# L5

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```					Current mirror
• Current mirror is a programmable current source
• What is the load current (IC2)?

15V

IC2

IC1

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Twin - matched pair
• Q1 and Q2 are matched pair (TWIN)
– they have exactly the same performance because
they are fabricated on the same chip, at the same
temperature and have the same level of doping

• Since Q1 and Q2 are matched, and have the same VBE
– IC2 = IC1

• We can use the resistor in Q1 to control IC2
– IC2 = IC1 = (15V - 0.6V) / 15kW ~ 1mA

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Current mirror with ratios other than 1:1
• Quiz: What is the current ratio? Why?

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Current mirror with ratios other than 1:1
• Quiz: What is the current ratio? Why?

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Temperature effects
• IC increase with the temperature

– if VBE remains constant,
• IC grows at about 9% / oC

– if IC remains constant,
• VBE falls at 2mV / oC
• a smaller VBE is enough to provide the same IC
because of the higher temperature

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Temperature effects
• The great thing about mirror is that IC2 is not affected
by temperature change !

• If we heat up Q1, will IC1 be changed?
15V-VBE
I C1 =
14.4k
~14.4V/14.4k
~1mA
– VBE can only vary from 0.5 to 0.8V
– IC remains more or less fixed at 1mA

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Negative feedback
• A deeper reason why IC1 is constant
– temperature    => IC1

– IC1           => voltage drop across 15k resistor

=> VBE1

=> IC1

• This feedback mechanism keeps IC1 constant

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Temperature compensation for amplifier
• If VBE is fixed, what would happen if we heat up the
transistors in the following circuits?

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Auto-bias with feedback
• Left-hand circuit (without feedback resistor RE)
– extremely sensitive to temperature change
– IC    at a rate of 9%/oC, bias changed

• Right-hand circuit (with feedback resistor RE)
– temp              , IC            , VE   , VBE   (as VB is fixed),
– sqeezing VBE so that IC

• Another reason why IC is constant
VE VB  VBE VB  0.6
– IC ~                 ~
RE    RE      RE
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• The price to pay
– lower gain ( = -RC/RE) in exchange for stability

• Can we have both stability (constant DC bias) and high
(ac) gain?
– Yes !!

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Requirements of output driver
• High current output
– need large current to drive the mechanical voice coil
of the speaker’s driver
• Voltage gain is not important
– because signal has already been amplified

• An emitter follower is perfect
– VIN = VOUT
– small current in (IB), large current out (IE)
– Power out > power in => a power amplifier

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Output driver circuit
• A driver that produces large current over range +/-15V

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Output driver circuit
• Q1 and Q2 are both emitter followers

• What is the use of the current source in Q1?
– To replace RE

• Problem with RE
– If RE is too small => small input impedance
– If RE is too large => small IC => small output impedance

• Active load (the current source)
– it has  impedance => large input impedance
– You choose large IC so that output impedance is small
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Use a constant current source instead of resistor
• Q2 is a power transistor, needs large base current
– large resistor gives only small current
– ideal constant current source has infinitely large
impedance and can provide large current at the
same time
– Clever !

• Current source is an example of active load
– Used extensively in integrated circuits because it is
easier to fabricate transistors than resistors

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Output driver circuit
• DC offset voltage
– 0.6V (the VBE) for a simple follower

• Zero DC offset voltage in this circuit !
– Q1 (pnp) and Q2 (npn) are opposite type of transistors
– Note the 0.6V voltage drops of VBE in Q1 and Q2
cancels each other out
– So that 0V DC input gives 0V DC output

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Current source and current sink
• Current source
– The ability of giving out current (flow out of the circuit)

• Current sink
– The ability of absorbing current (flow into the circuit)

• Consider Q2
– Range of VE = +/- 15V
– If VE > 0, current flowing out to load (source)
– If VE < 0, current flowing in from load (sink)

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Current source and current sink
• Max source current (when Q2 is fully on)
– If Q2 is fully ON (VCE=0)
• max source current = 15V/8W ~ 2A

• Max sink current (when Q2 is fully off)
– IR = ISINK + IE
– ISINK is max when IE = 0 (zero current from Q2)
– this happens when VE= -15V
– Max ISINK = 15 / 8 A

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Max output power of the amplifier
• Max output voltage swing
• +/- 15V peak-to-peak

• V  V peak sin wt  V peak sin 
2
2   V 2
0    (V peak sin  ) 2            2
V peak
average power  0                       d                            d 
R                 R                      2R
2
Vpeak
• Average power dissipated on a 8W load =                                    14 W
2R

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Root-mean-square (RMS) voltage
V peak
• Define RMS voltage as VRMS         ~ 0.7V peak
2
VRMS
– For ac voltage, the average power = R

•   Hong Kong’s power supply : 240V
– It is a RMS voltage

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Quiescent (rest state, no signal) power dissipation
• At rest state, 0V input gives 0V output

• power dissipated in resistor = (0- -30)2/8 ~ 110W !!

