[Laser] TX Circuit for high speed data.

[Paul B. Webster VK2BZC]

On Mon, 2007-07-02 at 17:53 -0500, James Whitfield wrote:
> It is true that the voltage drop is not a problem.  The shunt circuit,
> which looks like a regulator but is not used as such for this application,

  A regulator and a modulator are almost indistinguishable (feedback).

> need only "steal" enough current from the laser diode to quench the laser
> action.

  In fact, it will absolutely be not a problem.  The diode drop is at
most 0.7V, and unlike transistors, FETs have no "saturation" voltage,
only a resistance - if you use enough of them, you can get the voltage
pretty much as low as you like.

  The laser threshold on the other hand, is somewhere upward of 1.5V, so
juggling the source resistance would enable you to turn it off quite
effectively - or not.

> It may still have sufficient current flowing through the laser diode for
> it to continue to emit light, just not with the intensity of the laser.

  And if you really want, you should be able to pull down below this
threshold also.

> I do not see how the diode would prevent gate to drain breakdown.

  It doesn't.  Well, actually it does block the effect of an excessive
*positive* excursion of the gate voltage, but perhaps not a negative
one.  The diode on the gate itself would however block this.

  But that is not the point - the concern about such a breakdown is that
the excess current will be fed to the laser diode and overload it; the
diode blocks this possibility, only allowing the FET and associated
components to shunt current *from* the laser.

  Of course, 74HC07s contain input limiting diodes, but it is certainly
good practice in a critical circuit like this, to provide external
diodes limiting gate excursion to the supply rails, so those other two
diodes should be used with a 74HC07 also.

> of course the "real" devices will have some small resistance in in the
> trace conductors and measurable voltage drop across each output FET
> which will vary with load current.

  The FETs have a significant "on" resistance - that's why you want to
parallel them in the first place.  Structurally, about half of that
resistance appears on the source end of the channel, so it effectively
has the balancing resistors "built-in".

  So do bipolar transistors.  Where they differ, is that bipolar
transistors have thermal runaway - the hotter they get, the lower their
threshold voltages and the harder they conduct.  FETs are the opposite,
and tend to self-balance.

> I presume that Paul has more experience than I with modern logic
> devices.

  Well, the above is how I understand it to be - from basic component
theory.

-- 
  Cheers,
    Paul B.