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Double VCF with AS3350 based on Rhodes Chroma - a project
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author Double VCF with AS3350 based on Rhodes Chroma - a project
ixtern
As the AS3350 double VCF is on the production line and I have received two pre-production chips, I came up with an idea to build Rhodes Chroma - based VCF with AS3350 as I think it is a quite interesting thing.
Here is the schematic I have drawn so far - electronic switches were replaced with mechanical, encoders added. This is an idea, not proofed yet, there may be errors in schematic, so beware. (One mistake I've found so far is 74148 encoder - should be 74HC148).
I am curious what is your opinion about that project.

Regards,
ixtern
mskala
I don't understand the levels and gain staging in this circuit. You've got input buffers with gain of -1; with typical Eurorack, those will produce +-5V. The AS3350's inputs seem to be typical transconductance amplifiers expecting input in the range of tens of millivolts. Then on the output you have amplifiers with voltage gain 100. So maybe it's not Eurorack. Is this meant to take input at a very low level and boost it as well as filtering?

In Filter 2 the input buffer op amp is basically shorted to ground at DC, through the 10-ohm resistor R56, but in Filter 1 that DC path is blocked by a capacitor. That looks like a drawing error; were the two filters intended to be the same? What are the 10-ohm resistors meant to do in either case?

Is a rotary switch, a chip to convert it to a digital code, and then a makeshift digital-to-analog converter, really how you want to control resonance? It seems like just using a pot would be cheaper and more playable. If you really want to use a switch and only allow eight quantized steps for the resonance level, then you could just have the switch choose positions along a resistor chain and eliminate the priority encoder chips.

I think the separate digital and analog grounds concept is usually a bad idea anyway, but if you eliminated those two digital chips, then you would no longer have a reason to want to mess with the grounds. Even without eliminating them, the digital chips are not switching any faster than you can turn the rotary-switch knobs, so they aren't going to be producing the kinds of spikes that lead people to try to separate digital and analog grounds.
ixtern
mskala wrote:
I don't understand the levels and gain staging in this circuit. You've got input buffers with gain of -1; with typical Eurorack, those will produce +-5V. The AS3350's inputs seem to be typical transconductance amplifiers expecting input in the range of tens of millivolts. Then on the output you have amplifiers with voltage gain 100. So maybe it's not Eurorack. Is this meant to take input at a very low level and boost it as well as filtering?

In Filter 2 the input buffer op amp is basically shorted to ground at DC, through the 10-ohm resistor R56, but in Filter 1 that DC path is blocked by a capacitor. That looks like a drawing error; were the two filters intended to be the same? What are the 10-ohm resistors meant to do in either case?

Is a rotary switch, a chip to convert it to a digital code, and then a makeshift digital-to-analog converter, really how you want to control resonance? It seems like just using a pot would be cheaper and more playable. If you really want to use a switch and only allow eight quantized steps for the resonance level, then you could just have the switch choose positions along a resistor chain and eliminate the priority encoder chips.

I think the separate digital and analog grounds concept is usually a bad idea anyway, but if you eliminated those two digital chips, then you would no longer have a reason to want to mess with the grounds. Even without eliminating them, the digital chips are not switching any faster than you can turn the rotary-switch knobs, so they aren't going to be producing the kinds of spikes that lead people to try to separate digital and analog grounds.

Thanks for your comments. There are errors in the schematic indeed. Both filters should have the same connections. Basically the signal should be attenuated at 10 Ohm resistors to tenths of mV levels, I have forgotten to draw resistors from audio amplifiers and some other connections are bad... Will fix it and upload again.
As for the switches and encoders it was to include original Chroma circuit as much as I could but it had gone against simplicity perhaps. But it is only first iteration of the project.
ixtern
Changes to v2: fixed signal path, signal divider added, encoder removed - instead set of resistors connected to the resonance switch. +5V regulator added.
mskala
Okay, that makes the schematic easier to understand.

