early effect & bjt biasing - penn engineeringese319/lecture_notes/lec_4_bjtbias… · dc...
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ESE319 Introduction to Microelectronics
12009 Kenneth R. Laker, update 11Sep12 KRL
Early Effect & BJT Biasing● Early Effect● DC BJT Behavior● DC Biasing the BJT
ESE319 Introduction to Microelectronics
22009 Kenneth R. Laker, update 11Sep12 KRL
VBE
VCE
IC
IC
VCE
Saturation region
0
-VA
VBE1
VBE2
VBE3
VBE4
Forward-Active region
Ideal NPN BJT Transfer Characteristic
Early Effect
ESE319 Introduction to Microelectronics
32009 Kenneth R. Laker, update 11Sep12 KRL
Early Effect - ContinuedCollector voltage has some effect on collector current – it increases slightly with increases in voltage. This phenomenonis called the “Early Effect” and is modeled as a linear increasein total current with increases in vCE:
iC= I S evBE
V T 1 vCE
V A is called the Early voltage and ranges from about 15 V to 150 V.
NMOS transistorn=
1V A
V A=V CE
I CI C
ESE319 Introduction to Microelectronics
42009 Kenneth R. Laker, update 11Sep12 KRL
Early Effect - Continued
-VA
VCE
VBE = ...
VBE = ...
VBE = ...
VBE = ...IC
VCEVBE
IC
Saturation region
Fwd-Active region
15 V ≤ VA ≤ 150 V
Observed by James Early from BTL
slope=1/ ro
ESE319 Introduction to Microelectronics
52009 Kenneth R. Laker, update 11Sep12 KRL
Early Effect - Continued
iC= I S evBE
V T 1 vCE
V A Total (bias+signal) quantities:
iC= I C vBE=V BE vCE=V CE
I C= I S eV BE
V T 1V CE
V A = I C' 1V CE
V A
Consider dc (bias) condition (signal = 0):
Let's call the idealized collector bias current (no Early Effect) I'C, i.e.
I C' = I S e
V BE
V T
iC= I Cic vBE=V BEvbe vCE=V CEvce
ESE319 Introduction to Microelectronics
62009 Kenneth R. Laker, update 11Sep12 KRL
We shall define: ro=V A
I C'
I C= I C'
V CE
ro
I C= I C' 1V CE
V A = I C'
V CE
V A
I C'
Rearranging slightly:
Early Effect - Continued
The dc current due to both VBE
and VCE
is:
MOS transistor
ro=V A
I DI C' = I S e
V BE
V T
I D=12
k n' W
LV GS−V t
2
=> ro = f(V
BE)
ro
ESE319 Introduction to Microelectronics
72009 Kenneth R. Laker, update 11Sep12 KRL
Although the bias current is better modeled by including the Early effect
I C= I C'
V CE
ro
We – almost always – will ignore the second term above in hand calculations and use our ideal expression for the bias current:
I C≈ I C' = I S e
V BE
V T
Early Effect - Continued
ESE319 Introduction to Microelectronics
82009 Kenneth R. Laker, update 11Sep12 KRL
The Early term adds ro to the large signal model:
Early Effect - Continued
IC
IC' = I S evBE /V T
VCEVBE VCEro
V CE= I C−I C' ro
I C=I C'
V CE
ro
VBEro
ESE319 Introduction to Microelectronics
92009 Kenneth R. Laker, update 11Sep12 KRL
For typical operating conditions:V A≈50−100 V.
I C' ≈1mA.
ro=V A
I C' ≈
100 V10−3 A
=100 k
We usually can ignore ro since, in practice, ro is in parallel with other resistors, which are much smaller than . For the time being, you will be specifically told if you must include ro in your circuit analyses and designs.
Early Effect - Continued
100 k
ESE319 Introduction to Microelectronics
102009 Kenneth R. Laker, update 11Sep12 KRL
Simulation Results
ro = ∞ro ≠ ∞
slope = -1/R
C
IC (mA)
VCE
(V)
12
10
8
6
4
2
00 2 4 6 8 10 12
Early Effect
load-line
Note: ro is in parallel with R
C.
