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Lecture 7. Bipolar Junction Transistor (1). Dr Tarek Abdolkader Dr Tarek AbdolkaderDr Tarek Abdolkader Dr Tarek Abdolkader. Dr Tarek Abdolkader Dr Tarek AbdolkaderDr Tarek Abdolkader Dr Tarek Abdolkader. By: Dr Tarek Abdolkader. OUTLINE. What is a Bipolar junction transistor? - PowerPoint PPT PresentationTRANSCRIPT
Dr T
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المملكة العربية السعوديةوزارة التعليم العالي - جامعة
أم القرىكلية الهندسة و العمارة
اإلسالميةقسم الهندسة الكهربائية
802311-4 ELECTRONIC DEVICES
KINGDOM OF SAUDI ARABIAMinistry of Higher Education
Umm Al-Qura UniversityCollege of Engineering and Islamic Architecture
Electrical Engineering Department
Dr T
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Lecture 7
By: Dr Tarek Abdolkader
10/1/1434 Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader Dr T
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OUTLINE
•What is a Bipolar junction transistor?
•Construction and symbol
•Active mode of operation
•Cut-off and saturation modes
•BJT characteristics
•BJT specifications
•BJT as an amplifier
•BJT as a switch
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10/1/1434 Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader Dr T
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Dr T
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der What is a Bipolar junction transistor?
• A transistor is a semiconductor device used to amplify and switch electronic signals.
• The name transistor is a combination of the words transfer and resistor - a transferred resistor
• This is because it can amplify electrical signals by transferring a current I from a low to a high-resistance circuit.
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• Bipolar junction transistor (BJT) is one of the most famous types of transistors.• Bipolar: current is carried by two types of
carriers (electrons and holes).• Junction: current passes through pn-junctions
• BJT is a three terminal device.
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Current increases with vi but vo has the same amplitude as vi (no amplification)
Amplification of signals
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• There are two main types of BJT:
1. npn transistor 2. pnp transistor
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• The arrow on the emitter is in the direction from p to n.
Construction and symbol
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In the normal operation of the transistor (when it is used as an amplifier), The Base-Emitter junction (BE) is forward biased and the Base-Collector junction (BC) is reverse biased.
Active mode of operation
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p
EfEf
n p p
n p
at equilibrium BE is forwarded and BC is reversed
n p n
n p n
Active mode of operation
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• The emitter is made heavily doped to emit large number of carriers into base.
• The base is made very thin so the least number of carriers injected from the emitter is lost by recombination before reaching the collector.
• The collector collects the minority carriers from the emitter through the base. The collector usually has the lightest doping concentrations of the three regions.
• The main source of current collected at the collector is the current sent from the emitter. The collector own current is very small (reverse-biased junction).
• The collector current is controlled by the voltage in a different circuit (VBE). Small changes in VBE leads to large changes in collector current (IC).
• The main difference between transistor and p-n junction is that the current is collected in a different circuit with different resistance (transfer resistor).
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p n p
1
2 3
4
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3. Some of the holes injected from the emitter are recombined with some of the electrons diffusing from the base (small portion because the base is thin).
4. Most of the hole current injected from the emitter is collected by the collector.
5. Minority holes are drifted by the reverse-bias field of the BC junction.6. Minority electrons are drifted by the reverse-bias field of the BC
junction.
1. Hole diffusion in the forward-biased BE junction.
2. Electron diffusion in the forward-biased BE junction (it is less than hole diffusion because base is lightly doped while emitter is heavily doped).
Active mode of operationTransistor currents:
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Transistor equations in active mode:
• The common-emitter dc current gain βDC: βDC >> 1
CDC
B
I
I
• The common-base dc current gain αDC : αDC < 1
CDC
E
I
I
• From KCL: IE = IC + IB
IC = βDC IB
IE = (1 + βDC ) IB
• Remember that for forward-biased Si pn junction, VBE = 0.7 V
• All above equations are valid only in the active mode.
1DC
DCDC
1
DCDC
DC
Active mode of operation
VBE = VBC + VCE
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Determine all currents and voltages in the circuit shown. βDC = 150
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Active mode of operation
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Cut-off mode:
BE is reverse-biased and BC is also reverse-biased
• No current will be in both circuits except the reverse saturation currents (can be neglected).
IB = 0, IC = 0, IE = 0
n p n
n p n
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Saturation mode:BE is forward-biased and BC is also forward-biased
• The field in the BC junction is no longer sufficient to collect all the current injected by the emitter.
• In this case, increasing the base current will not result in an increase in the collector current. We say that the current is saturated. In this case IC < βDC IB .
• Saturation starts before the BC junction is completely forward-biased (VBC 0.5 ~ 0.7 V , VCE(sat) 0 ~ 0.2 V).
n p n n p n
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VBE
VBC
active
saturation
Cut-off
inverse
VBE
VBC
active
saturation
Cut-off
inverse
npn
pnp
As long as BC junction is reverse biased (VCE > VCE(sat) ), the collector is able to withdraw αDC IE (βDC IB) and the current will depend only on IB and not on VCE.
Active region
IC = βDC IB
saturation
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Sketch an ideal family of collector curves for the circuit shown for IB = 5 A to 25 A in 5 A increments. βDC = 100
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BJT characteristics
0.3 V
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• Determine IB from the input circuit. (Use VBE = 0.7 V for forward-biased BE junction).
• Draw the load line and determine the operating point.
Load line analysis:
VCC = ICRC + VCE
• Note that the onset of saturation for all base currents is nearly the same. This value is called VCE(sat) ~ 0 → 0.2 V
IC = βDC IB
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Determine the mode of operation of the transistor of the circuit shown. Assume VCE(sat) = 0.2 V, βDC = 50
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1. Draw the collector characteristics and the load line. Then determine the Q-point.
2. Assume the transistor is in the active mode (IC = βDCIB), then find VCE. If VCE > VCE(sat), then the transistor is really in active mode. Otherwise, it is saturation.
3. Assume the transistor is in the saturation mode (VCE = VCE(sat) ), then find IC. If IC < βDCIB , then the transistor is really in saturation mode. Otherwise, it is active.
There are three methods to solve this problem:
Operating point and load line
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BJT data sheet:
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• Amplification is the process of linearly increasing the amplitude of an ac signal.
• The transistor in the active mode amplifies the base current linearly (Ic = β Ib )
• Large changes in collector current leads to large changes in the collector voltage Vc.
• Voltage gain is the ratio Vc /Vin.
Vin
Ib
Ic
Vc
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• Input voltage is zero (low).• Transistor is in cut-off mode.• Base current is zero.• Collector current is zero.• CE junction is open-circuited.• VCE = VCC (high)
• Input voltage is +VBB (high).• Transistor is in saturation mode.• Base current is high.• Collector current is IC(sat) (high) .• CE junction is nearly short-
circuited.• VCE = VCE(sat) ~ 0.2 V (low)
Transistor switch is called an inverter because low input results in high output and vice versa.
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For the circuit shown, (a) Find VCE for VIN = 0, (b) Find the maximum value of RB to drive the transistor into saturation with VIN = 5 V, βDC = 200. Neglect VCE(sat).
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BJT as a switch