Hi everyone,
I built another bor for a friend just for the dirty part (Distortron), following the scheme,
https://tinyurl.com/y7nw7zr3 (https://tinyurl.com/y7nw7zr3)
everything is ok, the components are always the same but the volume is slightly lower than the original bor ( maybe 2 dB ) and thinking about it, the first one I built also has the same problem.
Could it depend on the width of the tracks or the geometry, for example curves that create inductances or on the square shape of the PCB or capacitors between neighboring tracks ?
Thx
Quote from: menstrel on September 23, 2023, 04:04:32 AMCould it depend on the width of the tracks or the geometry?
No. But 2 dB is within tolerance for a pot (~20%).
How are your (drain) voltages looking?
Quote from: menstrel on September 23, 2023, 04:04:32 AMCould it depend on the width of the tracks or the geometry, for example curves that create inductances or on the square shape of the PCB or capacitors between neighboring tracks ?
For audio work, we can pretty much say that this is *never* the problem. While layout issues do sometimes cause problems, the width of the tracks or the shape of the PCB or capacitance between tracks is virtually never the problem unless you've done some really extreme PCB layout.
Quote from: FiveseveN on September 23, 2023, 04:32:47 AMHow are your (drain) voltages looking?
drain 1 5.91v
drain 2 4.49v
drain 3 4.57v
Here is an audio sample, you hear first the clone and then the original bor, then the tail of the note, passing from the clone to the original etc..
https://voca.ro/15aAJC6v8TuJ (https://voca.ro/15aAJC6v8TuJ)
The original one has more volume, brighter and more distortion.
The controls ( level, tone and drive) are at maximum.
The voltage on Q1 drain, with drive control at max , should be the lowest of the three (if the three mosfet were identical) .
If not , there is an issue in the first stage ( faulty or mis-biased Q1)
Could we have a schematic diagram??
Quote from: antonis on September 23, 2023, 03:14:54 PMCould we have a schematic diagram??
https://tinyurl.com/yc7em5dk (https://tinyurl.com/yc7em5dk)
Quote from: lcv on September 23, 2023, 03:01:04 PMThe voltage on Q1 drain, with drive control at max , should be the lowest of the three (if the three mosfet were identical) .
If not , there is an issue in the first stage ( faulty or mis-biased Q1)
in the original bor
Q1 4.22v
Q2 4.47v
Q3 4.63v
in truth, at the beginning the difference in volume was present but it was not as evident as now but I had not measured the voltages
To get the right drain voltage for Q1, just choose a higher value for R5 or R19.
Quote from: Clint Eastwood on September 23, 2023, 04:58:45 PMTo get the right drain voltage for Q1, just choose a higher value for R5 or R19.
I think the gain potentiometer is a bit faulty, its lugs move a bit and if I apply light pressure with a screwdriver, sometimes the volume suddenly rises with the drain voltage going from 5.01v to 4.88v
I had the same problem with el cheapo potmeters, lugs that make bad contact. Lesson learned, bad way to try save money.
Quote from: Clint Eastwood on September 24, 2023, 06:43:07 AMI had the same problem with el cheapo potmeters, lugs that make bad contact. Lesson learned, bad way to try save money.
I've always used Alpha potentiometers without problems, but never say never. I now tried a 10k both linear and log but as I imagined the excursion is not correct. It bothers me to buy another one...and if I wanted to use a linear or logarithmic 10k, what changes should I make to the circuit ?
You should use Rdrain=Potvalue.
The bias will be fine, but other things will change (gain, bandwidth, noise), not necessarily to the worse.
Quote from: lcv on September 24, 2023, 11:26:05 AMYou should use Rdrain=Potvalue.
The bias will be fine, but other things will change (gain, bandwidth, noise), not necessarily to the worse.
Are you saying to replace the R4=1M with a 10k linear or log ?
https://tinyurl.com/yc7em5dk (https://tinyurl.com/yc7em5dk)
No , you see in the original schematic potvalue=5K, Rdrain= R5 + R19 = 1.2K+3.9k = 5.1k
I'm saying if you replace potvalue with 10k , then you would need R5+R19=10k e.g. if you keep R19, then replace R5 with 6.8k
I solved the volume gap by replacing the BS170s with 2N7000s. Regarding your suggestion, I apologize, looking quickly for a moment on the drain I saw R4.
thx
Quote from: lcv on September 24, 2023, 12:24:23 PMNo , you see in the original schematic potvalue=5K, Rdrain= R5 + R19 = 1.2K+3.9k = 5.1k
I'm saying if you replace potvalue with 10k , then you would need R5+R19=10k e.g. if you keep R19, then replace R5 with 6.8k
Instead of R5+R19 =5.1k I put R=10k unfortunately nothing changes with both linear and log pot 10k, I got the same result leaving 5.1k. Maybe the 10k pot needs to be antilogarithmic like the original C5k one ?
a 10k pot is 10k. changing the taper does not affect the resistance. the circuit doesn't care if the pot is A, B or C. or W. it doesn't know anything but the resistance value. 10k.
Quote from: duck_arse on September 25, 2023, 11:41:58 AMa 10k pot is 10k. changing the taper does not affect the resistance. the circuit doesn't care if the pot is A, B or C. or W. it doesn't know anything but the resistance value. 10k.
I followed the lcv suggestion but the problem is that the excursion in all cases is never gradual, that is the sound remains clean for almost the entire rotation of the pot and when this is almost at maximum the gain increases quickly. If, however, I leave the resistance on the drain unchanged and put a 10k log pot and rotate it 180°, the excursion is gradual but to increase the gain I have to rotate counterclockwise and vice versa.
ah, please excuse me. it was not clear to me the problem was feel/using the pot. and yes, if a log feels right when backwards, an antilog/C is what you want. probably.