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(A V(diff) / A V(cm) ) where A V(diff) is the voltage gain for differential signals and A V(cm) is the gain for common-mode signals. A perfect diff-amp would have A V(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 (an attenuation) CMRR = 20 Log(A V(diff) / A V(cm) ) = 20 Log(100 / 0.001) = 20 Log(100,000) = 20 Log(10 5 ) = 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(A V(diff) / A V(cm) ) So A V(cm) = A V(diff) / Log -1 (CMRR/20) = 10 / Log -1 (80/20) = 10 / 10 4 = = So at the output there will be 10 mV of signal and 1 mV of noise

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

Instrumentation Amps Advantages of instrumentation amplifiers are: Gain set by one resistor High CMRR High Z in 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 = i out / V in The value of gm is proportional to a DC bias current: gm = K  I B OTAs have relatively wide bandwidth. OTAs have high output impedance (Z out ). 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 = I C / I F × 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) = (n 2 / n 1 )  f XTAL