I don't think that the switches will work as you intend because you're just putting different resistances in series with the base of the transistor to generate the resonance control voltage. The emitter-follower transistor is going to try to go all the way to the positive rail on all but the highest of those resistances. My suggestion instead would be to make a chain of resistors - that is, all of them in series from +5V down to 0V (or whatever voltages you decide on), and then have the switch select different points on the chain. That way you are bringing the transistor base to different voltages (which it will then buffer at its emitter) instead of bringing it to 5V in all cases, through different resistances. Think about what's going on in this circuit: the switch is trying to generate an adjustable voltage, and the transistor is acting as a buffer to create a better output impedance at the switch's voltage. I wonder about all the impedances and levels around these control-voltage buffers because they seem to be mixing high and low impedances (e.g. 47k and 100ohms into the same node) but I haven't done the math for how that subcircuit will respond, whether the modulation will really be V/oct, etc.

Your new signal voltage divider design balances a variable resistance of 1k to 2k against 10ohms. That's going to put a lot of load on the input buffer op amps. It also doesn't give you much adjustment range for the input level: only a 2:1 ratio. My suggestion would be to cut out the 1k series resistors (labelled R?) and reduce the feedback resistors on the input buffers (R27 and R28). That way the buffers will have less voltage gain and can do part of the work of reducing the level. Then you can use a less aggressive voltage divider ratio, and larger component values in the voltage divider. With the series resistors removed you'll get more adjustment range on the level.

Your level pots are going to respond in the reverse of what people normally expect from level pots (clockwise -> greater resistance -> lower level), unless you get reverse-log pots or change how they're used in the voltage divider. Maybe a different idea would be to use the level pots as the feedback resistors for the input buffer op amps; then it should be possible to make them respond as people expect, and it'd neatly solve the issue of the high feedback resistances.
ixtern
mskala wrote:
Okay, that makes the schematic easier to understand.

I don't think that the switches will work as you intend because you're just putting different resistances in series with the base of the transistor to generate the resonance control voltage. The emitter-follower transistor is going to try to go all the way to the positive rail on all but the highest of those resistances. My suggestion instead would be to make a chain of resistors - that is, all of them in series from +5V down to 0V (or whatever voltages you decide on), and then have the switch select different points on the chain. That way you are bringing the transistor base to different voltages (which it will then buffer at its emitter) instead of bringing it to 5V in all cases, through different resistances. Think about what's going on in this circuit: the switch is trying to generate an adjustable voltage, and the transistor is acting as a buffer to create a better output impedance at the switch's voltage.

Thanks for your valuable comments.
Let me quote original circuit description from http://www.rhodeschroma.com/?id=circuitdescriptions

"The three resonance control bits from the data latch for each channel are used to encode the setting of the Resonance parameter. When the parameter is 0, the three bits are all logic 1 (5 volts), and when the parameter is 7, the three bits are all at zero. Resistors R47, 49 and 51 (or R48, 50 and 52) combine these in a binary weighted manner. The purpose of the transistor is to provide a resonance boost when the parameter is set to 7. Normally (that is, when the filter isn't being used as an oscillator) at least one of the resonance control bits is high. This turns the transistor on hard, so that the current that flows into the 100 ohm summing resistor R53 (or R54) is thus equal to the sum of the control currents fed into the base of the transistor and the current through the 33k resistor on the collector. As the resonance parameter is increased, less current flows through the base, but the transistor remains saturated. When the resonance parameter makes the final transition from 6 to 7 and all three bits go low, the transistor shuts off and the current through the emitter drops sharply to zero, causing a large increase in Q. This guarantees that all filters will oscillate when set to 7, but not when set to 6. The added 33k resistor (R115 or R116) causes the resonance to be increased slightly at higher frequencies, to overcome a slight reluctance to oscillate."

And here is a original Rhodes Chroma resonance control circuit:
ixtern
mskala wrote:
Okay, that makes the schematic easier to understand.
Your new signal voltage divider design balances a variable resistance of 1k to 2k against 10ohms. That's going to put a lot of load on the input buffer op amps. It also doesn't give you much adjustment range for the input level: only a 2:1 ratio. My suggestion would be to cut out the 1k series resistors (labelled R?) and reduce the feedback resistors on the input buffers (R27 and R28). That way the buffers will have less voltage gain and can do part of the work of reducing the level. Then you can use a less aggressive voltage divider ratio, and larger component values in the voltage divider. With the series resistors removed you'll get more adjustment range on the level.

