Download Diode-connected BJT

Diode-connected BJT
Lecture 17-1
Current Mirrors
• Current sources are created by mirroring currents
• Example: with infinite β , Io = IREF
Diff Amp
VCC
IREF
Io
Lecture 17-2
Current Mirrors
• Example: with finite β
Diff Amp
VCC
IREF
Io
• What is the other reason for IREF ¦ Io?
Lecture 17-3
Output Resistance of Current Source, R
• What is the small signal output resistance of this current source, and why do
we care?
Diff Amp
VCC
IREF
Io
Lecture 17-4
Simple IREF Model
• Select R to establish the required reference current
Diff Amp
VCC
R
IREF
Io
Lecture 17-5
Widlar current source
• For a given Vcc, you need large resistor R values to obtain small current!
• Lagre resistors are expensive, Widlar current source uses smaller resistor in
emitter to reduce achive the same current?
IoRE =VBE1- VBE2
VCC
R
IREF
Io
VBE1=VT ln(IREF/Is)
VBE2=VT ln(Io/Is)
Q1
Q2
VBE1- VBE2=VT ln(IREF/Io)
RE
IoRE =VT ln(IREF/Io)
Lecture 17-6
Widlar current source vs. ordinary current mirror
• Let us say we need Io = 10µA
• Assume that for I - 1mA VBE = 0.7
VCC
R
IREF
Io
VCC
R1
Q1
IREF
Io
Q2
Q1
Q2
RE
IoRE =VT ln(IREF/Io)
Lecture 17-7
Widlar current source - output resistance
• If we neglect R || re1 , base of Q2 is on ac ground
VCC
vox
IREF
R
Io
+
R
Q1
Q2
RE
r e1 v π
-
gm vπ
rπ
ro
RE
• Presence of RE is increases output resistance to (1+gm RE||rπ)ro.
(Read Sec. 6.4 in the textbook!)
Lecture 17-8
Current Steering
• With an IREF established, steer and/or scale the reference value
V+
Io
R
IREF
Io
2Io
Lecture 17-9
Reading IC circuit schematics
• Find a path between + power supply and - power supply which sets the reference
current (very often there is only one even in a large circuit): Only VBE and resistors
are in this path.
• Type of transistor will tell you the expected direction of current: npn - current sink,
pnp - current source.
• Identify current mirror configurations (Widlar, Wilson, etc.) and respective emitter
areas.
• Proceed from the reference current branch an calculate subsequent currents
independently
V+
V+
V+
V+
R1
R
IREF
R2
V+
Lecture 17-10
Beta Dependence
• When “steered” to several points, the Io dependence on β can be a problem
VCC
IREF
VCC
Io
Lecture 17-11
Simple Opamp Example
• First stage is used to reject common mode voltages
• The 2nd diff amp and level shifting stage provide the gain
• The input diff amp also provides the large input resistance
• Why is Q6 designed to be 4x larger than Q3?
+
R4
2300Ω
VCC
15V
R3
3E3Ω
R1
20E3Ω
+
R2
20E3Ω
VEE
-15V
Q7
Q4
Q2
Q1
+
Q5
VI
SIN
Q8
R7
28600Ω
Q9
Q3
4x
Q6
R5
157E2Ω
R6
3E3Ω
Lecture 17-12
Differential Amplifiers: Active Loads
• IC resistors are impractical
• Active loads provide current-source-like loads, hence large small signal gains
VCC
VCC
vout
+
vd
_
I
VEE
Lecture 17-13
Differential Amplifiers: Active Loads
• The output in this example is single-sided, but behaves sort of differentially
• The output is a current, proportional to vd --- transconductance amplifier
VCC
VCC
iout
vout
_
1
+
2
vd
+
iout
Gm
vd
_
I
VEE
• Assuming infinite β , what is the output current when vd = 0 ?
Lecture 17-14
Common Mode Gain
• If all of the parameter values are exactly matched to one another, and β =100,
will there be any common mode gain?
VCC
VCC
iout
vout
I
VEE
• Will there be any dc offset?
Lecture 17-15
Small Signal Gain, Gm
VCC
VCC
iout
+
vd
_
I
VEE
Lecture 17-16
Small Signal Gain, Gm
Lecture 17-17
Transconductance Stage of Opamp Model
• The voltage gain of stage 1 depends on the output impedance of stage 1 and
the input impedance of stage 2
_
1
+
2
vd
–µ
Gm
stage 1
+
vd
_
+1
+
vo
_
stage 2
+
Gmvd
Ro1
Ri2
v o2
_
Lecture 17-18
Transconductance Amplifier Voltage Gain
• Active loads are often designed to maximize Ro
stage 1
+
vd
_
stage 2
+
Gmvd
Ro1
Ri2
v o2
_
Lecture 17-19