lecture 8 bjt_1
DESCRIPTION
Lecture Slide for Chapter Bipolar Junction Transistor(BJT).edt2012TRANSCRIPT
Lecture 8: Lecture 8:
BIPOLAR JUNCTION BIPOLAR JUNCTION TRANSISTORSTRANSISTORS
Semester II2010/2011
Code: EEE2213
BJT STRUCTUREBasic structure of the bipolar junction transistor (BJT) determines its operating characteristics.
Constructed with 3 doped semiconductor regions called emitter, base, and collector, which separated by two pn junctions.
2 types of BJT;
(1)npn: Two n regions separated by a p region
(2)pnp: Two p regions separated by an n region.
BIPOLAR: refers to the use of both holes & electrons as current carriers in the transistor structure.
Base-emitter junction: the pn junction joining the base region & the emitter region.
Base-collector junction: the pn junction joining the base region & the collector region.
A wire lead connects to each of the 3 regions. These leads labeled as;
E: emitter
B: base
C: collector
BASE REGION: lightly doped, & very thin
EMITTER REGION: heavily doped
COLLECTOR REGION: moderately doped
Standard BJT Symbols
BASIC BJT OPERATIONFor a BJT to operate properly as an amplifier, the two pn junctions must be correctly biased with external dc voltages.
Figure: shows a bias arrangement for npn BJTs for operation as an amplifier.
In both cases, BE junction is forward-biased & the BC junction is reverse-biased. called forward-reverse bias.
Look at this one circuit as two separate circuits, the base-emitter(left side) circuit and the collector-emitter(right side) circuit. Note that the emitter leg serves as a conductor for both circuits. The amount of current flow in the base-emitter circuit controls the amount of current that flows in the collector circuit. Small changes in base-emitter current yields a large change in collector-current.
The heavily doped n-type emitter region has a very high density of conduction-band (free) electrons.
These free electrons easily diffuse through the forward-based BE junction into the lightly doped & very thin p-type base region (indicated by wide arrow).
The base has a low density of holes, which are the majority carriers (represented by the white circles).
A small percentage of the total number of free electrons injected into the base region recombine with holes & move as valence electrons through the base region & into the emitter region as hole current (indicated by red arrows).
BJT operation showing electron flow.
When the electrons that have recombined with holes as valence electrons leave the crystalline structure of the base, they become free electrons in the metallic base lead & produce the external base current.
Most of the free electrons that have entered the base do not recombine with holes because the base is very thin.
As the free electrons move toward the reverse-biased BC junction, they are swept across into the collector region by the attraction of the positive collector supply voltage.
The free electrons move through the collector region, into the external circuit, & then return into the emitter region along with the base current.
The emitter current is slightly greater than the collector current because of the small base current that splits off from the total current injected into the base region from the emitter.
Transistor CurrentsThe directions of the currents in both npn and pnp transistors and their schematic symbol are shown in Figure (a) and (b). Arrow on the emitter of the transistor symbols points in the direction of conventional current. These diagrams show that the emitter current (IE) is the sum of the collector current (IC) and the base current (IB), expressed as follows:
IE = IC + IB
BJT CHARACTERISTICS & PARAMETERS
Figure shows the proper bias arrangement for npn transistor for active operation as an amplifier. Notice that the base-emitter (BE) junction is forward-biased by VBB and the base-collector (BC) junction is reverse-biased by VCC. The dc current gain of a transistor is the ratio of the dc collector current (IC) to the dc base current (IB), and called dc beta (DC).
DC = IC/IB
The ratio of the dc collector current (IC)to the dc emitter current (IE) is the dc alpha. α DC = IC/IEα DC = IC/IE
Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.
Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.
7350
65.3
A
mA
I
I
B
CDC
IE = IC + IB = 3.65 mA + 50 μA = 3.70 mA
986.070.3
65.3
mA
mA
I
I
E
CDC
The collector current is determined by multiplying the base current by beta.
Thus, IC= βDC * IB
Analysis of this transistor circuit to predict the dc voltages and currents requires use of Ohm’s law, Kirchhoff’s voltage law and the beta for the transistor;
Application of these laws begins with the base circuit to determine the amount of base current. Using Kichhoff’s voltage law, subtract the VBE =0.7 V, and the remaining voltage is dropped across RB .
Thus, VRB = VBB - VBE
.
