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Unformatted text preview: ELEC202 – Electronic Circuits II BJT Bipolar Transistor (BJT)
Symbol (npn) Ideal Structure (npn) Actual Structure in IC (npn) Collector C
C
n
B
Base IC IB p E IE
E C n+
p
n− B n B NEmitter > NBase > Ncollector Emitter E Modes of Operation
VBC
VCE =VCB +VBE
Reverse
E/B reverse
C/B forward
VBC(on)
Cutoff
Both reverse Note: VCE(sat) = −VBC(on) + VBE(on)
≈ −0.5V + 0.7V
≈ 0.2V Saturation
Both forward 0 VBE(on) VBE
Active Forward
E/B forward
C/B reverse Typical values
VBE(on) = 0.6V  0.7V
VCE(sat) = 0.2V 0.3V Amplifier  BJT operates at Active Forward Region
Switch  BJT operates between Cutoff and Saturation Regions BJT in Active Mode
Operations:
C
VCB
>0
B
VBE
>VBE(on) Injected
holes Collected
n
n−
electrons
p
p
diffusing
electrons
n
n+ W C/B junction
Reverse bias
E/B junction
Forward bias Good BJT transistor requires:
(1) doping concentration in
Emitter (ND) >> in Base (NA)
(2) very narrow base width W
(3) large EmitterBase junction AE E Last updated: 8Sep03 1 ELEC202 – Electronic Circuits II BJT Equations: + VD −
V
I D = I S exp D nV
T Recall ID Collector Current:
V I C = I S exp BE , n = 1
V T A qD n
IS = E n i
N AW
A
∝E
N AW IC is a function of VBE and independent of collector voltage
as long as C/B junction is reverse biased. 2 AE = E/B junction area, Dn = diffusion constant
NA = Base doping concentration, W = base width
ni = intrinsic carrier concentration = 1.5×1010 carrier/cm3 Base Current:
IB is a factor of IC.
I
IB = C where β current gain (commonemitter current gain)
 function of W and ND/ NA ratio β β ↑ with W ↓ and/or ND/ NA ↑ typical values of β is between 50  200
Emitter Current:
I E = IC + I B
I
β +1
= IC + C =
IC β or iC = α i E β where α = β
β +1 or β= α
1−α α – commonbase current gain
100
for large β, say 100, α =
≈ 0.99 ≈ 1
101
20
β = 20, α =
≈ 0.95
21 Last updated: 8Sep03 2 ELEC202 – Electronic Circuits II BJT Large Signal Model
→ IB B
Common Emitter
Configuration
( πmodel ) ← IC +
VBE C V
I S exp BE
V
T − ← IC (IS /β) B → IB βIB Size of this diode = (IS /β)
E β = commonemitter
current gain E C α = commonbase
current gain C V I S exp BE VT Common Base
Configuration
( Tmodel ) C αIE
B B (IS /α) IE (IS /α) E E VoltageControlled Current Source CurrentControlled Current Source pnp transistor
Symbol (pnp) Ideal Structure (pnp) Actual Structure in IC (pnp) Collector E
E E
p
B
Base n IE IB
B p IC
C p+ C B p+
n
psubstrate In general, βpnp < βnpn Emitter C Last updated: 8Sep03 3 ELEC202 – Electronic Circuits II BJT Voltage Polarities and Current Flow
npn BJT pnp BJT C VCB
>0 IB IC B
VBE
> VBE(on) E IE
E VBE
> VBE(on) IB IE B IC VBC
>0 C V I C = I S exp BE V T
IC = β I B V I C = I S exp EB V T
IC = β I B IC = α I E IC = α I E Replace
all VBE
by VEB DC Analysis:
Example 1: Find VC , VB , VE , IC , IB and IE. Assume β = 100, VBE = 0.7V at IC = 1mA.
10V Assume the BJT operates in active region.
