Terminator III

(Originally published in EDN Magazine, April 27, 2006)
 

My last column showed how the capacitance of a receiver loads an end-termination resistor, preventing the termination from doing its job. I used an isolation resistor (Figure 1), to partially decouple the effects of receiver capacitance from the termination and obtained a modest benefit. For now, ignore components T₁ and L₁. Assume that the external 50Ω end-terminating resistor connects directly to the terminating bias voltage, V_T. The FPGA-circuit model includes a short length of BGA-substrate trace plus the 9-pF capacitance of the die. Isolation resistor R₂ brings into play a fundamental trade-off. Make the value too small, and the receiver capacitance, C_IN, loads down the termination, which results in massive reflections. Increasing the value of R₂ reduces the reflections but at the cost of degrading the signal rise time.

This time, I want to force the apparent termination impedance to equal precisely 50Ω, with minimum degradation of the received-signal rise time.

Using reciprocal impedances accomplishes this trick. In the context of a constant signal impedance, Z₀, every impedance, A(f), has a reciprocal impedance that you can define as B(f)=Z₀²/A(f). For example, in a 50Ω circuit, a 9-pF capacitor and a 22.5-nH inductor have reciprocal impedances.

Assuming that A and B are reciprocal, check the following general relation:
  Z₀=(Z₀+A)||(Z₀+B).

In this equation, the symbol || implies a parallel connection. The proof involves only simple algebra.

The equation says that, if you have any impedance, Z₀+A, you may stabilize that impedance by placing it in parallel with the special impedance, Z₀+B, where B is the reciprocal of A. The impedance of the resulting parallel combination precisely equals Z₀ at all frequencies.

In Figure 1, the input signal strikes two parallel paths. The lower path comprises an impedance, Z₀+A, where Z₀ represents the 50Ω resistor, R₂, and A represents the transmission line and capacitor circuit within the FPGA.

The upper branch presents impedance Z₀+B, where Z₀ represents the external 50Ω termination resistor and B is the combination of T₁ and L₁.

Here is where the magic happens. If T₁ in the upper branch and the BGA trace in the lower branch both share the same characteristic impedance, Z₀, both lines have the same delay, and L₁=Z₀²C_IN, then impedances A and B will always be reciprocal. At that point, you meet the requirements for the application of the equation, so your overall terminating impedance equals precisely 50Ω at all frequencies.

Second-order parasitics associated with the inductor and vias in the design surely affect circuit performance, but, to first order, this beautifully compensated termination works perfectly.

All Publications by Dr. Howard Johnson except as noted.
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