Determining the current for the base with this information is a matter of applying of Ohm’s law. VRB
/RB = IB
What we ultimately determine by use of Kirchhoff’s voltage law for series circuits is that, in the base circuit, VBB is distributed across the base-emitter junction and RB in the base circuit. In the collector circuit we determine that VCC is distributed proportionally across RC and the transistor(VCE).
BJT Circuit AnalysisThere are three key dc voltages and three key dc currents to be considered. Note that these measurements are important for troubleshooting.
IB: dc base current
IE: dc emitter current
IC: dc collector current
VBE: dc voltage across
base-emitter junction
VCB: dc voltage across
collector-base junction
VCE: dc voltage from
collector to emitter
When the base-emitter junction is forward-biased,
VBE 0.7 V≅
VRB = IBRB : by Ohm’s law
IBRB = VBB – VBE : substituting for VRB
IB = (VBB – VBE) / RB : solving for IB
VCE = VCC – VRc : voltage at the collector with respect to the grounded emitter
VRc = ICRC
VCE = VCC – ICRC : voltage at the
collector with
respect to the emitter
The voltage across the reverse-biased collector-base junction
VCB = VCE – VBE where IC = βDCIB
Ex 4-2 Determine IB, IC, IE, VBE, VCE, and VCB in the circuit of Figure. The
transistor has a βDC = 150.
Ex 4-2 Determine IB, IC, IE, VBE, VCE, and VCB in the circuit of Figure. The
transistor has a βDC = 150.
When the base-emitter junction is forward-biased,
VBE 0.7 V≅
IB = (VBB – VBE) / RB
= (5 V – 0.7 V) / 10 kΩ = 430 μA
IC = βDCIB = (150)(430 μA) = 64.5 mA IE = IC + IB = 64.5 mA + 430 μA = 64.9 mA
VCE = VCC – ICRC = 10 V – (64.5 mA)(100 Ω) = 3.55 V VCB = VCE – VBE = 3.55 V – 0.7 V = 2.85 V
Since the collector is at a higher voltage than the base, the collector-base junction is reverse-biased.
Gives a graphical illustration of the relationship of collector current and VCE with specified amounts of base current. With greater increases of VCC , VCE continues to increase until it reaches breakdown, but the current remains about the same in the linear region from 0.7V to the breakdown voltage.
Collector Characteristic Curves
Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5 μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.
Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5 μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.
IC = βDC IB
IB IC
5 μA 0.5 mA10 μA 1.0 mA15 μA 1.5 mA20 μA 2.0 mA25 μA 2.5 mA
CutoffWith no IB , the transistor is in the cutoff region and just as the name implies there is practically no current flow in the collector part of the circuit. With the transistor in a cutoff state, the full VCC can be measured across the collector and emitter(VCE).
Cutoff: Collector leakage current (ICEO) is extremely small and is usually neglected. Base-emitter and base-collector junctions are reverse-biased.
SaturationOnce VCE reaches its maximum value, the transistor is said to be in
saturation.
Saturation: As IB increases due to increasing VBB, IC also increases and VCE decreases due to the increased voltage drop across RC. When the transistor reaches saturation, IC can increase no further regardless of further increase in IB. Base-emitter and base-collector junctions are forward-biased.
DC Load LineThe dc load line graphically illustrates IC(sat) and cutoff for a transistor.
DC load line on a family of collector characteristic curves illustrating the cutoff and saturation conditions.
Active region of the transistor’s operation.
Ex 4-4 Determine whether or not the transistors in Figure is in
saturation. Assume VCE(sat) = 0.2 V.
mAk
VV
R
VVI
C
satCECCsatC
8.90.1
2.010
)()(
Ex 4-4 Determine whether or not the transistors in Figure is in
saturation. Assume VCE(sat) = 0.2 V.
First, determine IC(sat)
mAmAII
mAk
V
k
VV
R
VVI
BDCC
B
BEBBB
5.11)23.0)(50(
23.010
3.2
10
7.03
Now, see if IB is large enough to produce IC(sat).
Thus, IC greater than IC(sat). Therefore, the transistor is saturated.
Maximum Transistor RatingsA transistor has limitations on its operation. The product of VCE
and IC cannot be maximum at the same time. If VCE is maximum, IC can be calculated as
CE
DC V
PI (max)
Ex 4-5 A certain transistor is to be operated with VCE = 6 V. If its maximum power rating is 250 mW, what is the most collector current that it can handle?
mAV
mW
V
PI
CE
DC 7.41
6
250(max)
Ex 4-6 The transistor in Figure has the following maximum ratings: PD(max) = 800
mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?
