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 Forum index » DIY Hardware and Software » YuSynth
Yusynth Auto Bend
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helix_modular



Joined: Aug 31, 2017
Posts: 22
Location: Germany

PostPosted: Wed Mar 21, 2018 5:09 am    Post subject: Yusynth Auto Bend
Subject description: Output Levels too low [SOLVED]
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Hi,

biting the bullet of becoming increasingly unpopular around here, I have another issue with a Yusynth module. I've developed a stripboard version of the Autobend circuit (which I'm happy to share, if that's okay with Yves Usson). It's functioning, but the output levels are way too low. On the website (http://yusynth.net/Modular/EN/AUTOBEND/index.html) is stated, that with the values R10=R11=120kOhm one would obtain a range of +/-12 semitones, i.e. +/- 1V.

In my implementation, this is not the case: I'm getting something less than +/-50mV. Troubleshooting made me simulate the circuit in LTspice (see attached file), et voilà! Exactly as I've measured, same result in the simulation.

Has anyone experienced similar effects? Has anyone ever measured the output of their build?

Last edited by helix_modular on Mon Mar 26, 2018 2:21 pm; edited 1 time in total
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helix_modular



Joined: Aug 31, 2017
Posts: 22
Location: Germany

PostPosted: Wed Mar 21, 2018 5:25 am    Post subject: Reply with quote  Mark this post and the followings unread

Well, apparently, e-m blocks .asc files...

Anyway.

Looking at the schematic one can easily see, that the maximum output level is dominated by the charge of the cap C5. Following the signal path to the output, there are only voltage followers or inverters (U1c/d and U1b). By introducing R10 and R11 the voltage is even further reduced. Therefore, the charge of the cap must be too low (simulation confirms this).

Now, there are many ways to tackle this problem.

a) The cap is charged by Q2, with its current limited by R6. One could reduce R6 from 470Ohm to 180Ohm and charge the cap to about 1.1V. This charging current is fairly high and might pull on the powerrails... Still this only results in +/- 100mV at the output.

b) Another thing is to then reduce R10 and R11. Going as low as 1k for each gives +/-1V out. Yet the current that must be supplied by the opamps (U1a and U1b) is significantly higher (8uA vs 80uA).

c) My best guess is to reduce C5 to 470nF and increase P1 to 1M to maintain the same time constant (1uF*501kOhm = 501ms vs 470nF*1001kOhm = 470.5ms) and shorting R10 and R11. Then again there is the current issue stated in b).

Am I missing something here? Any suggestions?

Regards,
helix
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g.gabba



Joined: Nov 29, 2008
Posts: 463
Location: berlin
Audio files: 23

PostPosted: Wed Mar 21, 2018 8:21 am    Post subject: Reply with quote  Mark this post and the followings unread

there is no way of becoming increasingly unpopular around here by asking questions - everybody is free in the decision to invest time in answering Wink

iam assuming you have an oszi - what is the max voltage that C5 is charged? You can measure it directly at C5

How this charging behaves? when you
- short C3
- when u raise C3 to lets say 100n

so far for today
cheers bb
gabbagabi
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helix_modular



Joined: Aug 31, 2017
Posts: 22
Location: Germany

PostPosted: Wed Mar 21, 2018 11:29 am    Post subject:   Reply with quote  Mark this post and the followings unread

Hi g.gabba, thank you for your kind words.
But I'm feeling a bit like the student in class who claims to have spotted a mistake in the math book. That guy is seldom popular Very Happy

g.gabba wrote:
iam assuming you have an oszi - what is the max voltage that C5 is charged?

Yes, I do have access to some pretty high quality stuff. C5 goes up to about 1.5V.

g.gabba wrote:
How this charging behaves? when you
- short C3
- when u raise C3 to lets say 100n


The first one doesn't make sense to me, I did it anyways: depending on the gate frequency, it almost reaches V+ (which is reasonable). Yet, the very idea behind C3 is to extract a short positive pulse to make Q3 conducting. This leads us too...

I had the idea of increasing C3 as well. Dropping in 100nF in parallel to the 10nF makes the voltage at C5 shoot up to about 4.5V. Pretty neat. This results in some 750mV at the output which is not bad. Then again, this makes the rise very rounded off (much like longer attack compared to instantly jumping to the bended pitch). So, not really a viable option to me. Can't check it whether it's 'musically' since I don't have a VCO at hand atm.

