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Breath Controller with 3.5mm Jack to make 0 - 5 V Variable voltage (Control Voltage) on 6.5mm Jack

Thread starter #1
I am a beginner and wish to tackle this project...

I have an old Yamaha breath controller device that has a 3.5mm Jack and requires -9v on the tip (the sleeve is positive). It draws a current of approx 7mA to 20mA depending on how hard you blow it (from what I have managed to research but may be wrong). I want to now reimplement this device to control CV (control voltage) on my synthesizer. I would love some help to know how to achieve this... I was thinking to use a 9v battery to power it, then convert the voltage to a variable 0 to 5v so that I can plug it into my Sythesizer CV input (6.5mm Jack) and vary the voltage by blowing... Will a simple resistor in circuit achieve this? How would you go about it?

Attached is the circuit diagram of the Breath Controller in case it helps ;)

Thanks very much!
:joyful:
Nicholas
 

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cowboybob

Well-Known Member
Most Helpful Member
#2
Welcome to ETO, Nicholas D!

Found this site, with this link. Not sure if it's useful. Couldn't find what, exactly, the BC2's output voltages are, although from the schematic I came make a SWAG.

Do you know what the output value(s)?

Might be as simple as using a simple voltage divider circuit at the output of the BC2 to derive a voltage range suitable for your synth.

<EDIT> Just found this:
"BC INPUT
The BC input on the Kurzweil K2661 is compatible with Yamaha BC1, BC2, and BC3. It is a 3.5mm stereo jack that: provides -15.32 V on tip, expects a range of -0.5V to -8.5V from BC on ring, sleeve is ground. The suggested maximum current load for the BC input is 20mA..."

From this site.

So a voltage divider circuit sounds good, anyway...
 
Last edited:

alec_t

Well-Known Member
Most Helpful Member
#3
Fleshing out the voltage divider idea ......
BC-to-Synth.PNG
 
Thread starter #5
That's really awesome of you guys, thanks so much... I can easily rig it up and try it out with a voltmeter first and see if I get 0 to 5v on the output :)

alec_t - really helpful thanks.... please forgive my ignorance... So this circuit is showing 2 x 9v batteries in series... Also is this using the formula?:

Vout = Vin x (R2 / (R1 + R2))

Because with this I calculated that a 10k and an 8k resistor would yield a Vout of 5v, given 9v Vin. So just curious how you calculated 11k for R2.

e.g. 5 = 9 x (10 / R1 + 10), R1 = 8k

Wouldn't Vout equal more like 4.3V? Vout = 9 x (10 / (10 + 11)) = 4.3?

I did some research but yeah have really minimal knowledge so really appreciate as much info as possible here... Love to learn.

Thanks again!
 

alec_t

Well-Known Member
Most Helpful Member
#6
So this circuit is showing 2 x 9v batteries in series..
Yes. V1 provides +9V to the top of R1 and V2 provides -9V to the jack tip of the BC.
I didn't use any formula: I left that to LTspice :).
The sim assumes the output voltage from the BC varies from 0V to -9V (your mileage may vary).
With R1=10k and R2=11k the circuit output goes from +4.7V to +0.4V.
With R1=10k and R2=8k it goes from +4V to -1V (which is not the range you want).
With R1=R2=10k it goes from +4.5V to 0V.
Of course, all these voltages are nominal, since they will vary with battery voltage.
 
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cowboybob

Well-Known Member
Most Helpful Member
#9
Welcome to ETO, Attila Publik!

As you might have noticed, Nicolas D never did respond with his results, so we don't know how it went for him.

Perhaps you can apply the circuit suggestions and let us know if they work :cool:.
 
Thread starter #10
Hi guys yeah sorry about that - I did assemble as per the circuit diagram but was unsuccessful... I then got side-tracked... If I get it working, I will post results. Thanks all the smae everybody for chiming in.
 

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