[sdiy] Passive filters and impedances

[Donald Tillman]

> On Nov 3, 2015, at 2:38 PM, Tom Wiltshire <tom at electricdruid.net> wrote:
> 
> Ok, you've picked the other approach - Kirchoff's law. I haven't tried that yet, but I'm willing to give it a go. Was there any particular reason you chose to do it that way rather than the simplify-the-impedances method I tried?

Simplify-the-impedances is only useful for very simple circuits.  For anything more difficult, Kirchoff's Current Law is much easier, more straightforward, and can grow arbitrarily complex.


> 
>>  For the two pole circuit, let Va be the voltage at the junction of R1, R2, and C1.  The sum of the currents going into this junction must be zero.
>> 
>> (1/R1)(Vin - Va) - (1/sC1) Va + (1/R2)(Vo - Va) = 0
> 
> This is just the application of Ohm's law with impedances around the junction, right? I get this part.

Yes, the total current into a node should be zero.  That's the voltage difference divided by impedance, for every connection into the node.  The signs can be tricky.


> 
>> Rearrange that into the junction order, by the V's:
>> 
>> (1/R1)Vin + (-1/R1 - 1/R2 - 1/sC1)Va + (1/R2)Vo = 0
> 
> Ok, I see what you've done here, and now you've done it, I see what you meant by "in junction order, by the V's", although I found it unintelligible on it's own!

Yes.  You want to rearrange it into the form of parameters for each V, and they all sum to zero.


> 
>> Next, do the same for junction Vo, and you're well on your way.
> 
> I think I can see how I'd do the same summing-the-currents process around Vo, which will presumably give me another equation somewhat similar to the one you've just derived. But then what?

This first equation relates Vin, Va and Vo.

The second equation relates Va and Vo.  This lets you write Va in terms of Vo.

Substitute that into the first equation and you should be good to go.

  -- Don

--
Don Tillman
Palo Alto, California
don at till.com
http://www.till.com