Building Blocks 1 Current Mirrors

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Analogue Building Blocks

Part 1: Current Mirrors

for

Electronic Circuits

http://cktse.eie.polyu.edu.hk/eie304

by

Prof. Michael Tse

September 2004



C.K. Tse: Analogue Building

Blocks

2

A Quick Motivation

Let’s look at a typical operational amplifier circuit. This circuit was presented

to you in BE as a “triangle” having ideally infinite input resistance and

infinitely large voltage gain.

What is actually inside this triangle?

How to design such a circuit?

I will introduce in this course some basic building blocks; later you will find

them useful in constructing many electronic circuits including this magic

“triangle”!

–

+

input 1

input 2

output

Rin = ∞

gain = ∞




Blocks

3

Design Problems:

–

+

input 1

input 2

output

Rin = ∞

gain = ∞

Input:

How can we make sure the inputs draw little current?How can we make sure the amplifier does not read excessive

noise (in the form of common-mode signals)?

Gain:

How can we make sure the gain is high enough?

Output:

How can we make sure the output can deliver current without

being ‘loaded down’?

Differential

amplifier serving asinput

Common-emitter

stage, with mirrorserving as active load

Class A output stage

to ensure low output

resistance.




Blocks

4

Solution Preview

+

–

Mirror/active loads

differential amp

CE gain stage

output stage




Blocks

5

What we are going to study

The first series of lectures in this course will introduce the essential building

blocks of analogue circuits. I will choose three basic types of building blocks

to focus on. They are:

Types Functions

Current mirrors Used as current sources and active loads

Differential Amplifiers/ Provides common-mode rejectioninput stages

Power Amplifiers / Delivers current to load, i.e., reduces output

output stages resistance




Blocks

6

Current Mirrors

Primary purpose: to make a current source

Ideally, a current source maintains a fixed current I

regardless of its load!

i.e., the current is always equal to I even when R

changes.

Questions:How to make such a current source out of transistors?

How to set the current magnitude?

I

R

V+




Blocks

7

Simple Current Mirror

Let’s consider a simple configuration. Obviously, we have

+Vcc

I REF R

bias

Note that this current will not change if R is fixed.

So, logically speaking, if we can “copy” this current

to another path where we wish to make a current

source, then we have effectively created a current

source of value I REF in that path.

Now the problem is HOW TO COPY CURRENT.

Hint: the collector current of a BJT is dependent

on its V BE only.




Blocks

8

Simple Current Mirror

+Vcc

I REF R

Idea: twins!!!

If two identical BJTs are having the

same V BE, they must have the samecollector current.

I ≈ I REF

It does not matter what R L is! R L

Effectively we have a current source.

R L

+Vcc

I ≈ I REF provided the base current is small

compared to I.




Blocks

9

Simple Current Mirror= BJT twins connected back to back

+Vcc

I REF

R

I ≈ I REF

Rload

Rload R

+Vcc

I REF

I ≈ I REF

Use PNP for grounded load




Blocks

10

Main Problem of the Simple Current Mirror

Try it out in the lab! We have I REF equal

Rload15kΩ

+15V

I REF

I ≈ I REF

Take three measurements:

1. When Rload = 15 kΩ, we get I very close

to 0.9533 mA!

2. When Rload = 8 kΩ, we get I about 1.01

mA, slightly (still significantly) greater

than 0.9533 mA!

3. When Rload = 6 kΩ, we get I even bigger!

So, this current mirror is not a very good current source because it does not

stay constant for all loads.

and expect I to be the same as I REF

.




Blocks

11

What is the Problem?Let’s look at the ideal case.

Rload15kΩ

+15V

I REF

I ≈ I REF

Q1 Q2

Q1

0.7V 15V

I C1

0.9533mA

V CE1

BUT, why is I C2 different from I C1 in the real circuit, giving I ≠ I REF?

I C2 I C1

E

C

B

Note: I C1≈ I REF

Q2

They have

samecharacteristic

because they

have the

same VBE.

15V

I C2

0.9533mA

7.37V V CE2

So, I C2 should be same as I C1! Hence, I ≈ I REF.




Blocks

12

The Devil is Early!The BJT characteristic is not flat in active region.

Rload15kΩ

+15V

I REF

I ≈ I REF

Q1 Q2

Q1

0.7V 15V

I C1

0.9533mA

V CE1

Clearly, I will vary with Rload, i.e., I ≠ I REF.

I C2 I C1

E

C

B

Note: I C1≈ I REF

Q2

They have

samecharacteristic

because they

have the

same VBE.

15V

I C2

> 0.9533mA

7V V CE2

So, I C2 cannot be same as I C1, depending on Rload.

