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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
• 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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posted:10/31/2011
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