V CC
I C=1
RCV CC−V CE
RC
load-line
dictated by circuit
V CE
I C
ESE319 Introduction to Microelectronics
112009 Kenneth R. Laker, update 11Sep12 KRL
Active Mode Conditions
Base-emitter diode forward-biased:
Base-collector diode reverse-biased:
V BE≥0.7 V
V BC=V BE−V CE≤0.5V
−V CE≤0.5−V BE⇒V CE≥0.2V
V CE≥0.2V
Forward-Active (ideal cond.)
VBE
> 0V
BC < 0
iE = i
C + i
B
vCE
= vCB
+ vBE
ESE319 Introduction to Microelectronics
122009 Kenneth R. Laker, update 11Sep12 KRL
Amplifier Biasing GoalsWe wish to set a stable value of IC so that we can apply asignal voltage or signal current to the emitter-base circuit andobtain an amplified (undistorted) version of the signal betweenthe collector and ground.
The transistor cannot saturate during operation, i.e.
vCE0.2V.
And it cannot cut off during operation, i.e.iC0 mA.
ESE319 Introduction to Microelectronics
132009 Kenneth R. Laker, update 11Sep12 KRL
Amplifier DC Bias Problem
iC= I Cic
vBE=V BEvbe
vCE=V CEvce
TimevI = v
BE
vO = v
CE
RC
VCC
iC
vO
VCE
VCC
0
Time
vi = v
be
VBE
Timev
o = v
ce
VCEsat
= 0.2 V
vI
0.5 1.51.0
Q
cutoff saturation
fwdactive
slope = Av
ESE319 Introduction to Microelectronics
142009 Kenneth R. Laker, update 11Sep12 KRL
Amplifier Action ● Base current source: ● A small ac change in base current
results in a large ac collector current ( ).
● This yields a large change in the ac collector voltage v
ce.
● Base voltage source:● A small ac change in base voltage
results in a large change in the ac collector current (ic = IS exp(vbe/VT)).
● This yields a large change in the ac collector v
ce voltage.
ib
vC
iC
iB
Source vBE
VCC
RC
(ac + dc)
vCE
ESE319 Introduction to Microelectronics
152009 Kenneth R. Laker, update 11Sep12 KRL
Voltage Source Input With Collector Load
Solution of the simultaneousequations exists where the twocurves: the exponential (iC,vBE) andthe straight line (iC,vCE) intersect:
iC= I S evBE
V T
iC=V CC−vCE
RC
V CC−vCE
RC=I S e
vBE
V T
Load Line
BJT
Circuit
(ac + dc)
vCE
ESE319 Introduction to Microelectronics
162009 Kenneth R. Laker, update 11Sep12 KRL
Scilab Plot of NPN Characteristic //Calculate and plot npn BJT collector//characteristic using active mode modelVT=0.025;VTinv=1/VsubT;IsubS=1E-14;vCE=0:0.01:10;for vBE=0.58:0.01:0.63 iC=IsubS*exp(VTinv*vBE); plot(vCE,1000*iC); //Current in mA.endVCC=10;Rc=10000;vLoad=0:0.01:10;iLoad=(VCC-vLoad)/Rc;plot(vLoad,1000*iLoad);
ESE319 Introduction to Microelectronics
172009 Kenneth R. Laker, update 11Sep12 KRL
NPN Transistor Load Line
Vce (V.)
Ic (mA.)
0 1 2 3 4 5 6 7 8 9 100.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Plot Outputv BE=0.63V.
v BE=0.62V.
v BE=0.60V.
Load Line
iC=V CC−vCE
RC
V CC=10VRC=10k
vBE=0.04V
vCE≈7V
iC (mA)
vCE
(V) =vC
ESE319 Introduction to Microelectronics
182009 Kenneth R. Laker, update 11Sep12 KRL
Amplifier ActionNote that as vBE varies from about 0.59 V to 0.63 V, vCE varies from about 1 V to 8 V!
