[sdiy] Filter Structure Question

[Scott Gravenhorst]

Magnus Danielson <cfmd at bredband.net> wrote:
>From: Scott Gravenhorst <music.maker at gte.net>
>Subject: Re: [sdiy] Filter Structure Question
>Date: Fri, 16 Jul 2004 13:15:50 -0700
>Message-ID: <200407162015.i6GKFoZ11026 at linux6.lan>
>
>Scott,
>
>> Pictures being worth many more than a thousands words, here is what I have envisioned:
>> 
>> http://home1.gte.net/res0658s/CMOS_SVFb.gif
>> 
>> The 4007 power pins are both open.  All of the inverters come from the same 4069UB
>> package.  The ground symbol is the signal and CV ground, whereas the 0volts and
>> 5volts terminals are just the regulator outputs.  0volts is not connected to
>> ground.  Resistor values are SWAGs (SWAG = scientific wild ass guess).  From what I
>> read below, you would suggest 100 ohm as a SWAG for each power supply feed resistor.
>> 
>> Am I close?  Or are you talking about a more specialized power supply system?
>
>Yes, you are bloody darn close!
>
>First of all, I where considering +5V/-5V supply. When driving the ground like
>that you need the negative side, and also, the signal will go there.

Thinking out loud (so to speak, since I'm writing...):
I'm not at all sure why we need the -5v supply, or the normal ground junction between the
two supplies.  I would think you would ignore that ground and wind up with a 10volt
supply, the inverter would provide that actual signal ground and you'd see approx. +/- 5
volts to each rail with respect to the signal ground created by the inverter.

With what I drew, if you put your DVM black probe to the signal ground and measured
either rail, you'd see a bipolar (approx.) +/- 2.5v supply.  0v and 5v were notations
indicating how to connect a single regulator to the circuit.

>
>Then, beware that the MOSFETs only act properly for one current-direction, so
>you need the compelements, so you should be using the spare inverter to invert
>hthe CV signal accordingly. 

Ah, the P transistors should be paralleled and the gates tied together then.

>Then, you have marked the HP output as output, so this is a HP filter as it is
>drawn, but it should be simple enought for you to draw the other outputs 
>aswell.

Hmm, I don't get this.  According to my "Active-Filter Cookbook" (Lancaster) pp. 177, LP
comes from the output of the right most integrator, the HP output comes from the output
of the DC summer (leftmost inverter used as a linear amp) and BP comes from the output of
the middle iverter used as an integrator.

What am I missing?

>Also beware that the drive-levels for the MOSFET VCRs can't bet that great.
>Not the usual +/- 5V or more as normally seen. 

I'll have to see what happens here, I assume you're talking about parasitic diode
conduction, but from my studying of the internal schematic of the 4007, this won't happen
if pins 14 and 7 are left open.  Or perhaps that schematic isn't showing everything?


>
>You might consider some diode-limiters in the feedback path. Then again, you
>might not. ;O)

We shall see, if I can get this thing straightened out schematic-wise, I will fart about
with it and see if they are needed.  If the distortion sounds good without, then out they
stay!


>Cheers,
>Magnus
>

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-- Scott Gravenhorst | LegoManiac / Lego Trains / RIS 1.5
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