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From: George Herold on 19 Mar 2010 01:58
On Mar 18, 8:14 pm, John Fields <jfie...(a)austininstruments.com> wrote:
> On Tue, 16 Mar 2010 13:14:36 -0500, John Fields
>
>
>
>
>
> <jfie...(a)austininstruments.com> wrote:
> >Another way is to use a resistor of a known value in series with the
> >unknown cap or coil:
>
> >                                         +-------+
> >      +-----[10K]----+<----[X10 PROBE]---|VERT   |
> >      |              |                   |       |
> >      |              |                   |       |
> > [FUNCTION GEN]    [DUT]                 | SCOPE |
> >      |              |                   |       |
> >      |              |                   |       |
> >      +--------------+<------------------|GND    |
> >                                         +-------+
>
> >I've gotta go do some chores, but I want to get this off instead of
> >waiting until later to send the whole thing, so I'll finish it when I'm
> >done...
>
> OK.
>
> For the cap, set up your equipment like this:
>
>       E1                                 +-------+
>       |<---------------------------------|VERTA  |
>       |              E2                  |       |
>       +-----[10K]----+<----[X10 PROBE]---|VERTB  |
>       |              |                   |       |
>       |              |                   |       |
>  [FUNCTION GEN]    [DUT]                 | SCOPE |
>       |              |                   |       |
>       |              |                   |       |
>       +--------------+<------------------|GND    |
>                                          +-------+
>
> Using the sine wave output of your function generator, set its output,
> E1, to any convenient level and then tune it to the frequency which
> causes E2 to be 1/2 the voltage of E1.
>
> Then solve:
>
>              1  
>      C = ----------
>           2pi f Xc
>
> where C will be the capacitance in farads,
>       2pi is 6.28,
>       f is the frequency which causes E2 to be 1/2 of E1, in Hz,and
>       Xc is 5800 ohms.
>
> Why 5800 ohms?
>
> That's the capacitive reactance you need in order to get E2 equal to
> half of E1 at any frequency.
>
> Ergo, with a constant reactance, capacitance will go inversely with
> frequency.  
>
> JF- Hide quoted text -
>
> - Show quoted text -

If you've got a 'scope you can also just hit the RC or RL with a
square wave and find the time constant. (The point where the voltage
falls (or rises) by 1/e).

To the OP, If you want to do more than run stero speakers you can also
buy power opamps. I like the OPA544 (I think that's the number.)
( It can do at least 2 amps... more if you don't ask it to dissipate
too much power. Hmm, Just checking digikey and they are more
expensive than I remember ($18 for one) and also out of stock.

George H.
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