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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

9/28/2021 [email protected]

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C

B

E

IB

IC

IEIB

IE

IC

NPNPNP

BIPOLAR TRANSISTOR


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9/28/2021 [email protected] 3

BIPOLAR TRANSISTORFebruary 20, 2021




Using DMM to test the Transistor

TESTING NPN TRANSISTOR



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]




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



• 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





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


Saturation

Cut Off

Active

BIPOLAR TRANSISTOR

C

B

E

IB

IC

IE





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



BIPOLAR TRANSISTOR Mode Operations

1. Amplifier2. Saturation3. Cutoff


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)



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]



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



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



Capacitor Reactance XCXC= 1/( 2 x f x C )Where:XC=Ohm = 3.142 f= hertz C= Farad



DC model



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



THE COMMON – EMITTER AMPLIFIER

Example

Vin=10mVrms

5 k



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



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



THE COMMON –COLLECTOR AMPLIFIERAnother name EMITTER FOLLOWER AMPLIFIER

• Input applies to the Base• Output from Emitter• C1 and C2 DC Blocking• Input / output INPHASE




• 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




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




Example

=175



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




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




Example

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THE E N D


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