Download Current Source Biasing

Current Source Biasing
• Integrated circuits have transistors which are manufactured
simultaneously with the same device parameters (parameters from chip
to chip will vary)
• As a result, different bias techniques are employed than in discrete
designs
• One common technique is current source biasing, which allows the
designer to take advantage of matched devices
• We will begin by looking at some simple current source circuits
• A current source is not a “naturally” occurring device. It can be
simulated by a network of transistors and circuit elements.
The voltage across RE is approximately constant.
∴ IE is held at a constant value
IE =
−VEE − VBE
RE
Problem: For the previous circuits find the bias
values IC and VCE for each transistor
Solution
Assume D1 and D2 forward biases (I1 > IB2)
VB = VEE + 2VF = − 10 V + 2 ( 0.7 V) = − 8.6 V
Using KVL around loop A
2VF = VBE 2 + I E 2 R2
Since VBE 2 ≅ VF
V
0.7 V
IE2 = F =
≅ 3.7 mA
R2 180 Ω
Since
IC ≅ I E
IC1 = IC 2 = 39
. mA
Check that D1 and D2 are forward biased for a worst
case minimum βF = 20
I B2 =
I1 =
IC 2
≅ 019
. mA
βF
VCC − VB 10 V − ( −8.6 V )
=
= 0.37 mA
50 kΩ
R1
. mA)(1kΩ ) = 61
.V
VC1 = VCC − IC1 RC = 10 V − (39
VE 2 = VB − VBE 2 = − 8.6 V − 0.7 V = − 9.3 V
VE1 = VC 2 = 0 V − VBE1 = − 0.7 V
. V − ( −0.7 V) = 6.8 V
VCE1 = VC1 − VE1 = 61
VCE 2 = VC 2 − VE 2 = − 0.7 V − ( −9.3 V) = 8.6 V
Current Mirrors
• Current mirrors also take advantage of matched transistors but require
a minimal number of resistors. They are also well suited for circuits
with more than one stage.
Basic BJT Current Mirror
IA =
VCC − (VCE1 + VEE ) VCC − VBE1 − VEE
=
RA
RA
I A = I REF + I B1 + I B 2
IF β F is large I B1 << I REF
I B 2 << I REF
I REF ≅ I A
Problem
For the following circuit IC3 = 3 mA and
VCE = 5.4 V. Find the quiescent (DC bias)
power dissipated in each transistor.
IC 3 ≅ Io = I REF ≅ I A
IA =
0 V − (VF + VEE )
RA
RA =
−VF − VEE −0.7 V − ( −10 V)
=
= 31
. kΩ
3mA
I REF
VE = − 0.7 V; To achieve VCE = 5.4 V, VC = 4.7 V
VCC − VC = 10 V − 4.7 V = 5.3 V
RC =
5.3 V
= 18
. kΩ
3 mA
The DC power in each transistor is given by:
PQ = I CVCE + I BVBE ≅ ICVCE
PQ1 = (3 mA)( 0.7 V) = 0.2 mW
PQ 2 = (3 mA)[−0.7 V − ( −10 V)] ≅ 28 mW
PQ3 = (3 mA)(5.4 V) = 16 mW
MOSFET Current Mirror
Advantage:
IREF = IA
Gate current is negligible
Widlar Current Source
• The basic current mirror requires that the bias current and reference
current be equal
• The wildar current source sets the mirrored current to a value smaller
than IREF by using an extra resistor
• The widlar current source allows you to establish small bias currents
(µA) without using large resistor values
I REF ≅
VCC − V EE − V BE1
RA
VBE1 = VBE 2 + I E 2 R2
V BE
 VBE η V

