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BIPOLAR TRANSISTORSeptember 26, 2020
•Transistor Symbol and Formula
NPN Transistor PNP Transistor
IE = IC + IB IC @ IE IB << IC
α = IC/IE β (beta) = IC/IB
C(collector)
B(base)
E(emitter)
IB
IC
IE
IB
IE
IC
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C
B
E
IB
IC
IEIB
IE
IC
NPNPNP
BIPOLAR TRANSISTOR
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BIPOLAR TRANSISTORFebruary 20, 2021
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BIPOLAR TRANSISTORFebruary 20, 2021
Using DMM to test the Transistor
TESTING NPN TRANSISTOR
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BIPOLAR TRANSISTORFebruary 20, 2021
Using DMM to test the Transistor
TESTING NPN TRANSISTOR
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BIPOLAR TRANSISTORFebruary 19, 2021
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DC BIAS
B
E
C
+
-
VBE0.7V
+ -
-VBB +IBxRB +0.7=0
IBxRB=VBB-0.7
IB=(VBB-0.7)/RB
VRC = Ic x RC
VC= VCC - ICxRC
CE= VC - VE
IE = (1+ ) x IB
IC = x IB
IB= IE/ (1+ )9/28/2021 [email protected]
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BIPOLAR TRANSISTORFebruary 19, 2021
Example
Determine IB, IC, IE, VBE, VB, VE, VC, VCE in the circuit shown, the transistor has a beta=150 ( =150 ). Note: VBE=0.7V (diode voltage)
B
E
C
a) VBE= __________VDCb) IB= __________mAc) IC= __________mAd) IE= __________mAe) VB= _________VDCf) VE= _________VDCg) VC= _________VDCh) VCE=_________VDC
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• Cutoff: Diode from B to Emitter is Off
• Active: Diodes from Base to Emitter is ON and Diode
from Base to Collector is Off
• Saturation: Both diodes are ON and VCE= 0 vdc
BIPOLAR TRANSISTORFebruary 20, 2021
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BIPOLAR TRANSISTORFebruary 20, 2021
Example
For the transistor circuit showna) What is VCE when VIN=0 Vb) What minimum value of IB is required to saturate
this transistor if beta is 200c) Calculate the maximum value of RB when VIN=5 V
Solutiona) when VIN=0 V transistor is off VCE=VCC= 10 Vb) Since transistor is saturated VCE= 0 V then IC(sat)= VCC/ RC= 10V/1k= 10mA
therefore IB= IC(sat)/beta= 10mA/ 200=50 microAc) When the transistor is on voltage across RB is VRB= VIN-VBE= 5V – 0.7= 4.3 VDC
then RB(max)= VRB/ IB(min)= 4.3v/50uA = 86 K
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Saturation
Cut Off
Active
BIPOLAR TRANSISTOR
C
B
E
IB
IC
IE
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BIPOLAR TRANSISTORFebruary 20, 2021
How to draw DC load line1. Point A at saturation I(sat)=VCC/RC= 10V/1k=10mA2. Point B at cutoff, VCE=VCC3. Draw a line A to B
A
B
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BIPOLAR TRANSISTOR Mode Operations
1. Amplifier2. Saturation3. Cutoff
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Emitter Bias
=100
Example:For the circuit shownFind:(a) IB (mA)(b) IC (mA)(c) IE (mA)(d) VE (V)(e) VB (V)(f) VC (V)(g) VCE (V)
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VE= VBB-0.7= 4.3VIE= VE / RE= 4.3/2.2k = 1.95 mA
VC= VCC- ICxRC= 15 – 1.95x1kVC= 13.05V and VCE=VC-VE9/28/2021 [email protected]
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Example:For the circuit shown. Find:(a) VBB(b) VE(c) IE (mA)(d) IB (mA)(e) IC (mA)(f) VRC(g) VC(h) VCE
VBB
VE
VC
