Fundamentals of Linear Electronics Integrated _ Discrete

Document Sample
Fundamentals of Linear Electronics Integrated _ Discrete Powered By Docstoc
					CHAPTER 15


 Special
  ICs
              Objectives

Describe and Analyze:
• Common Mode vs. Differential
• Instrumentation Amps
• Optoisolators
• VCOs & PLLs
• Other Special ICs
               Introduction

• This chapter examines some important op-amp
  related topics such as common-mode rejection.
• It also examines some non op-amp linear circuits
  such as Voltage Controlled Oscillators (VCOs) and
  Phase-Locked Loops (PLLs)
  Single-Ended vs. Differential
A signal applied between an input and ground is called
  a single-ended signal.
A signal applied from one input to the other input is
  called a differential signal.
    Differential Amplifier




Resistances must be symmetric for a diff-amp.
       Common-Mode Signals
• Ground-referenced signals applied simultaneously
  to both inputs of a diff-amp are common-mode
  signals.
• Electrical noise and interference often appear as
  common-mode signals.
• Signals from transducers are usually differential.
• To extract small differential signals out of a “soup” of
  common-mode noise, a diff-amp requires a high
  common-mode rejection ratio (CMRR).
          Definition of CMRR
• The common-mode rejection ratio (CMRR) of a diff-
  amp is defined as:
            CMRR = 20 Log(AV(diff) / AV(cm))
• where AV(diff) is the voltage gain for differential
  signals and AV(cm) is the gain for common-mode
  signals.
• A perfect diff-amp would have AV(cm) equal to zero,
  so it would have infinite CMRR.
• Real diff-amps have CMRRs in the range of 90 dB
  to 110 dB or better.
       Example Calculation 1
• Find the CMRR required so that differential signals
  have a gain of 100 and common-mode signals have
  a gain of 0.001 (an attenuation)
            CMRR = 20 Log(AV(diff) / AV(cm))
                   = 20 Log(100 / 0.001)
                   = 20 Log(100,000)
                   = 20 Log(105)
                   = 20  5
                   = 100 dB
CMRR is less if the external resistors are not matched.
       Example Calculation 2
• A diff-amp has a gain of 10 and a CMRR of 80 dB.
  The input is a differential signal of 1 mV on top of
  1 Volt of common-noise. How much signal voltage,
  and how much noise voltage, will be at the output of
  the diff-amp?
CMRR = 20 Log(AV(diff) / AV(cm))
So AV(cm) = AV(diff) / Log-1(CMRR/20)
          = 10 / Log-1(80/20) = 10 / 104 = 10-3 = 0.001
So at the output there will be 10 mV of signal
                              and 1 mV of noise
   Instrumentation Amps




Except for Ri, all the above can be on one chip.
        Instrumentation Amps

Advantages of instrumentation amplifiers are:
•   Gain set by one resistor
•   High CMRR
•   High Zin on both input pins
•   Work well with most transducers
         Transconductance Amps
• Operational transconductance amplifiers (OTAs) look like other
  op-amps, but the output is a current instead of a voltage.
• Gain is a transconductance (mutual-conductance)
                          gm = iout / Vin
• The value of gm is proportional to a DC bias current:
                            gm = K  IB
• OTAs have relatively wide bandwidth.
• OTAs have high output impedance (Zout).
• The gain control by a current allows one signal to multiply
  another.
           Optoisolators




 An LED and a phototransistor in one package
current cannot pass from one side to the other.
               Optoisolators

Some important parameters:
• Isolation voltage (typically thousands of Volts)
• Current Transfer Ratio (CTR = IC / IF × 100%)
• Speed (how fast can transistor turn on and off)
Voltage-Controlled Oscillators




 Output frequency is proportional to input voltage.
              VCO Applications
Some applications:

•   Frequency modulator
•   Adjustable carrier-oscillator for a radio transmitter
•   Adjustable signal source
•   Analog-to-digital converter
•   Building block for Phase-Locked Loops (PLLs)
Phase-Locked Loops




 Used in communications circuits.
                       PLLs
• The VCO is set to run at a center frequency.
• The VCO output is compared to the input in a phase
  detector circuit. The bigger the phase difference
  between the two frequencies, the higher the voltage
  out of the phase detector.
• The output of the phase detector is fed through a LPF
  and becomes the control signal for the VCO. That
  closes the feedback loop.
• The VCO will eventually “lock on” to the input signal
  and “track” it as the input frequency changes. The
  VCO frequency will match the input frequency.
PLL as an FM Demodulator
PLL Frequency Synthesizer




      f(out) = (n2 / n1 )  fXTAL

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:9
posted:9/9/2011
language:English
pages:20