VB = 0
VE = VB – VBE = 0 − 0.7 = −0.7V
V − (−10) − 0.7 + 10
IE = E
=
= 0.93mA
RE
10k
100
β
IC = α I E =
IE =
0.93m = 0.9208mA
β +1
101
VC = 10 − ICRC = 10 − (0.7208m)(5k) = 5.396V
0.9208m
I
IB = C =
= 9.208µA
100
β RC = 5kΩ IC
VC IB VE
RE = 10kΩ IE −10V
Check: VC >VB , ∴BJT operates in active region More Accurate Calculation:
Given: VBE = 0.7V at IC = 1mA
Now IC = 0.9208mA ⇒ VBE = ?
V V
Since I C1 = I S exp BE1 and I C 2 = I S exp BE 2
V V T
T Last updated: 8Sep03 4 ELEC202 – Electronic Circuits II BJT V I S exp BE 2 V T = exp(V − V ) / V
=
BE 2
BE1
T VBE1 I S exp VT Therefore IC 2
I C1 or I
V BE 2 − VBE1 = VT ln C 2
I C1 where VT = kT
= 25.9mV @ 300k
q 0.9208m So, IC = 0.9208mA ⇒ V BE = 0.7 + 25.9m ln = 0.6979V 1m V + 10 0 − 0.6979 + 10
IE = E
=
= 0.9302mA
10k
RE
% error in VBE = 0.7 − 0.6979
= 0.3%
0.6979 0.9302 − 0.93
negligible
= 0.02%
0.9302
Now, assume the calculated IC = 10mA (10 times larger than the rated current)
0.7596 − 0.7
= 7.9%
VBE = 0.7 + 25.9m ln(10) = 0.7596V ⇒ % error =
0.7596
0.9240 − 0.93
10 − 0.7596
IE =
= 0.9240mA
⇒ % error =
= 0.6%
Still very small
10k
0.9240
% error in IE = Conclusion: We always assume VBE is a constant in DC analysis, VBE(on) = 0.7V
Example 2: Find VC , VB , VE , IC , IB and IE. Assume β = 100, VBE(on)  = 0.7V.
Again, you need to assume the BJT operates in active region.
VB = 0
VE = VB – VEB = 0 + 0.7= 0.7V
10 − V E 10 − 0.7
IE =
=
= 0.93mA
RE
10k
100
β
IC = α I E =
IE =
0.93m = 0.921mA
101
β +1
VC = −10 + ICRC = −10 + (0.921m)(5k) = −5.40V
I
0.921m
IB = C =
= 9.21µA
β
100
Check:
Note: VC < VB, ∴BJT operates in active region.
Usually keep 3 significant figures in the calculation. Last updated: 8Sep03 10V
RE = 10kΩ
IB IE
VE VC
RC = 5kΩ IC −10V 5 ELEC202 – Electronic Circuits II BJT Example 3: Find VC , VB , VE , IC , IB and IE. Assume β = 100, VBE(on) = 0.7V.
VCC = 10V IB ≠ 0 ⇒ VB ≠ 0
KVL: VEE + IE RE +VBE(on)+ IB RB = 0
But I E = ( β + 1) I B
− V EE − V BE ( on )
10 − 0.7
IB =
=
= 8.38µA
(β + 1)RE + RB (101)(10k ) + 100k
∴ I C = β I B = 0.838mA
I E = ( β + 1) I B = 101(8.38µ ) = 0.846mA
VB = 0 − IBRB = −(8.38µ)(100k) = −0.838V ≠ 0
VE = VB – VBE(on) = −0.838 − 0.7 = −1.508V
VC = VCC – ICRC = 10 − (0.838m)(5k) = 5.81V RC = 5kΩ
RB = 100kΩ RE = 10kΩ
VEE = −10V Again, need to check if VC > VB.
Example 4: VCC = 15V Find VB and IE. Assume β = 100, VBE(on) = 0.7V.
RB 2
50k
× 15 = 5V
VB =
VCC =
50k + 100k
R B1 + R B 2
VE = VB – VBE(on) = 5 − 0.7 = 4.3V
V − 0 4 .3
IE = E
=
= 1.433mA
RE
3k
I
1.433m
IB = E =
= 14.19 µA
β +1
101
Since IB ≠ 0, more current will be flowing through RB1
and VB should be lower than the above calculated value.
