recall lecture 8
DESCRIPTION
Recall Lecture 8. Clipper Step 1 : Find the clip value by doing KVL at the output branch Step 2 : Set the conditions to know whether diode is on or off – sketch your output waveform Clamper - PowerPoint PPT PresentationTRANSCRIPT
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Recall Lecture 8• Clipper
– Step 1: Find the clip value by doing KVL at the output branch
– Step 2: Set the conditions to know whether diode is on or off – sketch your output waveform
• Clamper– Step 1: Calculate value of VC by knowing which cycle it is
charged to (what is the peak value of that cycle) then confirmed the polarity of the capacitor
– Step 2: performed KVL as diode is now reverse- biased to obtain equation of Vout . Sketch your output waveform
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Multiple Diode CircuitsMultiple Diode Circuits
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Final Exam SEM I 2013/2014
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DIODE ID VD
OFF 0 VD < V
ON ID > 0 VD = V
REMEMBER THAT:
A pn junction diode will conduct when the p-type material is more positive than the n-type material
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OR GATE
V1 V2 VO
Vo = voltage across R
D1 and D2 off; no current flow, 0 0 0
D1 off, D2 on, current flow,
Vo – V2 + V = 0
0 5V ( 1 ) 4.3V
D1 on, D2 off, current flow,
Vo – V1 + V = 0
5V ( 1 ) 0 4.3V
Both on, using both loops will give the same equation
5V ( 1 ) 5V ( 1 ) 4.3V
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V1 V2 VO
Both on, using both loops will give the same equation
0 0 0.7
D1 on, D2 off 0 5V ( 1 ) 0.7
D1 off, D2 on 5V ( 1 ) 0 0.7V
Both are off; open circuit no current flowing through R since no GND destination
5V ( 1 ) 5V ( 1 ) 5V
AND GATE Vo = node voltage
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Chapter 4Chapter 4Bipolar Junction TransistorBipolar Junction Transistor
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REMEMBER THIS
Current flow in the opposite direction of the electrons flow; same direction as holes
e e e
I
h h h
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Transistor StructuresTransistor Structures
The bipolar junction transistor (BJT) has three separately doped regions and contains two pn junctions.
Bipolar transistor is a 3-terminal device.
Emitter (E)
Base (B)
Collector (C)
The basic transistor principle is that the voltage between two terminals controls the current through the third terminal.
Current in the transistor is due to the flow of both electrons and holes, hence the name bipolar.
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Transistor StructuresTransistor Structures There are two types of bipolar junction transistor: npn and
pnp.
The npn bipolar transistor contains a thin p-region between two n-regions.
The pnp bipolar transistor contains a thin n-region sandwiched between two p-regions.
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ActiveOperating range of the amplifier.Base-Emitter Junction forward biased.Collector-Base Junction reverse biased
CutoffThe amplifier is basically off. There is voltage but little current.Both junctions reverse biased
SaturationThe amplifier is full on. There is little voltage but lots of current.Both junctions forward biased
3 Regions of Operation3 Regions of Operation
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OPERATIONS - npnOPERATIONS - npn
The base-emitter (B-E) junction is forward biased and the base-collector (C-B) junction is reverse-biased,.
Since the B-E junction is forward biased, electrons from the emitter are injected across the B-E junction into the base IE
Once in the base region, the electrons are quickly accelerated through the base due to the reverse-biased C-B region IC
ACTIVE MODE
C
B
E
Some electrons, in passing through the base region, recombine with majority carrier holes in the base. This produces the current IB
+
-VBE
iB
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C B E
TO ILLUSTRATE
-
VBE
+
•Imagine the marbles as electrons
•A flat base region with gaps where the marbles may fall/trapped – recombine
•A sloping collector region represents high electric field in the C-B region
•Hence, when enough energy is given to the marbles, they will be accelerated towards to base region with enough momentum to pass the base and straight ‘fly’ to the collector
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MATHEMATICAL EXPRESSIONSC
B
E
+
-VBE
IE
IC
IB
IE = IS [ e VBE / VT -1 ] = IS e VBE / VT Based on KCL: IE = IC + IB
No. of electrons crossing the base region and then directly into the collector region is a constant factor of the no. of electrons exiting the base region
IC = IB
No. of electrons reaching the collector region is directly proportional to the no. of electrons injected or crossing the base region.
IC = IEIdeally = 1, but in reality it is between 0.9 and 0.998.
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Based on KCL: IE = IC + IB IC = IB IC = IE
IE = IB + IB = IB( + 1)
= [ / + 1 ]
IE = IB( + 1)
NowWith IC = IB IB = IC / Hence, IE = [ IC / ] ( + 1)IC = IE [ / + 1 ]
Comparing with IC = IE
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B
C
E
-
+
VEB
OPERATIONS - pnpOPERATIONS - pnpFORWARD ACTIVE MODE
The emitter – base (E- B) junction is forward biased and the base-collector (B- C) junction is reverse-biased,.
IE = IS [ e VEB / VT -1 ] = IS e VEB / VT
**Notice that it is VEB
IE
IC
IB
Based on KCL: IE = IC + IB
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pnp Transistor- Active mode
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SUMMARY: Circuit Symbols and SUMMARY: Circuit Symbols and ConventionsConventions
npn bipolar transistor simple block diagram and circuit symbol.
Arrow is on the emitter terminal that indicates the direction of emitter current (out of emitter terminal for the npn device)
pnp bipolar transistor simple block diagram and circuit symbol.
Arrow is on the emitter terminal that indicates the direction of emitter current (into of emitter terminal for the pnp device)
Based on KCL: IE = IC + IB
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EXAMPLE 4.1 Calculate the collector and emitter currents, given the base current and current gain. Assume a common-base current gain and a base current of . Also assume that the transistor is biased forward in the forward active mode.
Solution: The common-emitter current gain is
The collector current is
And the emitter current is
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BJT: Current-Voltage BJT: Current-Voltage CharacteristicCharacteristicIICC versus V versus VCECE
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The Emitter is common to both input (base-emitter) and output (collector-emitter).
Since Emitter is grounded, VC = VCE
With decreasing VC (VCE), the junction B-C will become forward biased too. The current IC quickly drops to zero
because electrons are no longer collected by the collector
Common-Emitter Configuration Common-Emitter Configuration - npn- npn
Node B
0V
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Characteristics of Common-Emitter Characteristics of Common-Emitter - npn- npn
NOTE: VEC for PNP
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Examples
• EXAMPLE 1• Given IB = 6.0A and
IC=510 A
Determine , and IE
• EXAMPLE 2• NPN Transistor • Reverse saturation
current Is = 10-13A with current gain, = 90. Based on VBE = 0.685V, determine IC , IB and IE
EXAMPLE 3 PNP Transistor = 60, IC= 0.85mA Determine , IE and IB