[time-nuts] LVPECL logic for dummies (that would be Moi)
ehydra
ehydra at arcor.de
Sun Jan 23 18:06:59 UTC 2011
I think you have chosen the worst output possible.
CMOS is good but better is CML. With CML you generate constant-current
load at the VCC supply. So lower back-induced noise! And suppression of
the symmetrical noise to the next stage. This next input is a
differential transistor stage. CML to CMOS translator chips are
available too.
CML is also more immun to shorts. Good for breadboarding :-)
- Henry
--
ehydra.dyndns.info
Michael Baker schrieb:
> Hello, TimeNutters--
> Part deux of the continuing saga of the SiLabs
> oscillator chip that has no output....
> I did not realize that LVPECL chips were such a
> hassle... Next time, I am going insure that such
> devices I use are CMOS and not LVPECL. This
> particular SiLabs Si595 chip has complementary
> LVPECL output.
> As you may recall, I wired up dead-bug legs
> on the 110MHz Si595 VCXO and hooked the
> circuit up on a breadboard and got no output.
> Sadly, the chip behaved as if it were dead.
> (Sniff....)
> Poking around (in-between bouts of sniffling)
> I came across the following in an app note
> by another manufacturer of LVPECL logic chips:
> Why can't I get any output from an ECL Output Connector,
> and how should I measure it?
>
> It has been shown that that ECL outputs are open emitters. Without
> pull-down resistors, the outputs are turned off and therefore, there is
> no output voltage. Even if the output has an internal pull-down
> resistor,
> it may still not be possible to measure the true output signal either,
> unless the measurement device is impedance-matched to the ECL
> output structure. The reason for this problem is that the internal
> connection between the output ECL device pin and the output connector
> is most likely a "long line," and neither the scope probe nor the high
> impedance scope input represents an impedance match to the ECL
> output structure.
>
> If one was to connect the ECL output directly to a 50 Ohm oscilloscope
> input, there would no output either, because the output emitter will be
> turned off by the ground-referenced 50 Ohm input, even if the output
> has
> a 200 Ohm pull-down resistor. However, AC coupling an ECL output with
> an internal 200 Ohm pull-down resistor to a 50 Ohm input instrument is
> OK
>
> So much for not being able to measure an ECL signal, now we shall
> show how it can be measured using an ECL Terminator.
>
> ECL/PECL output circuits are designed to drive 50 Ohm loads
> terminated into a terminating voltage V[TT]= V[CC]-2 V.
> For ECL, V[CC] = 0 V, and V[TT] = -2 V. For PECL, V[TT] = +3 V.
> If the input of a measurement instrument is made to look just like a
> 50 Ohm/V[TT] termination, then all should be well. In fact, that is
> exactly
> what an ECL or PECL Terminator is.
>
> An ECL Terminator is basically a biased 50 Ohm microwave attenuator.
> The input has an equivalent 50 Ohm/-2 V termination, and the output is
> suitable for driving a ground referenced 50 Ohm input instrument.
> Similarly,
> the input of a PECL Terminator has an equivalent 50 Ohm/3 V
> termination.
> In order to protect sensitive instruments, however, a properly
> designed
> ECL/PECL terminator should have a near ground level output
>
> For measuring a differential ECL output either an instrument with a
> differential input and the proper termination or a differential to
> single-ended converter is required.
>
> Caution! Do not connect the output of a PECL device to an ECL
> terminator
> or to a ground-referenced 50 Ohm input instrument. This could spell
> instant
> disaster for the PECL device or the instrument Although connecting an
> ECL output to a PECL Terminator may not destroy the ECL device, it
> could cause gradual degradation of the output emitter follower, due to
> possible excessive reverse bias voltage developed across the base
> to emitter junction.
> It is also shown that the collectors of the ECL output emitter
> followers are
> connected to V[CC]. When V[CC] is ground, shorting the emitter to
> ground
> merely turns off the emitter follower and no damage will occur.
>
> This is not the case when V[CC] is = +5 V. The transistor output
> current
> is limited only by b times its base current, which is supplied by R[1]
> or
> R[2] connected to V[CC]. In most cases, the device is destroyed
> instantly.
> In fact, connecting a PECL output device to a ground-referenced 50 Ohm
> load often destroys the device instantly as well.
> -----------------------------------------------------------------------
> ------------
>
> Now-- back to the breadboard to see if I can get this ornery LVPECL
> oscillator to show me some output... (next time, I am going to make
> sure
> such chips I use are CMOS !!)
>
> Mike Baker
> Micanopy, FL USA
> _______________________________________________
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