First, find IB so that you can determine IC.
The voltage drop across RC is.
PD = VCE(max)IC = (15V)(19.5mA) = 293 mW
VCE(max) will be exceeded first because the entire supply voltage, VCC will be dropped across the transistor.
VRc = ICRC = (19.5 mA)(1.0 kΩ) = 19.5 VVRc = VCC – VCE when VCE = VCE(max) = 15 VVCC(max) = VCE(max) + VRc = 15 V + 19.5 V = 34.5 V
mAAII
Ak
VV
R
VVI
BDCC
B
BEBBB
5.19)195)(100(
19522
7.05
Ex 4-6 The transistor in Figure has the following maximum ratings: PD(max) = 800
mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?
Derating PD(max)
P D(max) is usually specified at 25°C.
At higher temperatures, P D(max) is less.
Datasheets often give derating factors for determining P D(max) at any temperature above 25°C.
Ex 4-7 A certain transistor has a P D(max) of 1 mW at 25°C. The derating factor is 5 mW/ °C. What is the P D(max) at a temperature of 70°C?
Transistor DatasheetRefer Figure 4-20 (a partial datasheet for the 2N3904 npn transistor).
The maximum collector-emitter voltage (VCEO) is 40V.
The CEO subscript indicates that the voltage is measured from collector to emitter with the base open. VCEO= VCE(MAX)
The maximum collector current is 200 mA.
* Other characteristics can be referred from the datasheet.
A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?
A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?
PD(max) = 800 mWVCE(max) = 15 VIC(max) = 100 mA.
VCC(max) = VCE(max) + VRc = 40 V + 19.5 V = 59.5 V
PD = VCE(max)IC = (40V)(19.5mA) = 780 mW
However at the max value of VCE, the power dissipation is
Power Dissipation exceeds the maximum of 645 mW specified on the datasheet.
THE BJT AS AN AMPLIFIERAmplification of a relatively small ac voltage can be had by placing the ac signal source in the base circuit.
Recall that small changes in the base current circuit causes large changes in collector current circuit.
The ac emitter current : Ie ≈ Ic = Vb/r’e
The ac collector voltage : Vc = IcRc
Since Ic ≈ Ie, the ac collector voltage : Vc ≈ IeRc
The ratio of Vc to Vb is the ac voltage gain : Av = Vc/Vb
Substituting IeRc for Vc and Ier’e for Vb : Av = Vc/Vb ≈ IcRc/Ier’e
The Ie terms cancel : Av ≈ Rc/r’e
Ex 4-9 Determine the voltage gain and the ac output voltage in Figure if r’e = 50 Ω.
The voltage gain : Av ≈ Rc/r’e = 1.0 kΩ/50 Ω = 20The ac output voltage : AvVb = (20)(100 mV) = 2 V
THE BJT AS A SWITCHA transistor when used as a switch is simply being biased so that it is in cutoff (switched off) or saturation (switched on). Remember that the VCE in cutoff is VCC and 0V in saturation.
Conditions in Cutoff & Saturation
C
satCECC
satC R
VVI )(
)(
DC
satC
B
II
)(
(min)
A transistor is in the cutoff region when the base-emitter junction is not forward-biased. All of the current are zero, and VCE is equal to VCC
VCE(cutoff) = VCC
When the base-emitter junction is forward-biased and there is enough base current to produce a maximum collector current, the transistor is saturated.The formula for collector saturation current is
The minimum value of base current needed to produce saturation is
Ex 4-10 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?
(b) What minimum value of IB is required to saturate this transistor if βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.
Ex 4-10 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?
(b) What minimum value of IB is required to saturate this transistor if βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.
AmAI
I
mAk
V
R
VI
DC
satCB
C
CCsatC
50200
10
100.1
10
)((min)
)(
kA
V
I
VR
B
RB
B 8650
3.4
(min)(max)
(a) When VIN = 0 VVCE = VCC = 10 V
(b) Since VCE(sat) is neglected,
(c) When the transistor is on, VBE ≈ 0.7 V.VRB = VIN – VBE ≈ 5 V – 0.7 V = 4.3 V
Calculate the maximum value of RB
Transistor ConstructionTransistor Construction
There are two types of transistors: • pnp • npn
The terminals are labeled: • E - Emitter• B - Base• C - Collector
pnppnp
npnnpn
4242
Transistor OperationTransistor OperationWith the external sources, VEE and VCC, connected as shown:
• The emitter-base junction is forward biased• The base-collector junction is reverse biased
4343
Currents in a TransistorCurrents in a Transistor
The collector current is comprised of two currents:
BICIEI
minorityCOI
majorityCICI
Emitter current is the sum of the collector and base currents:
4444
Common-Base ConfigurationCommon-Base Configuration
The base is common to both input (emitter–base) and output (collector–base) of the transistor.
