Q8: How to
Drive a Floating Differential ECL Input?
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
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
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
(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
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
respectively –0.8 V and –1.6 V. This translates to a current flow of
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
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
The following Logic Function