Early effect




Blocks

13

How bad?Let’s do some precise calculation. Assume that Q1 and Q2 are identical, and

they have an Early voltage of 100 V.

Q1=Q2

0.7V 15V

I C

I C1 = 0.9533mA

V CE

Early effect

–100V

Early voltage V A

=100V

≈7V

0.7 7

0.9533 I C2?




Blocks

14

Can we fix this problem?Solution to precise current source.

Problem review: V CE for Q1 is fixed, but V CE for

Q2 varies with Rload. Thus, I C2 varies with load.

Logical fix: Make V CE for Q1 and Q2 constant.

But they need not be equal! As long as V CE1 and

V CE2 do not change with load, the current I C2 can

be fixed.

+Vcc

I REFR

Rload

Q1 Q2

Q3

0.7V

0.7V

1.4V1.4V

Let’s fix V CE1 and V CE2 so that they have

a fixed difference, regardless of Rload.

This is the clever Wilson Mirror!




Blocks

15

Wilson Mirror+Vcc

I REF R

Rload

Q1 Q2

Q3

0.7V

0.7V

1.4V1.4V

Now V CE1 = 1.4 V and V CE2 = 0.7 V. So, the

current I C2 will not change even when Rload

changes.

Note that I REF ≈ I C1 and I out = I C3 ≈ I C2. I out

Q1=Q2

0.7V15V

I C

I C2

V CE

1.4V

I C1

Remember that I C1 ≠ I C2, but it does not matter as long as they don’t change. This current

source is still a good current source because once we have designed it to give a certain I out,

it keeps this I out for all Rload (of course, for as long as no transistors saturate!).

I C1 I C2




Blocks

16

Wilson Mirror for Grounded Loads

Rload I REF

Q1 Q2

I C2 I C1

E

C

B

Q3

For grounded loads, we use PNP transistors. The corresponding Wilson

construction is:

I out

Again, we have

I C1 and I C2 fixed, and

I REF ≈ I C1 and I out = I C3 ≈ I C2




Blocks

17

Widlar Mirror (for small current)Suppose we wish to build a current source which has a magnitude much

smaller than the reference I REF.

Idea:Make V BE2 smaller than V BE1 because I C is proportional to exp(V BE/V T), where

V T is thermal voltage (≈ 25 mV at room temp).

Clearly,

or

So, a small difference in V BE can give a large difference in I C.



C.K. Tse: Analogue BuildingBlocks

18

Widlar Mirror (for small current)

+Vcc

I REF

R

I out

Rload

R2

For this circuit, we have

From

we get

Thus, we can find I C2 numerically, which

is much smaller than I C1.

Q1 Q2




19

Widlar Mirror Example

+15V

I REF

14.3kΩ

I out

Rload

R2

Suppose we want I out = 10 µA.

First, we have

Then, using the formula

we can find R2:

Q1 Q2




20

Further ApplicationsCurrent Copier

Rload1

+Vcc

I REF

I REF

Q1 Q2

E

C

B

Rload2

I REF

Q3

Rload3

I REF

Q4

Note: if either one of Q1, Q2, Q3 saturates (e.g., due to disconnected load), its basewill draw current from the common base line, lowering the current supply to the other

normal loads. Can we solve this problem? Hint: supply current to the common base

line.

common base line




21

Further ApplicationsImproved Current Copier

Rload1

+Vcc

I REF

I REF

Q1 Q2

E

C

B

Rload2

I REF

Q3

Rload3

I REF

Q4

common base line




22

Further ApplicationsCurrent Copier/Source with Different Current Ratios

Rload

+Vcc

I REF

2 I REF

Q1 Q2

E

C

BQ3

Rload

+Vcc

I REF

0.5 I REF

Q1 Q2

E

C

BQ3




23

Further ApplicationsActive Loads

Consider the CE amplifier. The gain is gm R L.

If we want a high gain, we need a large R L.

But a large R L would need a large Vcc to

maintain the active operation!

IMPRACTICAL!

+Vcc

R L

vi

vo

Vcc

V CE

I C




24

Further ApplicationsActive Loads

Can we have a “super” load resistor

which has large R L at the operating

point, but can allow small Vcc?

Let’s imagine!

+Vcc

vi

vo

Vcc

V CE

I C

Just the transistor itself!

Q L

where ro is output resistance of Q L

Gain = gmro




25

1. Simple current mirrors2. Problems due to Early effect

3. Wilson mirror

4. Widlar mirror

5. Applications

• Current sources• Current copiers

• Active loads

Summary

We have studied

Building Blocks 1 Current Mirrors

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