Therefore, the above calculation is wrong. RC = 5kΩ RB1 = 100kΩ RB2 =50kΩ RE = 3kΩ Correct calculation:
VCC = 15V
RB1 = 100kΩ VCC = 15V
RC = 5kΩ RBB RC = 5kΩ where V BB = RB 2
VCC
R B1 + RB 2 =
RB2 =50kΩ Last updated: 8Sep03 50k
× 15 = 5V
50k + 100k RBB = RB1//RB2
RE = 3kΩ RE = 3kΩ = 100k//50k = 33.3kΩ 6 ELEC202 – Electronic Circuits II BJT Now use KVL: 0 + IE RE + VBE(on) + IB RBB = VBB
VBB − VBE ( on )
IB =
........(*)
(β + 1)RE + RBB
5 − 0.7
= 12.78µA
(101)(3k ) + 33.3k
I E = ( β + 1) I B = 101(12.78µA) = 1.291mA
VB = VBB – IB RBB = 5 − (12.78µ)(33.3k) = 4.57V, not 5V
= % error in the initial calculation:
1.433m − 1.291m
error in IE =
≈ 11%
1.291m
5 − 4.57
error in VB =
≈ 9%
4.57
In fact, from (*), the initial calculation is correct if (β+1) RE >> RB or (β+1) RE >> RB1, RB2.
However, RB1 & RB2 cannot be too small; otherwise, lots of energy will be lost in the “biasing
resistors” RB1 & RB2. Note: Example 5: Find the voltages in all nodes & current in all branches. Assume β = 100, VBE(on)  = 0.7V.
Similar to Example 4, we can find
IB1 = 12.78µA, IE1 = 1.291mA, and
IC1 = βIB1 = 1.278mA VCC = 15V Now there is a problem. We do not know
the voltage of VC1 as the current IRC1 ≠ IC1. IRC1
RB1 = 100kΩ In fact, IC1 = IRC1 + IB2.
So, how are we going to find out the voltage
of VC1?
Solution 1: Use Algebra
1 VCC − VE 2
I
⋅
I B2 = E 2 =
β +1 β +1
RE 2
= VCC − (VC1 + VEB ( on ) ) (β + 1)RE 2 = Q1 RE2 = 2kΩ
RC1 = 5kΩ
IE2
IB2
Q2
IC1
RC2 = 2.7kΩ RB2 =50kΩ RE1 = 3kΩ 15 − Vc1 − 0.7
.......(1)
101(2k ) But VC1 = VCC – IRC1RC1= VCC – (IC1 − IB2 ) RC1 ……………..(2)
Sub (1) into (2), VC1 = 15 − (1.278m − 14.3 − VC1
) ⋅ 5k
202k 5k 5k 1 +
VC1 = 15 − (1.278m)(5k ) + 14.3 ⋅
202k 202k ⇒ VC1 = 8.75V Once you fix VC1, you can find the rest. Last updated: 8Sep03 7 ELEC202 – Electronic Circuits II BJT Solution 2: Use approximation & iteration
Assume IB2 is very small compare with IC1.
VC1 = VCC – IRC1RC1 ≈ VCC – IC1RC1 = 15 − (1.278m)(5k) = 8.61V
VE2 = VC1 + VEB(on) = 8.61 + 0.7 = 9.31V
V − VE 2 15 − 9.31
I
2.85m
= CC
=
= 2.85mA ⇒ I B 2 = E 2 =
= 0.0282mA
2k
β + 1 101
RE 2 Then,
and I E2
∴ I B 2 0.0282m
=
≈ 2.2% error
I C1 1.278m and % error in I E 2 ≈ ⇒ VC1 we calculated has ≈ 2% error 2.85 − 2.775
× 100% ≈ 2.7% .
2.775 To get a more accurate answer, we can use iteration method.
Use the IB2 value in the above calculation.
IRC1 ≈ IC1 – IB2 = 1.278m − 0.0282m = 1.250mA
VC1 = 15 − (1.250m)(5k) = 8.75V ,
∴ VE2 = 8.75 + 0.7 = 9.45V I E2 = 15 − 9.45
= 2.78mA
2k ⇒ I B2 = 2.78m
= 0.0275mA
101 The new IRC1 = IC1 – IB2 = 1.278m − 0.0275m = 1.251mA, which is very close to the one we used in the
1st iteration. So, we conclude that VC1 ≈ 8.75V.
