From: Andreas Robinson (sleeper75se_at_yahoo.se)
Date: 2001-12-29 21:28:57
--- In buildcheapeeg_at_yahoogroups.com, "Joerg Hansmann"
<info_at_jhansmann.de> wrote:
> > here's my first and hopefully last design of the
> > input protection stage.
>
> I am not so sure ... ;-)
Well, you are right ... ? :-p
> If I understand Jim Meissners circuit right, it is a
> pure ESD input protection, using antiparallel BC-
> Diodes of two 2N3904 Si-NPN transistors.
Yes, though it clamps overvoltages coming from the
opamps (such as in a short-circuit) as well.
You can easily limit the short-circuit current through
the patient to a few mA regardless of the operating
voltage.
> Perhaps 2 shottkey diodes to the power rails and a
> resistor in series to the output of the opamp ?
That sounds reasonable, so I'll try that, if I can get
the rest (see below) to work.
What you wrote forced me to rethink a bit ... :-)
Ok, about connecting the emitters:
The opamp outputs won't work (as you pointed out)
because the gain is not 1, more like 150. So the
voltage potential there will be quite different from
the input, and the emitter-diodes won't work as
intended.
Perhaps the best place to connect all emitters to
would be the output of the reference input amplifier
(AMP0 in fig3, artikel7) Then the emitter and
collector would see the same potential and it would
always be very close (< 0.1V) to what the base sees,
under normal circumstances.
Short-circuit currents could pass through the
transistors as before and ESD-currents would go to the
reference amplifier.
> Further you use one reference electrode. Do you
> think this could be exchanged with the DRL output ?
Could you please clarify what you mean? The reference
is channel E0 in the biosemi-schematic and if it
fails, the current flowing in or out of the positive
input must go through the same safety circuitry as the
other channels use.
> BTW: Does the circuit really need +-6V or could it
> be operated with -2/3V as the modularEEG ?
It is supposed to be entirely passive. The threshold
voltage is determined by the transistors alone so I
don't see any reason why it should not.
I've attached the new attempt.
I hope this works better.
Regards,
Andreas
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