Hmmm - beware that the diode may well have an effective internal
resistance of more than 0.05 ohm, in which case, the majority of the
current will flow through your shunt even when the diode is present
during the fault condition. I don't think that there is any
alternative for protection here other than to make your overcurrent
detection faster or add enough mass so that the temperature will not
rise too much before overcurrent is detected and shut-down. By the way
- how are you shutting-down this current when there is a fault?

Schottky diodes are fundamentally different than normal PN junction
diodes. They can have significant leakage current - I know that in the
reverse-bias condition, at high temperature, they can pass 10s or 100s
of microAmps. I would expect even worse in the slightly forward-biased
condition. These current levels, though, do not seem relevant when
your sense resistor is only 0.05 ohm.

I searched for "Schottky diode iv curve" and it seems that they follow
the same exponential diode equation as PN junction diodes, although
with different values for the constants. This means that the forward
current is roughly exponentially-dependent on the forward voltage, so
that a given incremental increase in the forward voltage causes a
multiplicative increase in the forward current. Let's say, for
example, that a particular Schottky diode has a forward drop of 0.2V
at 1mA. Then, at 0.15V it might be 30 microAmps, and then at 0.1V it
might be 1 microAmp (a factor of 30 for every 0.05V change in the
voltage). This relationship no longer holds true, though, once the
current is around the "saturation current", which is highly
temperature dependent.

Sean


On Fri, Dec 16, 2011 at 10:45 PM, Forrest Christian <forrestc@imach.com> wr=
ote:
> So I'm looking for ideas, and information.
>
> I have 0.05 ohm shunt resistor which I need to protect from overcurrent
> (effectively overvoltage). =A0 It's measuring steady-state current, and
> also being used for overcurrent trip purposes. =A0In certain
> very-high-current 'dead short' situations, this poor component becomes
> less of a resistor and more of a fuse - I'd like to prevent that,
> bypassing some of the current long enough for the trip to occur - a few
> ms at most.
>
> The obvious cheap solution here is to do some sort of tvss-like device
> around the unit to bypass some of the current once the voltage across
> the resistor rises to a certain point. =A0 And looking at the curves of
> the B1100 Schottky diode which is a standard part around here, it looks
> like I might be able to just put one forward-biased across the resistor,
> and use it in sort of tvss-mode...
>
> My mental picture of a diode is that below Vf there is minimal if any
> current conducted. =A0Is this an accurate mental picture? =A0 What curren=
t
> will flow if you 'bias' a schottky diode at say 0.2V? =A0I'm not worried
> about capacitance more than what I'd call 'leakage current' (in the
> forward direction though).
>
> Is there something else I should be using instead?
>
> -forrest
>
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