L1
L2
Vcc---oooo-+-oooo--+
|
|
C1
=
Load
| |
Gnd--------+-------+
In order to attenuate the noise as much as
possable without having
to use excessivly large value components, I
intend to set the filter
at 2500Hz, 2
octaves below the target frequency, which should give
about -39dB of attenuation
((2 octaves*18db/octave)+3db at crossover
frequency), dropping the
noise to below 0.01mV (each -3db represents
a halving of the power of the
signal).
The equations I have for this
filter are as follows:
Rw
L1 = ------------------- *
10e3
1 1/3 * Pi *
Fc
Rw
L2 = ------------------- *
10e3
4 * Pi *
Fc
1
C1 = ------------------------ *
10e6
1 1/2
* Pi * Fc * Rw
Where Rw = impedance of the load, Fc =
Crossover frequency.
(The x 10e? parts simply change the units
to mH and uF)
I have no way to test the impedance of the
inverters inputs,
but the DC resistance measures as 5000
ohms. If this value
is off, I hope it is low, otherwise the
crossover frequency
will be shifted up.
With Fc = 2500 Hz and Rw = 5000 ohms, I
get:
L1 = 477mH
L2 = 160mH
C1 = 0.017uF
I don't know yet if I can find inductors of
that value small
enough to fit into the calculator, or if
the filter will even
work, given that the input impedance of the
inverter could be
much higher than 5000 ohms. I'll find
out answers to both of
those questions tomorrow.
Now, does anyone have any comments or
suggestions on how
I might be able to acomplish this better,
or do none of
you have any experiance with analog
electronics?
Thanks,
DK