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
• Advantage 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
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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
about 1mA each
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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 )
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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
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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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