| FAQs > ECL/PECL > Answer 8 | ||
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Q8: How to Drive a Floating Differential ECL Input? A8: Case I: The drivers are a pair of differential ECL devices. This type of input circuit can be represented by the configuration shown in Fig. 2B or 2C. Because it has an equivalent floating resistor, no current path exists for the driver open NPN emitters. Therefore, only differential output circuits with the "appropriate" pull-down resistors can drive this type of load. A pair of 300 Ω pull-down resistors can handle either 124 Ω or 75 Ω floating input.
Next, let's see what we mean by "appropriate." Let's take the input configuration shown in Fig. 2B and connect it to a pair of differential output emitter followers including pull-down resistors, R5 and R6, as shown in Fig. 4. Lets assume the "hi" and "lo" logic levels to be –0.8 V and –1.6 V, respectively, a lightly loaded condition. The difference of these voltages, which is the logic swing of 0.8 V, appears across the 124 Ω resistor, resulting in a current flow of 6.45 mA. This current must be sourced by the emitter follower in the "hi" state and sunk by the emitter follower in the "lo" state.
Knowing this, we can easily calculate the appropriate value for R5 and R6. As we mentioned above, the NPN emitter cannot sink current. Therefore, the current in the pull-down resistor, say R6, in the "lo" state must be made larger than 6.45 mA; otherwise, the emitter follower will be cut off before it reached the intended "lo" state. This requirement sets the condition that (5.2 V-1.6 V)/R6³ 6.45 mA. Solving for R6, it is seen that it must be less than 558 Ω. In another words, unless the output emitter followers have internal pull-down resistors with values less than 558 Ω, the drivers cannot drive this differential load properly. Chances are that the circuit will still work if the resistor value is higher than 558 Ω, but the frequency response and noise margin will suffer, because the "lo" level will not attain its full value of –1.6 V. Now let's look at the configuration in Fig. 2C; the resistors are assumed to be each 50 Ω. It is easily seen that this is almost identical to the case just described. The pull-down resistors need to be less then 450 Ω instead.
The next logical question to be asked is "What does one do if the ECL outputs are not provided with pull-down resistors and the load is a differential floating resistor?" One option is to add pull-down resistors at the output emitter followers, if one has access to the circuit. If not, then an ECL buffer circuit with a 50 Ω/-2 V input terminations and differential outputs with 200 Ω pull-down resistors would be needed( See PRL-430A data sheet ). Adding Pull-down resistors somewhere in between the open emitter outputs and the floating resistor input should be avoided, because this would cause an impedance mismatch. As discussed in the first example, a cable with a Characteristic Impedance matching the load termination must also be used. For this example, a 124 Ω shielded twisted pair with the proper connectors would be required. Case II: The Drivers are the complementary outputs from a 50 Ω-output generator. Since the floating resistor in a differential input configuration is not necessarily 50 Ω, one would need to use the equivalent circuit shown in Fig.6 to determine the proper pulse generator output level setting.
For example, let's assume the floating load resistor RL is 124 Ω. If the pulse generator open circuit output Hi and Lo levels are assumed to be E1 and E2, respectively, then the required voltage at V1 and V2 are, respectively –0.8 V and –1.6 V. This translates to a current flow of 6.45 mA through all three resistors. Working backwards, E1 should be set to –0.8 V + 6.45 mA x 0.05 kΩ =-0.478 V and E2 should be set to –1.6 V- 6.45 mA x 0.05 kΩ = -1.92 V. In this differential input configuration, the interconnecting cable can be either 50 Ω or 124 Ω depending on the choice of the interfacing connector arrangement. If a twinax or triax connector is used in the differential input end, then a 50 Ω input to 124 Ω differential output converter, such as the PRL-433, together with a 124 Ω cable can greatly simplify the interface. If the differential input uses two SMA or BNC connectors, then 50 Ω cables should be used. In either case, there is impedance match in one direction only, a topic to be posted later on the web. PRL Products using ECL Circuits: Our line of Logic Level
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