Note that the value is the same as the one calculated by 1st method.
Conclusion:
In DC Analysis, you can assume IB is small and calculate the voltage of each node. Then estimate
the values of IB and see if it is really small compare with other currents. The assumption is
acceptable if IB is less then 12% of the other currents.
Also, if β > 100, we can assume α = 1 or IC = IE in analyzing a big circuit. Graphical Representation
IC ICVBE characteristic IC ICVCE characteristic (ideal) V I C = I S exp BE V T VBE3 >VBE2 for VCB > 0 VBE2 >VBE1 Diode like IV char.
VBE Last updated: 8Sep03 VBE1 >VBE(on)
0 VCE
0.20.3V 8 ELEC202 – Electronic Circuits II BJT IC
Saturation Practical ICVCE characteristic Active VBE4 >VBE3
VBE3 >VBE2
VBE2 >VBE1 IC +
VCE +
VBE VBE1 >VBE(on) − − 0 −VA VA = Early Voltage
 due to the decrease in
effective base width Weff
 typical value 50100V
 depends on the process VCE(sat) VCE n n p
VBE VCE1
Weff1 p n VBE VCE2
Weff2 n VCE2 > VCE1 ⇒ Weff1 >Weff2
Modified Equation
V
I C = I S exp BE
V
T VCE
1 + VA Define ro ≡ output resistance looking into the collector ∂i
≡ C ∂VCE VBE = consta nt VBE I S exp V
T
= VA −1 ro −1 = VA
= ro
IC Note: ro is fixed with a given VBE (or a given IC). Last updated: 8Sep03 9 ELEC202 – Electronic Circuits II BJT Transistor as an Amplifier VCC iC ic RC
vC
iC +
vbe
−
+
VBE
−
Note: t IC VBE
0 vBE vBE = total signal
VBE = lare signal, DC signal
vbe = small signal, AC signal
t Now vBE = VBE – vbe
v V + v be
i C = I S exp BE = I S exp BE
V V
T T v
1 v
= I C 1 + be + be VT 2! VT I
= C
V
T iC 2 1 v + be 3! VT v be = g m v be where V = I S exp BE exp V T 3 + .... gm = IC
VT ≈ I c (1 + v be V
T v = I C exp be V T I v be
) = I C + C v be
V VT T for Vbe << VT = 25.9mV = transconducta nce gm ∝ IC , i.e. the small signal output current ic depends on the large signal DC biasing current IC.
iC ic Short segment of curve such that it is almost a
straight line.
t VBE 0 vBE t ∂i
gm = C ∂VBE Note: In general, we consider vbe < 10mV to be a small
signal.
v v
~5% error in exp be ≈ (1 + be ) for vbe < 10mV
V VT T = slope of the iCvBE characteristic curve at iC = IC.
ic = I C gm increases with the increase of IC. Last updated: 8Sep03 10 ELEC202 – Electronic Circuits II BJT Base Current & Input Resistance at the Base iC iB = = β I C + ic = I B + ib β rπ ≡ Rin looking into the base ≡ Define β rπ = β = g m v be β = v be β ⇒ gm vbe
β
=
ib
gm vbe
β
=
ib
gm I
1mA
gm = C =
≈ 40mA / V
VT 25.9mV If IC =1mA,
If β = 100, rπ =
Since rπ ∝ Note: ic = Base resistance (small signal) gm Eg. ib = where β
gm = rπ 100
= 2.5kΩ
40m 1
1
∝
, if the BJT operates at high current, it has a lower Rin = rπ.
gm IC Emitter Current & Input Resistance at the Emitter IE =
ie =
Define re =
Note: IC α ic α = g m vbe α ⇒ vbe α
=
ie
gm re ≡ Rin looking into the emitter ≡ α
gm
rπ = α
gm ≈ re 1
gm = Emitter resistance (small signal) β
gm , re = α
gm ⇒ β
rπ = α
re ∴ rπ = (β + 1) re i.e . rπ >> re for large β > 100. Voltage Gain vC = VCC – iC RC = VCC – (IC + ic) RC = (VCC – IC RC) – ic RC = VC – ic RC ⇒ vc – ic RC = −gm vbe RC = (−gm RC ) vbe
I
output voltage vc
=
= − g m RC ⇒ AV = − g m RC = − C RC
Voltage gain ≡
input voltage vbe
VT
Note that the voltage gain ∝ I C only, and independent of β.
