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From: George Herold on 9 May 2010 17:05
On May 9, 4:55 pm, George Herold <gher...(a)teachspin.com> wrote:
> On May 9, 7:11 am, Nemo <z...(a)nospam.nospam.nospam.nospam.co.uk>
> wrote:
>
>
>
>
>
> > George Herold writes
>
> > >There are three layers.  On the outside is ground, next is
> > >the shield, the center of the coax is the signal line from the
> > >resistor.  Driving the middle shield, reduces the capacitance between
> > >the inner signal line and outer ground.  However it adds the
> > >capacitance from the shield to the inner signal line.  I'll look up
> > >the AD8626.  And try adding a bit of R in the output line.  Do you
> > >have a TI app note number?  What sort of source impedance are you
> > >looking at?  100k looks OK, but 10k ohms shows gain peaking.
>
> > I am looking at a biased photodiode which is, I suppose, very high
> > impedance indeed. It strikes me that my topology is fundamentaly
> > different to yours - I am using two different coaxes, and driving the
> > shields of both with images of the signals on their cores (which is why
> > I use a dual op amp). That's why I assumed there is no capacitance from
> > inner core to outer shield, because I have two independent shields. I
> > then wrap a metal mesh round the pair of them to act as an RFI screen,
> > I'll connect that to 0V. This is all try-it-and-see stuff, I can't say
> > it works yet! Hope to improve leakage (mainly), noise (hopefully) and
> > see if I can push bandwidth up to 1MHz.
>
> > The points about phase inversion etc is very interesting. Eeek, hadn't
> > thought of that! This is going to be fun   8)
> > --
> > Nemo
>
> A photodiode is a whole different ball of wax.  You can look at the
> current and run it into the inverting 'virtual ground' input of an
> opamp.  The impdedance of a photodiode changes with the amount of
> current its generating.  It's kT/e (25mV @RT) divided by the current.
> (I think? I'm sure someone will correct me if I'm wrong.)  Since it's
> a current source you have to put the impedance in parallel with it.
> So at higher currents the impedance goes down.... which is not what
> you want for a current source.
>
> I was thinking I could also look at the current noise from a resistor
> (put it into the inverting input.)  But then how to do the driven
> shield?  Can I put TIA's in series?  like this,
>
>        |\
> i--->--- \
>        |  >--- to shield
>     +--+ /
>     |  |/
>     |
>     |  |\
>     +--- \
>        |  >--- to signal
>   GND--+ /
>        |/
>
> The current noises may add, but with fets and 1Meg ohm you hardly
> care.
>
> George H.- Hide quoted text -
>
> - Show quoted text -

Opps, I forgot the feedback resistors there....

> +--RRRR+
> | |\ |
> i--->+-- \ |
> | >-+- to shield
> +--+ /
> | |/
> +--RRRR-+
> | |\ |
> +--- \ |
> | >-+- to signal
> GND--+ /
> |/

Hmm that doesn't look like it will work... Sorry I should think more
before posting.

George H.
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