Lecture 17-1

Diode-connected BJT

Lecture 17-2

Current Mirrors

• Current sources are created by mirroring currents

• Example: with infinite , Io = IREFβ

IREF

VCC

Io

Diff Amp

Lecture 17-3

Current Mirrors

• Example: with finite β

IREF

VCC

Io

Diff Amp

• What is the other reason for IREF ¦ Io?

Lecture 17-4

Output Resistance of Current Source, R

• What is the small signal output resistance of this current source, and why do we care?

IREF

VCC

Io

Diff Amp

Lecture 17-5

Simple IREF Model

• Select R to establish the required reference current

IREF

VCC

Io

Diff Amp

R

Lecture 17-6

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?

IREF

VCC

IoR

Q1 Q2

VBE1=VT ln(IREF/Is)

VBE1- VBE2=VT ln(IREF/Io)

VBE2=VT ln(Io/Is)

IoRE =VBE1- VBE2

IoRE =VT ln(IREF/Io) RE

Lecture 17-7

Widlar current source vs. ordinary current mirror

IREF

VCC

IoR

Q1 Q2

IoRE =VT ln(IREF/Io) RE

IREF

VCC

IoR1

Q1 Q2

• Let us say we need Io = 10µA

• Assume that for I - 1mA VBE = 0.7

Lecture 17-8

Widlar current source - output resistance

IREF

VCC

IoR

Q1 Q2

RE

• If we neglect R || re1 , base of Q2 is on ac ground

vox

gmvπ

RE

rπR re1 ro

vπ

+

-

• Presence of RE is increases output resistance to (1+gm RE||rπ)ro. (Read Sec. 6.4 in the textbook!)

Lecture 17-9

Current Steering

• With an IREF established, steer and/or scale the reference value

IREF

V+

IoR 2Io

Io

Lecture 17-10

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

IREF

V+

R

V+

R1

V+ V+

R2

V+

Lecture 17-11

Beta Dependence

• When “steered” to several points, the Io dependence on can be a problemβ

IREF

VCC

Io

VCC

Lecture 17-12

Simple Opamp Example

Q1 Q2

R120E3Ω

Q3

Q4 Q5

Q6

Q7

R220E3Ω

+ SINVI

Q9

R728600Ω

Q8

R33E3Ω

R42300Ω

+ 15VVCC

+-15VVEE

R63E3Ω

R5157E2Ω

• 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?

4x

Lecture 17-13

Differential Amplifiers: Active Loads

• IC resistors are impractical

• Active loads provide current-source-like loads, hence large small signal gains

VCC

vd

VCC

vout

I

VEE

+

_

Lecture 17-14

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

vd

VCC

vout

I

VEE

+

_

iout

• Assuming infinite , what is the output current when vd = 0 ?β

vd

1

2+

_

Gm

iout

Lecture 17-15

Common Mode Gain

• If all of the parameter values are exactly matched to one another, and =100, will there be any common mode gain?

β

VCC VCC

vout

I

VEE

iout

• Will there be any dc offset?

Lecture 17-16

Small Signal Gain, Gm

VCC

vd

VCC

I

VEE

+

_

iout

Lecture 17-17

Small Signal Gain, Gm

Lecture 17-18

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

vo

vd

1

2+

_

+

_

Gm +1µ–

Gmvd Ro1 Ri2

+

_

vo2vd

+

_

stage 1 stage 2

Lecture 17-19

Transconductance Amplifier Voltage Gain

Gmvd Ro1 Ri2

+

_

vo2vd

+

_

stage 1 stage 2

• Active loads are often designed to maximize Ro