Fully differential amplifiers and transmission lines
By Ron Mancini -- EDN, August 7, 2003
Fully differential amplifiers aim at amplifying high-frequency signals while rejecting the noise that always exists in mixed-signal systems. You wire these amps on pc boards and in cabling schemes such that any injected noise is common-mode rather than single-ended. The amplifier-input stage rejects common-mode noise very well, so having two signal wires in close proximity carrying the signal to the amplifier inputs makes the best case for noise rejection.
At high frequencies, signal wires become transmission lines, and, in many cases, coaxial cable or twisted-wire pairs carry signals to preserve their integrity. The electromagnetic propagation of a high-frequency signal along a cable causes reflections from both ends of the cable. These reflections mix with the signal, causing distortion, potential overload, and noise. RS is the signal-source output impedance, RT is the termination impedance, and ZO is the cable characteristic impedance. When RS=RT=ZO, the reflection coefficient, ρ, becomes zero (see equations 1 and 2), and there are no reflections from the source or the termination.
EQUATION 1
EQUATION 2
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The signal couples into the noninverting input of an op amp, so the amplifier-input resistance is high, and you can neglect it (Figure 1). RS terminates the source in ZO, and RT terminates the cable in ZO; thus, RT=RS. The gain-setting resistors, RF and RG, connect to the inverting-op-amp input, where they have no effect on the termination-resistor calculations. The circuit-gain calculation follows:
EQUATION 3
The termination-resistance calculations of a fully differential circuit are more complicated than those for a single-ended circuit because the gain-setting resistors come into play (Figure 2). The only way to keep these resistors out of the calculations is to buffer the inputs, and that option is intolerable costwise and spacewise. First, the gain-setting resistors, R1 and R3, are in parallel with RT, lowering its effective value. Second, the source impedance can't equal zero; hence, it affects the overall circuit gain. These calculations get complicated; my next column will offer the solution to the termination/gain-setting-resistor problem.
| Author Information |
 Ron Mancini is staff scientist at Texas Instruments. You can reach him at 1-352-569-9401, rmancini@ti.com.
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