η VT
T
I E = I EO  e
− 1 ≅ I EO e




VBE = ηVT ln
IE
I EO
Assuming matched BJTs
ηVT ln
I E1
I
= ηVT ln E1 + I E 2 R2
I EO
I EO
I E1
I E 2 R2 = ηVT ln
IE2
I E1 ≅ I REF
Io =
•
•
I E 2 ≅ Io
ηVT I REF
ln
R2
Io
Equation difficult to solve in closed form. Use
successive iteration or trial and error
When you know the desired Io then IREF can be
found directly
I R 
I REF = I o exp o 2 
 ηVT 
Problem:
Using a widlar current source find the values of
RA and RB that will produce Io = 100 µA. Given
VCC = 10 V, VEE = -10 V, VF = 0.7 V and
η = 1.
Solution:
Select a value of R2 such that
I o R2 ≅ ηVT
To keep exponent from becoming too large
ηVT = 25 mV
Choose
I o R2 = 100 mV
∴ R2 = 1 kΩ
 (100 µA)(1kΩ ) 
I REF = (100 µA ) exp 
= 5.46 mA

0.025 V


I REF =
RA =
VCC − VEE − VF
RA
10 V − ( −10 V) − 0.7 V
= 354
. kΩ
5.46 mA
Wilson Current Source
• Refined Widlar source that can produce IO > IREF
•
The balance between VBE1 and VBE2 is set by the
ratio of R1 to R2
Assuming IC ≅ I E
VBE1 + I REF R1 = VBE 2 + I o R2
ηVT ln
I REF
I
+ I REF R1 = ηVT ln o + Io R2
I EO1
I EO 2
Assuming matched devices
Io =
ηVT I REF
R
ln
+ I REF 1
R2
Io
R2
Small Signal Modeling of Three Terminal Devices
• Incremental signals
• Piecewise linear models
• Incremental circuit models
– BJT
– FET
• Refinements to incremental model
– Output resistance
– Input resistance
– Alternative BJT representation
• Two - port representations
Small Signal Modeling of Three Terminal Devices
• Related to PWL concept in which the V-I characteristics are modeled
by a straight line tangent to the curve at a particular operating point
• With three terminal devices the relationship between the output port
and input port must be taken into account. This generally leads to a
PWL model with a linearly dependent source.
• Circuits containing small signal models can be analyzed using linear
circuit theory under proper conditions
• The terms small-signal and incremental will be used interchangeably
Incremental Signals
• Any transient, periodic or AC fluctuation in a voltage or current
• An incremental signal is small in magnitude compared to the bias
voltages or currents in the circuit
• Incremental signal carries the signal information processed by the
circuit
PWL Models of Three Terminal Devices
• Formation of small signal model begins with PWL model
• PWL model can be applied to three terminal device if the dependency
of the output port is considered
rBE
•
•
∂ vbe
=
∂ ib
VBE , I B
ηVT
=
IB
Model valid only in constant current region
If the circuit in which the BJT is connected
produces a signal as well as a bias component to iB
then:
ic (t ) = β ο ib (t )
•
•
where ic and ib are inc
remental signals and βο is the incremental current
gain
• Since βF is fairly constant it is possible to assume βο = βF in many
cases
• The symbols hFE and hfe are sometimes used instead of βF and βο when
using h-parameter analysis
Incremental Circuit Model
In analyzing the small signal performance of a
circuit it is customary to ignore the DC components
of the model once the bias conditions have been
established.
This can be accomplished by the following
procedure:
1. Find the DC bias point and determine an
appropriate PWL model
2. Set all bias values to zero by setting all DC sources
to zero (including those in the PWL model)
3. Solve the desired variables using linear circuit
theory
4. Superimpose the signal variables onto the
corresponding DC bias voltages and currents to
obtain the total voltage and current values
IB =
VBB − VF 1 V − 0.7 V
=
= 30 µA
RB
10 kΩ
IC = β F I B = (100 )(30 µA ) = 3 mA
VOUT = VCE = VCC − IC RC = 10 V − (3 mA)(1kΩ ) = 7 V
•
Transistor operates in constant current therefore we
can use PWL model developed earlier
rbe =
ηVT 1(0.025)
=
≅ 833 Ω
IB
30 µA
ib =
vs
RB + rbe
vo =
− β ο RC
−100 (1 kΩ )
vs =
v s ≅ − 9.2 v s
RB + rbe
10 kΩ + 0.833 kΩ
− β ο RC
RB + rbe
vo = − β ο ib RC
is the incremental or small - signal voltage gain
VOUT
= VCE
Total
= Bias + Incremental
Voltage
+ vo ( t )
Voltage
Signal
=
7 V − 9.2 v s ( t )
Problem:
For the following circuit find the incremental components of vc and ve.
Note: vs connection does not represent typical
amplifier design.
KVL around the input loop for incremental signal
R1
vs
= ib ( R1 R2 ) + ib rπ + ( β ο + 1) ib RE
R1 + R2
R
v s  1 ( R + R )