Your level pots are going to respond in the reverse of what people normally expect from level pots (clockwise -> greater resistance -> lower level), unless you get reverse-log pots or change how they're used in the voltage divider. Maybe a different idea would be to use the level pots as the feedback resistors for the input buffer op amps; then it should be possible to make them respond as people expect, and it'd neatly solve the issue of the high feedback resistances.


You are right that 1kOhm load may be too low for op amp outputs especially if applying full 10V p-p input signal. Lowering feedback resistors is a good idea.
2:1 ratio should be enough. 3350 expects about 50 mV p-p on the input, so 100:1 to 200:1 divide ratio should do (5 to 10V p-p on input) and we have also additional pots to attenuate on the inputs.

I don't understand your objections about divider trimpots (am I corectly understanding? Are you writing about trimpots, not input level pots?). In my opinion there is no expectations about working direction of the trimpots, as they are set once during calibration and forgotten.
mskala
Okay, yes, I missed that those trimmers in the voltage dividers are not the main level controls. But in that case, I'm not sure you need them at all - and if you do need them, a 2:1 control range still is probably not going to be enough.

On the resistors driving Q1: the important difference between the original circuit with resistors connecting to the digital outputs to form a crude DAC, and the simplified circuit with a mechanical switch, is that the digital outputs are driven to 0V or 5V with low impedance. Not only 5V. So when you have a given three-bit code on the digital outputs like say 101, you get a weighted average of 5V, 0V, and 5V, with the resistors determining the weights, and you get a voltage at the transistor base that is different for each code, somewhere between 0V and 5V.

With the mechanical switch circuit as you've drawn it, all the resistors go to 5V. There is no connection to 0V except when the switch is in the one position that has no resistor. So you don't get different voltages; there is always 5V being applied to the transistor, just through different resistances. That's going to make the transistor's behaviour significantly different from what it would be in the original.

If you used a series chain of resistors, this wouldn't happen - different points along the chain would have different voltages. I suppose another way of doing it would be to add a single resistor from the transistor base to 0V. Then your different resistances will form different-ratio voltage dividers with that fixed resistance depending on the switch position, and again you'll have different voltages for the base.
ixtern
mskala wrote:
Okay, yes, I missed that those trimmers in the voltage dividers are not the main level controls. But in that case, I'm not sure you need them at all - and if you do need them, a 2:1 control range still is probably not going to be enough.

On the resistors driving Q1: the important difference between the original circuit with resistors connecting to the digital outputs to form a crude DAC, and the simplified circuit with a mechanical switch, is that the digital outputs are driven to 0V or 5V with low impedance. Not only 5V. So when you have a given three-bit code on the digital outputs like say 101, you get a weighted average of 5V, 0V, and 5V, with the resistors determining the weights, and you get a voltage at the transistor base that is different for each code, somewhere between 0V and 5V.

With the mechanical switch circuit as you've drawn it, all the resistors go to 5V. There is no connection to 0V except when the switch is in the one position that has no resistor. So you don't get different voltages; there is always 5V being applied to the transistor, just through different resistances. That's going to make the transistor's behaviour significantly different from what it would be in the original.

If you used a series chain of resistors, this wouldn't happen - different points along the chain would have different voltages. I suppose another way of doing it would be to add a single resistor from the transistor base to 0V. Then your different resistances will form different-ratio voltage dividers with that fixed resistance depending on the switch position, and again you'll have different voltages for the base.

Thanks again for your comments. But this time I cannot agree with you. Too much focus on voltages when this is current-working circuit. Transistor in modes 1-7 is working in saturation mode and transistor voltage on the base is almost constant so the resistors connections to the 0V are not very important. In quoted circuit description it was mentioned that sum of the base currents is important. Perhaps there is some dependency on the 0V connection but this is crude circuit and I don't think that original designer accounted such dependency. Even my resistor values was taken from E24 series although precise values was little different (taking as base values original ones).
I can always return to the encoder smile I like the idea of discrete Q values as I didn't find very useful continuous Q control by pot in my VCFs. But this may be an (switchable) option in the next versions.
I am also thinking about restoring original multiplexers to have digital control but it is for the future perhaps.
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