i.e. independent to the characteristic of the BJT. Last updated: 8Sep03 11 ELEC202 – Electronic Circuits II BJT Smallsignal Equivalent Models VCC VCC
RC RC iC iB vC +
vbe
−
+
VBE
− RC IC
VC IB iE ic
vc ib
+
vbe
− IE
+
VBE
− ie DC Analysis
Large signal model AC Analysis
Small signal model
Note: All DC voltage sources
become AC signal
ground. Hybridπ Smallsignal Model b b c
+
vbe rπ gmvbe rπ ro e rπ =
ro = Tmodel (small signal)
c − re =
ro = e Last updated: 8Sep03 gm =
ro re β gmvbe = gm ib rπ
= βib e gm ro VA
IC
c gmvbe
+
vbe βib I
gm = c
VT − b c ib Ic
VT α
gm
VA
IC αie
ro b
ie re e gmvbe = gm ie re
= (β/rπ )iere
= αie 12 ELEC202 – Electronic Circuits II BJT Example: VCC =10V
RC = 3Ω Find the voltage gain of the amplifier.
Assume β = 100, VT = 25mV, VBE(on) = 0.7V. Vo RBB = 100kΩ ± Vi
VBB =3V VCC =10V
VBB =3V Step 1: Find the DC biasing current IC.
V BB − V BE ( on ) 3 − 0.7
IB =
=
= 23µA
R BB
100k
∴ I C = β I B = 100(23µ ) = 2.3mA Step 2: IC RB = 100kΩ Find out the smallsignal model parameters.
I
2.3m
gm = C =
= 92mA / V
VT 25m β rπ =
re = Step 3: RC gm = 100
= 1.087 kΩ
92m rπ
1.087 k
=
= 10.76Ω
101
β +1 Find out the AC voltage gain using one of the smallsignal model.
RBB = 100kΩ vi +
vbe − vo
rπ gmvbe vbe = RC =
3kΩ rπ
vi
rπ + RBB vo = − g mvbe RC =
⇒ − g m rπ
vi RC
rπ + RBB vo
− g m rπ
=
RC
vi rπ + RBB
1.087k
(92m)(3k )
1.087k + 100k
= −2.97V / V =− Last updated: 8Sep03 13 ELEC202 – Electronic Circuits II BJT Signal Waveforms VCC = 10V VCC = 10V
RC = 3kΩ
vO RBB = 100kΩ VO = VC RBB = 100kΩ with no ac signal
i.e. DC condition vi vI RC = 3kΩ
+
VBE
− VBB = 3V VBB = 3V From previous example, IB = 23µA, IC = 2.3mA
⇒ Vo = VC = VC – ICRC = 10 − (2.3m)(3k) = 3.1V and VBE = 0.7V Include the ac small signal,
vI 4V
ˆ
vi = 1V ˆ
Assume vi = 1V VBB = 3V
2V t vBE 0.711V
ˆ
vbe VBE =0.7V ˆ
vbe = 0.689V
t 0
vO 6.07V ˆ
vo
VO =3.1V
0.3V
0 rπ
ˆ
⋅ vi
rπ + RBB 1.087k
⋅ 1V
1.087k + 100k
= 10.75mV
= ˆ
ˆ
vo = AV vi
= −2.97 ×1V
= −2.97V 0.13V clipping t Two limitations:
1. Linearity: If vbe is too large (> 10mV), the output waveform will be distorted.
2. Saturation: If vO is too small, the BJT will go into saturation.
Note: VCE > VCE(sat) ≈ 0.3V when operates in active region.
The waveform will also be clipped when vO > VCC, i.e. IC ≈ 0 (BJT operates in cutoff region) Last updated: 8Sep03 14 ...
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This note was uploaded on 01/05/2010 for the course EENG 202 taught by Professor P.mok during the Spring '03 term at Arizona.
 Spring '03
 P.Mok
 Transistor

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