I b Rb  U be     1 I bU Re
Always calculate in the smallest current Ib !!
VRc  I c Rc
Vc  Vcc  I c Rc
Ve  I e Re
Vce  Vc  Ve
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BJT, base bias + emitter feedback
Load line is the loading of the transistor seen from Uce (>0V7)
Vcc, Rc and Re determines the;
“open voltage” and the “short circuit current” 33
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BJT, base bias + emitter feedback example
Calculate;
I b, I c
URc, Uc, Ue, Uce
Draw output caracteristic
Ib = 6,2 uA, Ic = 0,74 mA,
URc = 8,9 V, Uc = 7,1 V, Ue =-0,9 V, Uce = 8,0 V
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BJT, base bias + emitter feedback example
Calculate;
I b, I e
URe, Ue, Uce
Draw output caracteristic
Ib = 24 uA, Ie = 2,9 mA,
URc = 3,5 V, Ue = -2,5 V, Uce = 2,5 V
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BJT, base bias + collector/emitter feedback
If Ic >
then Uc <
then Ib <
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If Ic >
then Uc <
and Ue >
then Ib <
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BJT, base bias + collector feedback
Vcc  U Rc  U Rb  U be
Vcc     1 I b Rc  I b Rb  U be
Always calculate in the smallest current Ib !!
The current through Rc is not Ic but Ic + Ib,
so (β+1)Ib !!!
If Ic rises for any reason, then Uc falls and
also Ib decreases, so then Ic decreases
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BJT, base bias collector feedback example
Calculate;
Ib, ß, Ic
Draw output caracteristic
Ib = 13 uA, ß = 196, Ic = 2,5 mA
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BJT, base bias collector/emitter feedback
Vcc  U Rc  U Rb  U be  U Re
Vcc     1 I b Rc
 I b Rb
 U be
    1 I b Re
Always calculate in the smallest current Ib !!
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BJT, base bias collector/emitter feedback ex
Calculate;
I b, I e
URc, Uc, Ue, Uce
Draw output caracteristic
Ib = 11,8 uA, Ie = 1,1 mA
URc = 5,2 V, Uc = 4,8 V
Ue = 1,3 V, Uce = 3,5 V
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BJT, voltage divider
Vb is a stable voltage - 0,7 V =
so Ve is a stable voltage
Ie is determined by Ve/ Re
Ic = Ie . ß/(ß+1)
Ic is very stable and nearly
independent to ß variation, as
long as ß is BIG in value
2 methods of calculating Ic
- neglegting Ib, use voltage divider
- not neglecting Ib and use thevenin
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BJT, voltage divider, neglect Ib
R2
Vb 
Vcc
R1  R2
Ve  Vb  0V 7
Ve
Ie 
Re
So neglegt Ib to R2,
or in general Ri >> R2
In practice 10 times bigger
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Ic 

 1
Ie
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BJT, voltage divider, exact, thevenin
Thevenin resistance
R1 // R2
62k // 9k1= 7k9
Thevenin voltage Vth 
R2
Vcc
R1  R2
9k1
Vth 
16  2V 0
62k  9k1
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BJT, voltage divider, exact, thevenin
7k9
2V0
Vth  I b Rth  Vbe     1 I b Re
2, 0  I b 7k 9  0, 7   80  1 I b 0, 68k
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Ib = 20 uA
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BJT, voltage divider, example
Thevenin resistance = 6k8
Thevenin voltage
= 3V1
Ib = 18,8 uA
Ic = 2,25 mA
re = 11,5 Ω
URc = 7V4
Uc = 10V6
Ue = 2V3
Uce = 5V1
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BJT, voltage divider, example
Thevenin resistance = 255k
Thevenin voltage
= 0V0
Ib = 14,3 uA
Ic = 1,9 mA
re = 14 Ω
URc = 17V3
Uc = 0V7
Ue = -3V7
Uce = 4V4
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BJT
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BJT
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