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Current mirror

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input have to be greater than 0.6V or less than -0.6V in order to make Q1 or Q2 conducting ... Class B: transistors conduct only half of the time (180o) ... – PowerPoint PPT presentation

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Title: Current mirror


1
Current mirror
  • Current mirror is a programmable current source
  • What is the load current (IC2)?

15V
IC2
IC1
2
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

3
Current mirror with ratios other than 11
  • Quiz What is the current ratio? Why?

4
Current mirror with ratios other than 11
  • Quiz What is the current ratio? Why?

5
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

6
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?
  • VBE can only vary from 0.5 to 0.8V
  • IC remains more or less fixed at 1mA

7
Negative feedback
  • A deeper reason why IC1 is constant
  • temperature gt IC1
  • IC1 gt voltage drop across 15k resistor
  • gt VBE1
  • gt IC1
  • This feedback mechanism keeps IC1 constant

8
Temperature compensation for amplifier
  • If VBE is fixed, what would happen if we heat up
    the transistors in the following circuits?

9
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

10
  • 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 !!

11
Requirements of output driver
  • High current output
  • need large current to drive the mechanical voice
    coil of the speakers 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 gt power in gt a power amplifier

12
Output driver circuit
  • A driver that produces large current over range
    /-15V

13
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 gt small input impedance
  • If RE is too large gt small IC gt small output
    impedance
  • Active load (the current source)
  • it has impedance gt large input impedance
  • You choose large IC so that output impedance is
    small

14
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

15
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

16
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 gt 0, current flowing out to load (source)
  • If VE lt 0, current flowing in from load (sink)

17
Current source and current sink
  • Max source current (when Q2 is fully on)
  • If Q2 is fully ON (VCE0)
  • 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

18
Max output power of the amplifier
  • Max output voltage swing
  • /- 15V peak-to-peak
  • Average power dissipated on a 8W load

19
Root-mean-square (RMS) voltage
  • Define RMS voltage as
  • Advantage
  • For ac voltage, the average power
  • Hong Kongs power supply 240V
  • It is a RMS voltage

20
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 110W56W

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

21
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

22
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

23
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

24
Class AB amplifier
  • Use diode to compensate for crossover distortion
  • R can be replaced by current source

0.6V
25
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

26
Differential Amplifier
  • A differential amplifier has two input terminals
  • only amplify the difference in signals
  • common signal is NOT amplified

27
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

28
Long-tail pair
29
DC Quiescent point
  • 0V input (because base is tied to ground)
  • VA -0.6V
  • ITAIL
  • 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

30
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

VA
DVIN/2
-DVIN/2
31
Differential gain
  • Because VA is fixed, only need to analyze the
    part of the circuit above A
  • just a common emitter amplifier

32
Differential gain
  • Common emitter amplifier gain
  • VOUT/VIN -RC/(REre)
  • In out case, input -DVIN/2
  • Differential gain

33
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
  • split the circuit in two
  • Common mode gain
  • GCM -RC/ (reRE2(RTAIL))

Vin
34
CMRR (Common-mode rejection ratio)
  • CMRR differential gain / common mode gain
  • 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

35
CMRR
  • Replace RTAILby constant current source
  • CMRR 100,0001 !!
  • 100dB !
  • dB 20log 105
  • 205 100

36
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)

37
Generator 2
Common signal
Generator 1
38
Vdiff Vcommon
Floating ground !
Vcommon
Vcommon (from gen 1)
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