MF-101 Control Voltage question

[haaa_haaa0]

What is the relationship between the control voltage of a input device (from 0 to +5V) and the resultant cutoff frequency? I can assume that 0V correlates to a 20 Hz cutoff, and +5V correlates to a 12k Hz cutoff; what is the relationship between the two? That is, is it linear, logarithmic, or some odd relationship? Thanks in advance; feel free to question for clarify!

[varice]

The Cutoff CV response is about 1V/octave, which I believe is logarithmic (it is non-linear). But, a 0 to +5V CV input only sweeps the cutoff frequency about half of the range and is added to the freq setting of the Cutoff knob. For example, if you set the Cutoff knob to minimum 20Hz and change Cutoff CV input from 0 to +5V, the filter cutoff frequency will sweep from about 20Hz to the midrange of about 500Hz. If the Cutoff knob is set to 500Hz, the sweep will be from about 500Hz to 12KHz.

Although the manual states that the Cutoff CV input range should be limited to 0 to +5V, I have found that my MF-101 will work with -5 to +5V into the Cutoff CV input. With the Cutoff knob set to midrange, a -5 to +5V signal will sweep the full range from about 20Hz to 12KHz. I discovered this while setting up my MF-101 to be used as an external analog filter module for my digital Nord Modular G2X synth. More info about that here:

http://electro-music.com/forum/topic-12266.html

[haaa_haaa0]

Ah, wonderful! Thank you so much. It'll be easier to just set the cutoff to min and fully control the sweep w/ my controller; whenever I finish it I'll be sure to mention that for anybody wishing to mod it.

[haaa_haaa0]

I just checked, and correct me if I'm wrong, but...isn't 20Hz - 12kHz about a 10 octave range? I have done some maths. I might be totally wonky in what I'm doing, but check it out.

N = Octaves above; F = Frequency
(2^N) x 10 = F Solve for N
N = ((Log F) - 1)/Log 2
Therefore:
N = ((Log 12,000Hz)-1)/Log 2
N = 10.229

If I need to put the full proof up, I'll do so; otherwise I can't be arsed to do so right now
But assuming a linear relationship between Voltage and cutoff octave, wouldn't that make it roughly a 1/2 V per octave?

[unfiltered37]

Gb (0) is 23.12hz and Gb (9) is about 11 839.84 hz, so about 9 octaves. BTW can somebody explain to me what negative voltage is? Is is just a positive voltage with opposite polarity? No one has ever given me a clear answer.

[varice]

I just checked, and correct me if I'm wrong, but...isn't 20Hz - 12kHz about a 10 octave range? I have done some maths. I might be totally wonky in what I'm doing, but check it out.

N = Octaves above; F = Frequency
(2^N) x 10 = F Solve for N
N = ((Log F) - 1)/Log 2
Therefore:
N = ((Log 12,000Hz)-1)/Log 2
N = 10.229

If I need to put the full proof up, I'll do so; otherwise I can't be arsed to do so right now
But assuming a linear relationship between Voltage and cutoff octave, wouldn't that make it roughly a 1/2 V per octave?

Only if you assume that a 0 to 5V CV will sweep the full cutoff range. But, as I stated in my reply, it does not. 0 to 5V only gets about half the range. I found that -5 to +5v (10V change) will (with the Cutoff knob set to midrange).

[Just Me]

It is exponential response. at 1V/Oct.
A negative voltage is a signal below 0. A positive is a signal above 0. A 5v pp sine wave signal would be between -2.5 and +2.5v. You can use an offset mixer to move the "0" point (effectively) so that you get a 0 to +5V or 0 to -5V or anywhere in between. (CP-251 mixer is great for this)
Some modules can also add gain to a signal and make your 5Vpp signal anything from 0pp to whatever it's upper limit is. (Most often 10Vpp in 5U modular synths.) The beauty of negative voltages is to drive a CV the other direction. You could drive the Cutoff downwards instead of upwards from it's set point. Invert the Env Follower and you can use it as a limiter for example.


(edited while a keyboard was available.)

[cliffman]

Gb (0) is 23.12hz and Gb (9) is about 11 839.84 hz, so about 9 octaves. BTW can somebody explain to me what negative voltage is? Is is just a positive voltage with opposite polarity? No one has ever given me a clear answer.

Yes, on the practical level it really is just opposite polarity. For example, you can make a crude +/- 9V supply by wiring one positive and one negative terminal of two 9V batteries to 'ground' and use the remaning two terminals as '-' and '+'. Likewise, a wall-wart style supply can be converted from '+12v' to '-12v' by rewiring the output jack and swapping the leads. (usually)

From what little i understand, it all has to do with electrons and holes, and which bit has the most holes (or electrons). Ask a physicist for more details. Hope this helps.

[varice]

... BTW can somebody explain to me what negative voltage is? Is is just a positive voltage with opposite polarity? No one has ever given me a clear answer.

Voltage is just a unit of measure of the electric potential (charge) between two points. That measurement can be positive or negative relative to what is regarded or specified as the reference point. And the decision as to what was originally specified and accepted in the past as a positive or negative electric potential was arbitrary. It eventually could have been either way

In fact (in our definition of the predominantly “normal” matter universe), electric voltage flow (current) is composed sub-atomic electron particles which also by our definition have a negative charge. So, when an electrically conductive circuit is completed between any two points of different electrical potential, electrons will flow from the negative point to the positive point (the negative point repels the like charged negative electrons which at the same time are also attracted to the oppositely charged positive point).

On the other hand, in a predominantly anti-matter universe, electric current could be composed of the anti-matter equivalent of the sub-atomic electron particles – positrons. And positrons would flow from positive to negative electric potential points – but only if we use the same definition of what is a positively and a negatively charged point in our “normal” matter universe

Oh crap – this might not be a clear answer after all – oh well, I gave my best

[mayidunk]

... BTW can somebody explain to me what negative voltage is? Is is just a positive voltage with opposite polarity? No one has ever given me a clear answer.

Simply put, electricity is made up of current and voltage, current being the flow of electrons in the circuit, and voltage being the force that moves the electrons, causing them to flow through the circuit.

Think of it this way... Current flow is like water flowing in a pipe, and voltage is like the pressure of that water flow. Of course, the amount of pressure will determine how much water will flow past a given point in a given amount of time. Small pressure, little water. Great pressure, lots of water. No pressure, no water. Under positive pressure, the water flows in one direction. Under negative pressure (vacuum), the water flows in the other direction. Same with electricity.

Current is the flow of electrons through a circuit, and they always flow towards the positive pole of a battery. Like varice said earlier, electrons are repulsed by a negative charge (the negative pole of the battery), and are attracted by a positive charge (the positive pole of the battery). How strong the negative/positive charges are will determine the voltage. The greater the negative/positive charges, the more forceful the electrons are repulsed/attracted by the charges, hence the higher the voltage or pressure.

If you connect a battery to a circuit one way, the current will flow in one direction (positive voltage). If you reverse the leads on the battery, the current will flow in the opposite direction (negative voltage)! Hence, negative voltage is just like reversing the leads on the battery.

To simply illustrate the effect of positive versus negative voltage, consider this. If the load on the circuit is a light bulb, then the polarity of the voltage will make no difference at all, since the bulb will still light regardless of which direction the current is flowing. However, if the load on the circuit is an electromagnet, then the result of reversing the voltage will be to reverse the poles on the magnet.

The rest is way over my head! I hope this helps.