=100
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Example:For the circuit shown. Find:(a) VBB(b) VE(c) IE (mA)(d) IB (mA)(e) IC (mA)(f) VC(g) VEC
(a) VBB= VEE x R1 / (R1+R2) (b) VE= 0.7 + VBB(c) IE= (VEE-VE)/RE (d) IB= IE / (+1)(e) IC= IB x ](f) VC= ICxRC(g) VEC= VE-VC
E
VBB
C
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Capacitor Reactance XCXC= 1/( 2 x f x C )Where:XC=Ohm = 3.142 f= hertz C= Farad
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DC model
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CE AMPLIFIER
• Input signal to the Base
• Output from Collector
• Input and Output 180 degree out of
phase
• C1= DC blocking=AC Coupling
• C2= DC blocking=AC Coupling
• C3= bypass
C1
C2
C3
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THE COMMON – EMITTER AMPLIFIER
Example
Vin=10mVrms
5 k
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DC circuit for the CE Amplifier
Determine DC values for the circuit shown:a) VB= (VCC x R2)/(R1+R2)b) VE= VB-0.7Vc) IE= VE / REd) IB= IE / ( + 1)e) IC= x IBf) VC= VCC – IC x RCg) VCE= VC-VEh) r’e = 25mV/IE
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AC circuit for the CE Amplifier
• r’e = 6.58 • Rin(base)= ac x r’e• Rin(total)= R1 //R2 //Rin(base)= Zin• Rc= output resistance=RC // RL= Zout• Av =voltage gain= Rc / r’e
Determine AC values for the CE Amplifier:a) r’eb) Rin(base)c) Rin(total)=Zind) Rc(output resistance)=Zoute) AV=voltage gain f) Vout
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THE COMMON –COLLECTOR AMPLIFIERAnother name EMITTER FOLLOWER AMPLIFIER
• Input applies to the Base• Output from Emitter• C1 and C2 DC Blocking• Input / output INPHASE
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THE COMMON –COLLECTOR AMPLIFIERAnother name EMITTER FOLLOWER AMPLIFIER
• Rin(base)= (r’e + Re)• Rin(base)= x Re• Rin(total)= R1 //R2 //Rin(base)• Re= output resistance=RE // RL• Av =voltage gain= Re/(r’e + Re)• Vout= Av x Vin
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THE COMMON –COLLECTOR AMPLIFIERAnother name EMITTER FOLLOWER AMPLIFIER
Example
=175
For the CC Amplifier circuit shown, determine:a) VB= (VCC x R2)/(R1+R2)b) VE= VB-0.7 Vc) IE= VE / REd) VC=VCCe) VCE= VC-VEf) r’e= 25mV / IEg) Re=(Zout)= RE // RL= (RE x RL)/(RE+RL)h) Rin(base)= x Rei) Rin (total)= R1 // R2 // Rin(base)j) Av (voltage gain)= Re / (r’e + Re)k) Vout= Av x Vin
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THE COMMON –COLLECTOR AMPLIFIERAnother name EMITTER FOLLOWER AMPLIFIER
Example
=175
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COMMON BASE AMPLIFIERBIPOLAR TRANSISTORFebruary 20, 2021
• Voltage gain AV= Rc / (r’e // RE)• Input resistance Rin= (r’e // RE)• Output resistance Rout= RC // RL• C1 & C2 DC Blocking
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BIPOLAR TRANSISTORFebruary 20, 2021
For the CB Amplifier circuit shown, determine:a) VB= (VCC x R2)/(R1+R2)b) VE= VB-0.7Vc) IE= VE/IEd) VC= VCC – IC x RCe) VCE= VC - VEf) r’e= 25mV / IEg) Rin= r’e // REh) Rc (zout)= RC // RLi) Av (voltage gain)= Rc / r’e
Example
=250
COMMON BASE AMPLIFIER
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BIPOLAR TRANSISTORFebruary 20, 2021
Example
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THE E N D
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