4545
Common-Base AmplifierCommon-Base Amplifier
Input CharacteristicsInput Characteristics
This curve shows the relationship between of input current (IE) to input voltage (VBE) for three output voltage (VCB) levels.
4646
This graph demonstrates the output current (IC) to an output voltage (VCB) for various levels of input current (IE).
Common-Base AmplifierCommon-Base Amplifier
Output CharacteristicsOutput Characteristics
4747
Operating RegionsOperating Regions
• Active – Operating range of the amplifier.
• Cutoff – The amplifier is basically off. There is voltage, but little current.
• Saturation – The amplifier is full on. There is current, but little voltage.
4848
EI
CI
Silicon)(for V 0.7BEV
ApproximationsApproximations
Emitter and collector currents:
Base-emitter voltage:
4949
Ideally: = 1 In reality: is between 0.9 and 0.998
Alpha (Alpha ())
Alpha () is the ratio of IC to IE :
EI
CIα dc
Alpha () in the AC modeAC mode:
EI
CIα
Δ
Δac
5050
Transistor AmplificationTransistor Amplification
Voltage Gain:
V 50kΩ 5ma 10
mA 10
10mA20Ω
200mV
))((RL
IL
V
iI
LI
EI
CI
iR
iViIEI
Currents and Voltages:
5151
250200mV
50V
iV
LVvA
Common–Emitter ConfigurationCommon–Emitter Configuration
The emitter is common to both input (base-emitter) and output (collector-emitter).
The input is on the base and the output is on the collector.
5252
Common-Emitter CharacteristicsCommon-Emitter Characteristics
Collector Characteristics Base Characteristics
5353
Common-Emitter Amplifier CurrentsCommon-Emitter Amplifier Currents
Ideal CurrentsIdeal Currents
IE = IC + IB IC = IE
Actual CurrentsActual Currents
IC = IE + ICBO
When IB = 0 A the transistor is in cutoff, but there is some minority current flowing called ICEO.
μA 0
BICBO
CEO α
II
1
where ICBO = minority collector current
5454
ICBO is usually so small that it can be ignored, except in high power transistors and in high temperature environments.
Beta (Beta ())
In DC mode:
In AC mode:
represents the amplification factor of a transistor. ( is sometimes referred to as hfe, a term used in transistor modeling calculations)
B
C
I
Iβ dc
constantac
CEV
B
C
I
I
5555
Determining from a Graph
Beta (Beta ())
108
A 25
mA 2.7β 7.5VDC CE
100
μA 10
mA 1
μA) 20 μA (30
mA) 2.2mA (3.2β
7.5V
AC
CE
5656
Relationship between amplification factors and
1β
βα
1α
αβ
Beta (Beta ())
Relationship Between Currents
BC βII BE 1)I(βI
5757
Common–Collector ConfigurationCommon–Collector Configuration
The input is on the base and the output is on the emitter.
5858
Common–Collector ConfigurationCommon–Collector Configuration
The characteristics are similar to those of the common-emitter configuration, except the vertical axis is IE.
5959
VCE is at maximum and IC is at minimum (ICmax= ICEO) in the cutoff region.
IC is at maximum and VCE is at minimum (VCE max = VCEsat = VCEO) in the saturation region.
The transistor operates in the active region between saturation and cutoff.
Operating Limits for Each ConfigurationOperating Limits for Each Configuration
6060
Power DissipationPower Dissipation
Common-collector:
CCBCmax IVP
CCECmax IVP
ECECmax IVP
Common-base:
Common-emitter:
6161
Transistor Specification SheetTransistor Specification Sheet
6262
Transistor Specification SheetTransistor Specification Sheet
6363
Transistor TestingTransistor Testing• Curve TracerCurve Tracer
Provides a graph of the characteristic curves.
• DMMDMMSome DMMs measure DC or hFE.
• OhmmeterOhmmeter
6464
Transistor Terminal Transistor Terminal IdentificationIdentification
6565