Is your module working properly with the proposed values? (I'm meaning +/-1V with 120k)
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g.gabba



Joined: Nov 29, 2008
Posts: 463
Location: berlin
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PostPosted: Wed Mar 21, 2018 12:19 pm    Post subject: Reply with quote  Mark this post and the followings unread

the purpose of this exercises was to see if the input circuit is able to charge C5 properly which seems is the case

ok so u have (with the 10n):

at U1b: 1,5V ?
and at U1a: -1,5V?

and u would like to archive +/-1V at U1c and U1d?

than i would try 25k forR10 and R11 if u use 50k for P2 and P3

i have never build that module

cu tomorrow

edit: useless inforamtion overhead removed
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helix_modular



Joined: Aug 31, 2017
Posts: 22
Location: Germany

PostPosted: Sun Mar 25, 2018 1:13 pm    Post subject: Reply with quote  Mark this post and the followings unread

Thank you very much g.gabba... I've tried this and apparently, it works!

However -- out of sheer curiosity -- how did you calculate these values?

EDIT: Apparently -- this doesn't. But I've figured it out altogether. Refer to my solution down below!

Last edited by helix_modular on Mon Mar 26, 2018 1:20 pm; edited 1 time in total
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g.gabba



Joined: Nov 29, 2008
Posts: 463
Location: berlin
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PostPosted: Mon Mar 26, 2018 8:25 am    Post subject: Reply with quote  Mark this post and the followings unread

iam glad it worked,

from 1,5V to -1,5V u have a "Potentialdifferenz" of 3V,

and you want to archive +/-1V on the output, thats a "potential difference" of 2V

and with the 50k we have been given a third value, and we can calculate

2V/50kOhm = 3V/x

x=3V*50Kohm/2V

x=75Kohm

with x is the new totalresistance we have 75k-50k= 25k= R10 and R11 thogether

that means 25kOhm/2 = 12,5kOhm

which then brings us to the conclusion that my 25K-tip must have been wrong,
but it brought you closer to the desired outputswing,

my corrected tip is now 12kOhm (or two 25k in parallel to have exact 12,5k) forR10 and R11,

i hope u accept my apologies

edit: then that 120k in Yves schematic make more sense, he has just mixed up 12k with 120k
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helix_modular



Joined: Aug 31, 2017
Posts: 22
Location: Germany

PostPosted: Mon Mar 26, 2018 1:45 pm    Post subject: Reply with quote  Mark this post and the followings unread

Hi g.gabba, thanks again for working with me on this. I'm actually quite amazed that you invest so much into a module you haven't even built yourself! Thanks anyway.

Actually I've had a similar idea. And you've been close. But for the sake of completeness here is my solution:

1. the overall resistance R_x from U1a to U1b is R_x = R10 + P1 || P2 + R11 (P1 and P2 are in parallel!)

2. U_x := U1a - U1b

3. Ohm's law: U_x = I_x * R_x <=> U_x / R_x = I_x (I_x is the therefore the current that runs from opamp to opamp)

4. the voltage across the pots should be 2V (+/-1V = 2 octaves = 24 semitones), but the current is still I_x, so we can determine I_x = 2V / 25kOhm = 80uA

5. If we want to kill U1b - 1V = U_d, we calculate the resistance R10 = U_d / I_x, which yields in my example R10 = 0.5V/80uA = 6.25kOhm = R11

Which is the same thing that I've verified in a simulation of the circuit.

The only issue here is that this is neither a standard value nor should one take the 50k of the pots for granted. In fact, I used quite affordable ones that have a resistance in parallel of about 21kOhm. (Which yields about 5.25kOhm) Therefore I'm somewhat in the range of 5k6 or 5k1 for R10 and R11 -- which happen to be standard values: convenient!


Hope this might be useful for others...

Cheers,
helix
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g.gabba



Joined: Nov 29, 2008
Posts: 463
Location: berlin
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PostPosted: Mon Mar 26, 2018 11:26 pm    Post subject: Reply with quote  Mark this post and the followings unread

Embarassed

or you use for every poti a pair of 12k
or you could use 100k potis and 12k for R11/12
or you could use 100k potis and 25k with the extra-resistor-pair-method

helping other people is good for your carma, although one will risk to embarrassed himself by giving wrong or inaccurate tips

you will learn something by looking into schematics and try to figure out what is wrong,
and it will keeps this community alive which i like so much

in this case you brougth this circuit back in my field of view, and it is only now that i understand the usefullness of it, and may i will design one in my one way

cu, gabbagabi
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