Friday, April 01, 2016

The worlds cheapest VCA?

This post is just about why the key switches use a negative voltage.

It's not vitally important to the whole project, but I wanted to understand the whole keying circuit and satisfy myself as to why it works that particular way. As I said in an earlier post I was looking for a better understanding of the circuits to help with the key switching problem.

Why are we forced to think about switching a negative voltages anyway?
The answer lies in way the simple square-waves are transformed into saw-tooth waves and the switching is woven into that circuit. Understanding this leads to the reasons for all this below 0 volts brow-furrowing.

The worlds cheapest VCA?

“I have to find the edge of the envelope and put my stamp on it.”

(Terry Pratchett, Raising Steam)

Your actual schematic of the keying drive

Each note (referred to from here on in as tone) has it's own amplitude envelope and this is what makes the SS30 almost polyphonic.  Each key is a switch to start and end an envelope and the envelope is what drives the amplitude of the tones that you hear. When each key is pressed down it's envelope goes through an attack phase, where it increases in amplitude, then a sustain phase when it's held and finally a release phase where the amplitude decreases. The odd thing about the SS30 circuit is that this envelope goes from -7V up to 0V and back down to -7 V.

Trace of envelope from simulated keying drive circuit

Let's be positive!

If we had an envelope generator that went from ground to positive - which would be more normal - how would we put that together with an amplifier to control the amplitude of the tones? I would probably reach for an op-amp.  I'd use the envelope as the control voltage in a simple VCA (voltage controlled amplifier) circuit. This machine was being designed in the mid-seventies though. Maybe op-amps were more expensive then. In which case I'd go for a transistor based design. But what if my boss is already worried about the costs of this thing and transistors tend to demand biasing and all kinds of ancillary components to get a good result. Can I come up with something better cheaper?

Negativity wins the day

There are lots of solutions to the problem which don't use op-amps or transistors but the one actually used is pretty neat - albeit one that introduces those fiddly negative voltages.

The solution lies in AC coupling, DC level-shifting and half-wave rectification.

The square-wave output of the frequency divider chips is AC coupled through a capacitor, the DC level of the signal is then shifted by the envelope and passed through a diode. The following attempts to explain why that's a good idea. 

Conscious coupling

Any AC signal fed through a capacitor will be output on the other side DC-balanced to whatever reference you choose, no matter what the DC offset it went in with.  This is called AC coupling or capacitive coupling. If the output side of the capacitor is tied to ground, or nothing, the signal will swing equally both positively and negatively. The signal's current simply alternates back and forth and with no other reference than ground will centre on 0V. It's DC balanced with equal positive and negative current alternation.
If you introduce a reference voltage though it will be shifted to centre on whatever voltage it's tied relative to. If you tie the other side of the coupling capacitor to a negative voltage, like -7 V,  you will get a signal which is centred on that voltage. You are adding the AC component to the DC component.

Shift up!

This is exactly what happens in the tone switching on the SS30. The square-wave output from the divider chips is passed through a coupling capacitor. The output of that capacitor output is then tied to the envelope generator output thus adding them together.
When the key is down, the envelope is open, the coupling output is tied to 0V and the sum of the voltages centres the signal around 0V. In the trace below you can see the effect of this.

AC coupling - tied to ground

The lower trace is the source square-wave from the divider chip. It is 9Vpp (peak-to-peak) and -4.5 V offset from ground.
The upper trace is the output on the other side of the coupling capacitor, which is summed with the  0V from the envelope. As you can see it's oscillating around 0V.

You will also have noticed that the wave has lost it's square shape. This trace is from a simulation and I'm not going to get into the whys and wherefores of that shape here.

The next trace shows what you get if the output is tied to -7V

AC coupling tied to -7V

In this figure the lower trace is still the pre-coupling signal, all below the ground rail.
The upper trace is the signal summed with the -7V from the envelope.

The effect of the envelope on the signal is to shift it's DC component either completely below or partially above the 0V. That's all very nice, but the signal is never attenuated or 'off'. The amplitude doesn't change at all in fact. What's the purpose of that shift?

Give us a (half) wave!

If you send a DC-balanced square wave through a forward biased diode you still get a square wave on the other side. The difference is that you only get the positive half because current only flows in one direction through a diode. This is called a half-wave rectifier as only half of the wave makes it to the other side of the diode.

By now you may have worked out that when the key switch is off, and the envelope signal is -7V, the tone will not get through the diode at all. And furthermore, when the switch is on and the envelope rises to 0V, a half-wave signal will make it through the diode.

The traces below show this half-wave rectification of our shifted signal.

Square wave coupled and rectified

The lower traces are as shown before and the top trace shows the half-wave rectified signal.

(Again, this is based on the sloping topped wave of my simulation. The shape depends on the value of the capacitor and I'll deal with that another time.) 

Putting this all together, we have a kind a amplifier controlled by the envelope for the price of a capacitor and a diode. Ker-ching! If you don't care about the shape of your signal too much then this is a very cheap way of creating a VCA. The cheapest possible I might suggest. 

Arise Circuitry!

This is a grossly simplified and cut-down diagram of the circuits which are employed for each tone. There's more to follow that diode but this is the essence of the shift and rectifier which makes up the quasi-VCA.

Simplified tone switch example
Vss is -7 volts and the square wave is as described above.

You gotta accentuate the negative(?)

That's nearly it, but there's is one more, unexplained, thing. If the output of the divider chips is passed through the decoupling capacitor then why are they negative going to start with? As I noted above it doesn't matter what DC offset you start with; the coupling capacitor balances the signal to whatever reference you choose on the other side. So, why not start with a positive going square-wave?

Perhaps it's because you need fewer power rails. The amount of wiring in the SS30 is already extreme so limiting the power cables around the place is probably a good idea. The G boards use a single -26V supply with on board regulators to step it down.

Progress update

Eurorack DIY is still dominating my time but I'm learning some relevant and interesting things and it will all help with the SS30-M project.

I'm also enjoying some imperial phase Ultravox on BBC4's Top of The Pops repeats from 1981 at the moment. The SS30 has been on stage for Vienna and All Stood Still so far.

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