• power dissipated in transistor Q2
– quiescent collector current = 30V/8 = 3.75A
– VCE= 15V, IC = 3.75A,
– power= VCE IC = ~ 56W

• total quiescent power dissipation = 110W+56W
= 166W (HOT !!)

• 14W amplifier gives out 166W heat (inefficient)

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Push-pull emitter follower (Class B amplifier)
• Problem with class A amplifier
– quiescent power dissipation is many times larger
than maximum output power
• push-pull follower

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Class B amplifier
• Q1 (NPN) conducts on positive cycle
• Q2 (PNP) conducts on negative cycle

• 0V input, none of the transistor is conducting
– No current, no power dissipation at Q1 and Q2
– No resistor, no power dissipation

• Pros
– Efficient, no wasted power
• Cons
– Cross-over distortion
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Cross-over distortion
• input have to be greater than 0.6V or less than -0.6V in
order to make Q1 or Q2 conducting
• Output = 0V when input is between +0.6V and -0.6V
– Distortion

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Class AB amplifier
• Use diode to compensate for crossover distortion
• R can be replaced by current source

0.6V

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Amplifiers classification
• Class A : transistors conduct all the time (360o)

• Class B: transistors conduct only half of the time (180o)

• Class AB: transistors conduct more than half of the time,
and less than full time

• Class C:           transistors conduct less than half of the time

Department of Information Engineering                            369
Differential Amplifier
• A differential amplifier has two input terminals
– only amplify the difference in signals
– common signal is NOT amplified

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Differential Amplifier
• The pair of input signals can be separated into a common
part and a differential part
– e.g. (3V, 3.1V) inputs = (0.05V, -0.05V) (differential)
(3.05V, 3.05V) (common)

• Ideal differential amplifier
– Large differential gain
• Amplify the difference in signal
– Zero common-mode gain
• common signal produces zero output

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Long-tail pair

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DC Quiescent point
• 0V input (because base is tied to ground)
– VA ~ -0.6V
– ITAIL = VA  (15)  VA  15 ~ 2mA
RTAIL     RTAIL RTAIL
– ITAILis pretty constant because
• large negative voltage (-15V)
• variation of VA is small in practice
• A simple constant current source
• Can be improved by using a real current source

• By symmetry (same VBE), IC on both transistors is

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Differential gain
• The key in understanding differential gain
– feed a DVIN/2 signal on one input, and a -DVIN/2 on
the other, VA does not move

DVIN/2                          VA
-DVIN/2

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Differential gain
• Because VA is fixed, only need to analyze the part of the
circuit above “A”
– just a common emitter amplifier

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Differential gain
• Common emitter amplifier gain
– VOUT/VIN = -RC/(RE+re)

• In out case, input = -DVIN/2

• Differential gain
RC        ΔVIN
ΔVOUT = -         (-      )
R E +re     2
ΔVOUT      RC
       =
ΔVIN    2(R E +re )

Department of Information Engineering   376
Common mode gain
• Feed a common signal to both inputs, VA goes up
– model the 7.5k tail
by two 15k in parallel

– I = 0 because of symmetry
Vin
– split the circuit in two

• Common mode gain
– GCM= -RC/ (re+RE+2(RTAIL))

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CMRR (Common-mode rejection ratio)
• CMRR = differential gain / common mode gain
2 RTAIL  RE  re

2( RE  re )
RTAIL
                     ( RTAIL  RE   and   re )
RE  re

• Ideal differential amp has
– large differential gain
– small common mode gain

• Therefore CMRR should be as large as possible
– RTAIL should be as large as possible

Department of Information Engineering                                378
CMRR
• Replace RTAILby constant current source

– CMRR ~ 100,000:1 !!
• 100dB !

• dB = 20log 105
= 20*5 = 100

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How to generate composite signal in the lab
• Composite signal = common-mode + differential signal
– use function generator 1 to generate the common-
mode signal (tied to true ground)

– use function generator 2 to generate the differential
signal (floating ground, ground is connected to the
output of generator 1)

– Make sure the ground of generator 2 is floating (not
connected to the main supply)

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Generator 2

Common signal

Generator 1

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Vdiff +Vcommon

Floating ground !

Vcommon                           Vcommon
(from gen 1)

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 views: 8 posted: 10/31/2011 language: English pages: 38