1
2 
ib =
( R1 R2 ) + rπ + ( β ο + 1) RE
(β ο + 1) RE  R1 ( R
v
s

1 + R2 )
ve = ( β ο + 1) ib RE =
( R1 R2 ) + rπ + ( β ο + 1) RE
R
β ο RC  1 ( R + R ) v s


1
2 
vc = − ( β ο ib ) RC = −
( R1 R2 ) + rπ + ( β ο + 1) RE
In the limit
R1 R2 << ( β ο + 1) RE
βο +1 ≅ βο
rπ << ( β ο + 1) RE
ve ≅
R1
vs
R1 + R2
vc ≅ −
RC R1
vs
RE R1 + R2
Incremental Model of MOSFET
∂ iD
gm =
∂ v GS
VGS , I D
∂
=
k (VGS − VTR )2
∂ vGS
[
]
V
,I
GS D
Assume constant current operation
VGS − VTR
I 
=  D
 k
1/ 2
gm = 2 k I D
Similar expression can be derived for JFET
= 2 k (VGS − VTR )
• An incremental description for a FET can also be defined for triode
(resistive) region
• It can be shown that the incremental model is as follows
rds =
1
2 k (VGS − VTR )
gm = 2 k VDS
Problem:
(A) Find the small signal componont of VOUT for the
2
following circuitd vs = 0.1 Sin ωt, k = 0.2 mA/V ,
VTR = -2 V.
(B) Find the Thevenin circuit between VOUT and ground
..
The bias values can be found to be
I D ≅ 0.47 mA VGS = − I D RE = − 0.47 V
Applying KVL to output loop
VDS = VDD − I D ( RD + RE ) = 10 V − ( 0.47 mA)(3k) = 8.6 V
•
Since VDS > (VGS −VTR) = 1.53 V the device
operates in the constant current region
The incremental transconductance gm is given by
gm = 2 k (VGS − VTR )
= 2 (0.2 mA / V 2 ) [ −0.47 V − ( −2 V)]
≅ 0.61 mA / V
• The signal component of vOUT can be found by substituting the PWL
model and setting all DC sources to zero
Applying KVL to the output loop
v gs = v s − id RE = v s − ( g m v gs ) RE
v gs =
vs
1 + g m RE
∴ Note feedback limits the fraction of v s that appears
as v gs
v OUT = − id RD = − g m v gs RD
v OUT =
aV =
− g m RD
vs
1 + g m RE
v OUT
− g m RD
− ( 0.61 mA / V)(2 kΩ )
=
=
= − 0.76
vs
1 + g m RE 1+ (0.61mA / V)(1kΩ )
v OUT = aV v s = ( −0.76 )( 01
. sin ωt) = − 0.076 sin ωt
• Since vOUT is computed with no load it represents the incremental open
circuit Thevenin voltage
• The incremental rth can be found by setting vs to zero and applying
vTEST
v gs = g m v gs RE
itest =
rth =
v test
RD
vtest
= RD
itest
can only be satisfied if v gs = 0