Current Mirror + Darlington

Islamic university –Gaza

Electrical Engineering

Electronic II Lab

Fall-2011iugaza2010.blogspot.com

#1

kZo

Avnl

21

151

kZin

kZo

Avnl

12

22

302

Quiz 1

For the two stage amplifier circuit shown in

figure find

(a) Av no load total.

(b) the value of RL to get a total loaded voltage

gain of 300

150

30* 15*12

1

*

*

)2( )1(12

2

)2()1()(

vnlvnl

oin

in

vnlvloadedtotvnl

AAZZ

Z

AAA

Quiz Sol.

4k RL

30030*2

*15*21

1

*

*

)2(2

)1(12

2

)2()1()(

L

L

vnloL

Lvnl

oin

in

vloadedvloadedtotvloaded

R

R

AZR

RA

ZZZ

AAA

Since Av(load) given is larger

than Av(nl) ,so there is no RL

Part One: Current Mirror

A current mirror is used to replicate the

current from a current source to other locations

Current mirror circuit consists of two back-to-back npn transistors .

The controlling(Ref.) current is the collector current of Q1.

Q1 and Q2 are assumed to be a “matched”

The controlling element is R. RII cR /)7.010(1

Calculations

If the collector and base of QREF are not shorted together, there will not be a path for the base currents to flow, so that Icopy is zero.

Note:

Practical

Build the circuit shown in figure

Measure the value of Iref and Iload

IloadIrefRL

100

1k

2.2k

3.3k

For Rx=3.3k

IloadIrefRx (+470)

100

1k

2.2k

3.3k

10k

For RL=1k

Part Two: Darlington Connection

This configuration gives a much higher current gain

and unity voltage gain.

Calculations

cc B1 B E2 B

E1 B2 E2B1

1 1 1 2

E2 1 2 B1 B1

1 2

V +I R 1.4 I R 0

I I II

I I I

Super Beta

D

D

DC AnalysisPractical

E

C

I =

I =

AC Analysis

vA =