A 0.04 V peak-to-peak swing of vBE results in an 7 V peak-to-peak swing in vCE - a voltage-gain ratio of 7/0.04, or about 175.
The input signal has two components: a dc one called the bias voltage, and an ac one called the (small) signal voltage. For proper operation, let:
V BE=V BIAS=vBE MAX vBE MIN /2=0.61Vvbe=vsignal=vBE MAX −vBE MIN /2=0.02 V peak
ESE319 Introduction to Microelectronics
192009 Kenneth R. Laker, update 11Sep12 KRL
Candidate Bias Configurations
Base voltagesource
Base currentsource
Emitter currentsource
ESE319 Introduction to Microelectronics
202009 Kenneth R. Laker, update 11Sep12 KRL
Drive Base With a Base Current Source
Assume: =100
I C= I B=100⋅5⋅10−6
I C=0.5 mA.
For this collector current:
V CE=V CC−RC I C
V CE=10−104⋅0.5⋅10−3=5 V
The transistor is almost right in thecenter of the desired operatingregion!
VCE
IC R
c = 10 kΩ
VCC
= 10 VI
B
I = 5 µAQ1
ESE319 Introduction to Microelectronics
212009 Kenneth R. Laker, update 11Sep12 KRL
Current Bias Beta DependenceUnfortunately, β is often poorly controlled and may easilyvary from 100 to 200. And β is also temperature dependent!
I C=100⋅5⋅10−6=0.5 mA.For β = 100:
The BJT with a VCE = 5 V
For β = 200:I C=200⋅10⋅5−6=1.0 mA.
The BJT is saturated!
Base current source biasing → BIAS POINT IS UNSTABLE.
V CE=10−104⋅0.5⋅10−3=5V V CE=10−104⋅1⋅10−3=0 VV CE=V CC−RC I C
ESE319 Introduction to Microelectronics
222009 Kenneth R. Laker, update 11Sep12 KRL
Drive Base with a Base Voltage Source
For an IC of 0.5 mA:
V BE=V T ln I C
I S
Given: I S=10−14 A
I C=0.5⋅10−3 Aand:
V BE=0.025 ln 0.5⋅1011
V BE=0.025⋅24.635=0.616 V
OK. Apply 0.616 volts to thebase and we have the desired collector current!
Since VCE
= 5 V the transistor is nearly at the center of the desired operating region!I C= I S e
V BE
V T
VCC
= 10 V
Rc = 10 kΩ
VBE
= 0.616 V
Q1
ESE319 Introduction to Microelectronics
232009 Kenneth R. Laker, update 11Sep12 KRL
Voltage Bias IS and VCE DependenceUnfortunately, IS is highly temperature-dependent, doublingfor every 5oC increase in temperature.
If the base-emitter voltage is chosen to give IC = 0.5 mA at 20oC (68oF),
it will be 2x at 25oC and 0.5x at 15oC.
IC is also highly sensitive to VBE. Consider two values IC and 10IC:
10 I C
I C=
I S eV BE10
V T
I S eV BE1
V T
V BE10−V BE1=V T ln 10
V BE10−V BE1=0.025⋅2.3025=0.058V.
Less than a 60 mV change in VBE voltage increases IC by anorder of magnitude (10X). BIAS POINT IS UNSTABLE.
ESE319 Introduction to Microelectronics
242009 Kenneth R. Laker, update 11Sep12 KRL
Emitter Current SourceThis holds collector current close to its desired value since:
I C= I E
Changes in IC due to variations in α in the range determined
by the extremes of β are negligible, i.e.
100200⇒ 100101
200201
⇒0.9900.995
There is considerable variation in base current, however, but this is usually of no consequence.
I B=I E
1⇒
I E
101 I B
I E
201
=
1
ESE319 Introduction to Microelectronics
252009 Kenneth R. Laker, update 11Sep12 KRL
ConclusionBiasing a BJT poses potential large bias stability prob-lems, since its characteristics are highly sensitive to temperature and since its electrical properties (princip-ally β) can vary widely from one device to another!
The next lecture sequence will cover some techniques for stabilizing the BJT bias.