Diodes going to the ground kill my signal?

Started by tenser75, October 18, 2016, 08:58:36 PM

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tenser75

I built a fuzz that has two clipping diodes that connect to the ground
but i don't understand how should it work, since my fuzz works only if i remove them... seem slike the signal is going to the ground as is... is it maybe my diodes are wrong type??


Jdansti

The diode arrangement shown on the schematic is common. Are the diodes in reverse polarity relative to each other?

If so, let us know what type of diodes you're using.

Also, when you remove the diodes, does the volume sound like its at a normal level, or is it very low?
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ashcat_lt

Are you sure there's no DC voltage at that point?

antonis

What ashcat_II said..

Maybe you've placed diodes before output cap..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

amptramp

The diodes don't conduct until you reach a voltage consistent with the material - about 0.25 volts for germanium, 0.65 volts for silicon, 1.1 volts for infrared LED's, 1.8 volts for red LED, 2.2 volts for green or yellow LED and more for blue or white LED.  Below the threshold voltage they are effectively not in circuit.  The voltage drops as temperature increases.

R.G.

Quote from: tenser75 on October 18, 2016, 08:58:36 PM
I built a fuzz that has two clipping diodes that connect to the ground
but i don't understand how should it work
It works because, as noted earlier, the diodes conduct when the voltage across them is greater than the diode's conduction voltage. This is specific to the semiconductor material of the diodes, as Amptramp says. They work because they do not conduct until the signal is bigger than one diode drop above ground (for positive-going signals) or one diode drop below ground (for negative going signals).

Quotesince my fuzz works only if i remove them... seem slike the signal is going to the ground as is... is it maybe my diodes are wrong type??
No, it is not because your diodes are the wrong type. It is something else. The guesses you've seen here are good ones. It is possible that the diodes have some DC across them due to being placed before the output capacitor, or that the output capacitor is inserted incorrectly.

A possibility that hasn't been mentioned is the biggest and most likely one - it is likely that the diodes are being shorted by soldering or wiring mistakes when putting them in. Maybe not, but decades of experience with beginner builders makes me suspect this.

There is a sticky message at the top of this board that you need to read and follow. It's titled "Debug thread: what to do when it doesn't work". This will get you a lot more help, and the help you get will be much more likely to be applicable to your problem.

R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

GibsonGM

Measuring for DC there would be nice (after checking for shorts...).

A leaky output cap, perhaps?  Just throwing my hat in the ring, too! ha ha
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tenser75

ok guys, I feel very stupid now but the problem was... the volume level was SO LOW, compared to other pedals that i built, that i thought there was no signal at all...

so this actually make me ask another question: why the diodes bring the volume so low, i mean very low (eg. if i use my 25W amp at "2" with other fuzzes, when i switch on this pedal at MAX volume the volume get svery low i have to turn the amp at 4!

maybe im doing something else wrong... but is normal then?

Rob Strand

Assuming there is no DC on the diode ...

Quote
so this actually make me ask another question: why the diodes bring the volume so low, i mean very low (eg. if i use my 25W amp at "2" with other fuzzes, when i switch on this pedal at MAX volume the volume get svery low i have to turn the amp at 4!

Maybe the simple answer is all your other fuzzes produce higher than normal output!

The circuit you originally posted has a possible output swing of about +/- 4.5V without the diodes because the output is connected directly to the transistor collector.   When you add the diodes the level is dropped dramatically, to something like +/- 0.6V.

If you look at say the fuzz-face then the output does not come from the collector it is tapped off a much smaller resistor.
http://fuzzcentral.ssguitar.com/fuzzface/fuzzfacepnpschematic.gif

Here the voltage swing is roughly +/- 4.5V * 470 / 8200 = +/- 0.26V.  Which is in the same realm as the diode clipper.

The apparent loudness vs pot position depends on the type of pots used for the output volume (ie log or linear).
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

merlinb

Quote from: tenser75 on October 20, 2016, 12:48:20 AM
why the diodes bring the volume so low, i mean very low      is normal then?

antonis

#10
Just adding some jolly touches on Merlin's artwork
(although I haven't his written permission..)  :icon_redface:







Any waveform higher than +600mV is "cut" from output (shunted to GND) via the right diode (cathode facing to ground)..
(the same stands for -600mV & left diode..)

Waveforms of lower amplitude are simply "ignored" by the (600mV forward voltage drop) diodes.. 
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

ashcat_lt

Quote from: antonis on October 20, 2016, 09:07:33 AM
..."cut" from output (shunted to GND) via the right diode...
You can say the current is shunted, yes.  The voltage, though, is actually dropped across the transistor stage.  ;)

anotherjim

Should point out that here it is a transistor stage which has a high enough output resistance that the current thru the diodes when they conduct is limited by that resistance. If an op-amp stage was used instead, then a series resistor is added  before the diodes to drop the signal volts Ashcat mentions - an op-amp has a very low output resistance and would be feeding almost a short circuit when the diodes conduct.
Such as here...

See the 10k resistor before the clipping diodes.
I mention this because an op-amp based distortion is usually the next step on the path to enlightenment.

antonis

Quote from: ashcat_lt on October 20, 2016, 01:07:37 PM
Quote from: antonis on October 20, 2016, 09:07:33 AM
..."cut" from output (shunted to GND) via the right diode...
You can say the current is shunted, yes.  The voltage, though, is actually dropped across the transistor stage.  ;)
Quite right only if we take in mind that Current leads Voltage through the output cap... ;)


P.S.
I think that tenser75 bothers more for things happen and less for actual way of happening... :icon_biggrin:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

tenser75

thanks for such thorough answers! i learnt a lot!

j0shua

If you are using 1n4041 diodes, check the amperage of those diodes, try to replace those diodes for 2 zenner diodes, zenner diodes are better to get more power or stabilization , i like to use Led diodes for test hehehehe before use another