Another Diode Clipping Q

[robmdall]

Something that I have not been able to understand (I have searched). In most diode clipping circuits (diodes to ground) there is a cap and resistor in series between the signal and the clipping diodes. If I understand correctly, the cap is there to remove any dc / pass only ac?

What is the purpose of the resistor?

Limiting current?  Why would that be an ideal idea in a distortion circuit?

Are there best practices as far as a formula that should be investigated?

Thanks in advance for any insight.

Rob

[Mark Hammer]

Diode characteristics vary as a function of the current applied, though I am in no position to be able to say anything useful about how much current gets you what sort of clipping for what sort of diode.

The other factor to consider is that often, in those cases where there is a diode pair to ground, there will also be a cap in parallel with the diodes to roll off such of the uglier treble components.  Of course, to function as a lowpass filter, there needs to be not only a cap to ground, but a resistor in series with it as well, such that a rolloff can be calculated.

[PRR]

> What is the purpose of the resistor?

Take it to an extreme.

My local power-plant hums low B flat, very mellow tone.

I put a couple diodes across it to clip and fuzz.

The diodes EXplOde. The mellow hum is unaffected.

OK, so I buy all the diodes in the whole world and melt them together into two _BIG_ diodes.

The local power-plant explodes.

On the other hand, if I put 10,000 ohms between the power-plant and a couple 10-cent diodes, the resistor is a bit warm but the power-plant hardly knows I'm messing with it, and my side of the resistor is suitably clipped/fuzzed.

All these shunt-to-ground schemes are Voltage Dividers.

The top resistor is critical.

As a quick guess: it should be large enough so the source is not harmed and maybe hardly-loaded, large enough so the diodes can clamp/clip it without strain, but small enough so the final load gets nearly full signal (when not clipping),

Using normal diodes, the exact value of resistor is not critical. Diode incremential response is nearly the same at any (healthy) current. 1K resistor instead of 10K will give barely-audible increase of clip-level, which you anyway normally turn-down to taste. The resistor value generally does impact other value choices, like those caps.

[robmdall]

Thanks guys, I may understand it?  A very simplistic explanation is that a diode can behave like a switch allowing current to flow (in my example to ground) once a certain amount of voltage is met (diode Forward Voltage). So, allowing too much signal to pass to the diodes will put undo strain on the diodes?

So, I want to choose a resistor value that:

1: Will not load down the preceding amplifier stage. (too large a value and the stage prior is strained).

2: Reduce current enough to easily push the diodes into clipping (too low a value and I stand the chance of allowing too much current to flow into the diodes). Does this slightly defeat the purpose of the amplifier stage before the diodes?

Still not clear but getting there, so best practice during breadboarding would be to use a variable resistor (100k) and tune to taste?

[anchovie]

Does this slightly defeat the purpose of the amplifier stage before the diodes?

Not at all. The amplifier amplifies voltage; that's what makes your guitar signal louder.

Still not clear but getting there, so best practice during breadboarding would be to use a variable resistor (100k) and tune to taste?

The best practice would be to use a 10K like in the Distortion+ and forget worrying about it. As PRR said, you won't get a noticeable change in clipping level by altering the resistor value.

[PRR]

> a diode can behave like a switch allowing current to flow (in my example to ground) once a certain amount of voltage is met (diode Forward Voltage).

Think of Resistor Voltage Divider. (This concept should be absorbed well.) Think of the diode as several resistors. Grossly: Under 0.5V it is an "infinite" resistor. Over 0.7V it is a "small" resistor standing. So signals up to 0.5V peak get hardly-any Voltage Division. Signals over 0.7V peak get brutally slashed. And there's some round-knee in the 0.5V-0.7V zone.

> So, allowing too much signal to pass to the diodes will put undo strain on the diodes?

"Undue".

And despite my example you can't strain common diodes with common small amplifiers. (If you go as big as LM386, and might hit zero resistance, it might be wise to use 1N400x "rectifiers" instead of small-signal 1N914/4248.... or just don't run less than 100 ohms, there's no need to go so low.)

Plagiarize. Peep many diode clippers and use resistors from the same drawer. This is the same way you learned music: you copy riffs songs and suites from radio, Bach, LedZep before you move on to totally original works.