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'[EE:] low pass filters'
2004\07\29@043926 by

Hi Guys
I came across a low pass filter which is intriguing me:

R1           R2
Vin ----/\/\/\---.---/\/\/\----- Vout
|
|
---
--- C
|
|
GND

to calculate the time constant do you add the two resistors or use R1*C ?

regards
Luis

{Original Message removed}
Kind of hard to understand the ASCII art but is the capacitor halfway
between R1 and R2 or is R2 a variable resistor and C is connected to the
center of the pot?

Luis Moreira wrote:

{Quote hidden}

>{Original Message removed}
> >Hi Guys
> >I came across a low pass filter which is intriguing me:
> >
> >              R1           R2
> >    Vin ----/\/\/\---.---/\/\/\----- Vout
> >                           |
> >                           |
> >                    ---
> >                          --- C
> >                     |
> >                     |
> >                    GND
> >
> >to calculate the time constant do you add the two resistors or use R1*C ?

1.    The response will depend on the input (generator) impedance and load
(output) impedance.
Assuming these both resistive, time constant = Reffective x C

Where Reffective = (Rin x R1)(Rout x R2)/(Rin + R1 + R2 + Rout)

2.    Vout depends on frequency and is

Vout = Vin x [   B x Rout / [  (Rin + R1 + B) x ( Rout + R2) ]  ]

where B = (R2 + Rout) x Xc / (R2 + Rout + Xc)

Where Xc = 1/(2 x Pi x frequency x C)

RM

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{Quote hidden}

My analog skills are a little rusty, but I'd have said the effective
resistance would be:

(R1+Rin || R2+Rout)

Expanding this woud give

(R1+Rin)(R2+Rout)/(Rin + R1 + R2 + Rout)

Mike

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So if we ignore the input and output impedance the time constante is R1 and
R2 in parallell* C ?
regards
Luis

{Original Message removed}
> So if we ignore the input and output impedance the time constante is R1
and
> R2 in parallell* C ?
> regards
>         Luis

No!
If you ignore the input and output impedances the answer is meaningless or
wrong :-(.

If you assume that Ri and Ro are both very low then what you say is true.
This is not ignoring them but making assumptions about them (which may or
may not be true).

In real circuits the usual rough approximation that you can make is that Rin
is zero low (output of a buffer etc) and Rout is infinite or high (input of
an amplifier etc). In such a case the effective components are R1 and C and
the circuit acts like a simple 1 pole RC low pass filter.

In practice R2 is usually added as shown to "impedance match". If Rout is
actually low then adding R2 makes the capacitor 'see' R2 rather than a near
short circuit. The effect on signal levels is as per my previous formula (id
i didn't make an error somewhere)(anyone checked?)

RM

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Thanks Russell
That's what I meant to say ( Rin near zero and Rout is near infinity.
regards
Luis

{Original Message removed}
> > So if we ignore the input and output impedance the time constante is R1
> > and  R2 in parallell* C ?

> That's what I meant to say ( Rin near zero and Rout is near infinity.

Then in that case time constant is R1 x C.
R2 is "waving in the breeze" as it has a high impedance (=Rout) at it's
right hand end.

RM

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>-----Original Message-----
>From: pic microcontroller discussion list
>[PICLISTMITVMA.MIT.EDU] On Behalf Of Russell McMahon
>Sent: 29 July 2004 13:22
>To: PICLISTMITVMA.MIT.EDU
>Subject: Re: [EE:] low pass filters

>In practice R2 is usually added as shown to "impedance match".
>If Rout is actually low then adding R2 makes the capacitor
>'see' R2 rather than a near short circuit. The effect on
>signal levels is as per my previous formula (id i didn't make
>an error somewhere)(anyone checked?)

Yes, see previous post (and quoted in Luis's reply).

Mike

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It's a low cost version of a "tee" filter.

In the real thing, the Rs would be replaced by Ls.
But Ls are relatively expensive, and in many cases not needed.
The resistor on the output side isn't doing much, if the load impedance is relatively high, or not very capacitive.

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