[sdiy] HP from LP ? -> Moog 904B

Gordonjcp gordonjcp at gjcp.net
Fri Aug 11 09:25:10 CEST 2017


On Fri, Aug 11, 2017 at 08:07:13AM +0200, Michael Zacherl wrote:
> 
> On 6.Aug 2017, at 14:01 , Andrew Simper <andy at cytomic.com> wrote:
> 
> > In the Moog 904B the transistors form a non-linearity which causes the cutoff to increase. This also happens in an MS20 mk1 high pass filter if you construct in "properly" by swapping the positions of the transistors and capacitors.
> 
> thanks for the elaborate explanation!
> what about a Steiner-Parker filter?
> 

The Steiner-Parker filter is a Sallen-Key filter, with the resistors
made variable.

Take a look here: 

http://sound.whsites.net/articles/active-filters.htm#s3

at the general form of the Sallen-Key lowpass filter, and notice how the
two resistors are identical and the feedback capacitor is exactly twice
that of the capacitor to ground - that gives it a Butterworth response
but by changing values a bit you can tweak the response.

As long as you keep the overall loop gain low, it's a nice stable
predictable filter and it's a very useful tool to have in the box!  For
example, all those little noise toys using PWM outputs from
microcontrollers can be massively improved by sticking a Sallen-Key
filter with a corner frequency of about 16kHz (about half the sample
rate) on the output, compared to the strings of RC or even LC filters
that are used "for simplicity".

Okay but that doesn't answer the question, does it?  Well, let's take a
look at another filter, from the Gakken SX150:

http://gjcp.net/media/sx150.jpg

Okay, the capacitor values are a bit different (mostly a function of the
impedances around the feedback circuit) but do you notice how it has two
transistors instead of resistors, C10 to ground and C9 as part of the
feedback path to the output of the opamp?  Resonance is controlled by
either allowing feedback from the opamp or not - crude, but it does
switch the Q from "quite high" to "very low" ;-)

If the two transistors were perfect and tracked perfectly with base
current, the cutoff would move but the Q would stay the same.  If
they're not exactly equal though, the Q will be reduced.  So what you'd
find is that as you swept the base current of the transistors you'd hit
a point where they matched nicely and you'd have a higher Q - possibly
quite a bit higher - and you might well end up with it breaking into
self-oscillation.

Right, but what about the Steiner-Parker filter, that doesn't use
transistors, that uses diodes, doesn't it?  Well yes, but for much the
same reason - and it does a clever trick, too.

The simple filter in the Gakken SX150 (and the early Korg MS20, Korg
Monotron etc) has an interesting side-effect - the base current is set
by the base-emitter voltage of the two "variable resistor" trannies,
isn't it?  So what happens when we vary the voltage on the emitter?  The
filter cutoff changes!  If you stick a slow enough sawtooth into the
filter you can actually hear the filter cutoff sweeping as the input
voltage changes, giving a very rough and dirty-sounding filter, lots of
lovely grit there.  If you stick a good hot squarewave then as you
increase the level you can actually see the resonance peak split in two
on a spectrum analyser with enough resolution.

https://www.cgs.synth.net/modules/pic/schem_cgs35v1-3_syntha_vcf.gif

Notice that instead of an opamp we have two transistors here but don't
worry about that, and the resonance is set by varying the negative
feedback around the amp thus setting its gain directly.

WTF is going on with all those diodes?

Just look at the left half of the chain of diodes, first.  The lowpass
input comes in, is fed to one end of the chain, then there's a 2n2
capacitor going back to the amplifier output, then another link in the
chain, then a capacitor to ground (via a 1k resistor, so it can be used
as a highpass input but we can ignore that) and off via a 100n coupling
capacitor (we can ignore that too, too big to affect our signal) to the
input of the amp.

Why are there two sides though?  That's the clever bit.  Notice how the
signal is fed into both ends of the chain, and the output is taken off
at the middle?  Let's look at what those two transistors and 2k2
resistors would do to it.

The supply is decoupled by that 390R resistor and 47u capacitor, then
there's a 1k resistor through which some bias current will flow, the
input, and then a pair of 2k2 resistors to the collectors of the
trannies and the ends of the diode chain.  You'll notice the emitters
share a common 3k9 emitter resistor (ignore the 82R resistor and
decoupling cap) - it's a long-tailed pair!  As you increase the current
through the left tranny's base it conducts harder pulling its collector
voltage down, but because current through the whole thing is limited the
collector voltage of the right-hand tranny must rise.  They should end
up rising and falling by the same amount, with the small input signal
being added equally to both so one half of the diode chain will tune
"up" and one will tune "down" as you increase the input voltage.  Since
from an audio perspective they're essentially in parallel, the
differences cancel out.

The problem still remains that if the collector voltages of the
transistors don't *exactly* balance and the diode's response to current
aren't *exactly* matched you'll end up with the Q drifting off as you
tune up and down because the resistance won't vary equally.

Here's a Sallen-Key filter designer site that's very handy:

http://sim.okawa-denshi.jp/en/OPseikiLowkeisan.htm

Try plugging in various values and knocking them "out of balance" to see
the effect it has on Q.

-- 
Gordonjcp MM0YEQ



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