Introduction: Passive Guitar Overdrive (Black Ice)
A friend of mine brought to my attention a passive guitar effect, called "Black Ice" which takes the place of the capacitor (or wired with the capacitor) on the tone knob, and when used, gives the guiatr's sound a "crunch" similar to an overdrive. He wanted to know if i could make it, as opposed to buying one for 30 dollars. After a bit of research, I found out is was simpler than I first anticipated, and a whole heck of a lot cheaper. Along with building the effect as it was originally supposed to be done, i threw in my own twist. I found the amount that the version one effect changed the sound was underwhelming, but my version, when wired in as i specify in the instructable, has a profound change to the guitars sound.
Version one is the original Black Ice circuit. It produces an overdrive-type crunch, though it is not exactly like a true overdrive. It is a little more "hummy" than an overdrive, and the tone is more mellow.
Version two, my version, has a smoother tone than version one, until it is turned up all the way, when it becomes more agressive of a crunch, and plenty of 'wanted unwanted' distortion. It also modulates the higher frequencies into the loswer ones. The tone is also less mellow. I prefer its more noticable and agressive tone as compared to the priginal Black Ice circuit, but it does require a more complex hookup.
Step 1: Parts
For version one, you need two diodes. Thats it. The type of diode does matter however. Schottky diodes are what the original circuit uses, but cat-whisker diodes can be used as well, as long as they have a very low voltage drop. I made a version using Schottkys, but for this instructable i will feature the cat whisker diodes. This version is the original Black Ice circuit.
For version two, you need 4 diodes. This is my version of the circuit.
Step 2: Circuit
For version one, the two diodes are wired in parallel with opposite polarity. In version two, they are wired as a full wave rectifier circuit.
ATTENTION!
My version was originally intended to be wire as seen in the schematic. But after some experimentation i found a better way to wire it. On step 4 there is a diagram showing how to wire in the effect. Us the diagram on this step to build it, the diagram in step 4 to wire it
Step 3: Finishing Touches
Once soldered together, you can then fix them in your guitar. Or, you can jazz them up a little. These are going to my friend, who will then be installing them. Since his soldering skills are....lacking, I bundled everything up with shrink wrap so the joints dont come apart during soldering. Longer leads were also added, to make installation easier. With the version two effect, it is polarity sensitive, so make sure you dont mix it up. I added a dot of paint on the postitive output from the circuit. The negative output is directly opposite it, and the inputs from the guitar pickups are the two leads adjacent to the marked one.
Step 4: Wiring It Up
For the first version, It is wired in place of the capacitor on the tone pot. Polarity is not an issue. If you want to use the cap along with the overdrive, see below:
The tone pot has three contacts. First, wire the signal input for the tone knob into the middle contact. Then, the capacitor is wired to one of the outside contacts and the back of the tone pot, and the overdrive effect to the other contact and the back of the pot. The back of the pot is then grounded as normal. This allows the tone to be anywhere from full capacitor, to half capacitor half overdrive, to full overdrive. This does however mean that the tone/overdrive will always be on to some extent, and cant be turned alost off like it can be when just a single effect is used.
To wire in the version two effect, nothing to the original guitar's wiring needs to be changed. (If you saw this before i made this change, disregard the original information). Follow the diagram. There are two ways to wire it. The basic wiring is the same, but the opposite lead from the marked lead can be left unatatched for a stronger overdrive effect. Do not touch any of the exhisitng wires in the guitar.
Step 5: Jam Session!
Now that the effects are built and installed, its time to play! The tone knob is used as normal, unless you wired both the capacitor and overdrive onto the tone pot. Then, in one direction it controls the overdrive, and in the other, it controls the tone cap. With version two, the tone pot not only controls the amount of crunch, but how rough the crunch is. The higher it is turned up, the rougher the crunch. The diodes also have the effect of modulating the higher frequencies with the lower ones. The nice thing is that this effect is different in sound than overdrive delivered by an amp, but it can be controlled on the guitar itself, and unlike pedals, goes with you everywhere on stage, seeing as it is built into the guitar.
There will be a video up soon demonstrating the sound of the effect.
Congratulations on the completion of your own passive overdrive effect!

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42 Comments
3 years ago on Step 2
Are you sure your diagram is right? It does not look right. Here is what I came up with.
Reply 3 years ago
Hello sir can you please tell me witch way I feel is the best way I'm sure there is several different ways to wire it together and different ways to in stall it.what I did was number three on my pic. I ran two wires off my tone pot from L2 and to the back of the pot positive on L2 neg on pot the wires are in place of the capacitor with clips on the end to make quick experimental changes I have the mod like pic 3 but I have the positive at the super ground and the negat on the other ground on bottom and I rapped the left and right side of mod with grounding tape what do you think
Reply 3 years ago
#2
Question 5 years ago
Is it possible to get a clearer diagram for the type 2? Thank you for your time.
7 years ago
Great! Is there a way to place a switch so that I can deactivate it if I won't use it?
7 years ago
Please someone tell me which diodes i have to use. i bought some 1N34A germanium diodes, but they dont work :-(
9 years ago on Introduction
Here's an analysis of your circuit. First, I labeled the diodes, D1 through D4. Then I redrew the schematic. Vd is the voltage at which any of the diodes will turn on. First thing I noticed, diode D4 is grounded on both ends. That means it is shorted out, and it performs no function. D4 can be removed from the circuit. Next, I looked at D3. On the positive half of the input signal, D3 will turn on at it's forward voltage (+Vd) any voltage over +Vd will be shunted to ground. However that remaining positive voltage will be blocked by diode D1 and will not appear at the output. Therefore, D3 performs no function. D3 can be removed from the circuit. Now looking at diode D1, on the positive half of the input signal, D1 will block the positive voltage, but on the negative half of the input signal D1 will block the signal until it reaches -Vd, and then the D1 will turn on and allow the signal to pass until absolute value of the input voltage drops below Vd, then the diode will turn off again. D1 does perform a function in this circuit. It can stay. Finally, looking at D2, it will attempt to shunt any positive voltage to ground. However D3 has already shunted any positive voltage higher than Vd to ground, and D1 has blocked any remaining positive voltage from reaching the output. Therefore, D2 performs no function. D2 can be removed from the circuit. So, you can see, the whole circuit can be replaced by a single diode.
Reply 9 years ago on Introduction
Your analysis is correct.
Reply 9 years ago on Introduction
If you remove D3 from the circuit then you can't use it's function as justification for removing D2
Reply 9 years ago
But D2/D4 is not connected to ground in the corrected schematic. It is functioning as a bridge rectifier.
Reply 8 years ago on Introduction
Both positive and negative peaks of the input sine wave will be clipped with this circuit.
Reply 8 years ago
So does that mean that both the + and - go to the output jack?
Reply 8 years ago
Yes
Reply 9 years ago on Introduction
OK... this is a correct bridge rectifier circuit. Both postive and negative peaks higher than 2 x Vd will be let through as Negative pulses (the way you have this wired...)
Reply 9 years ago on Introduction
No positive voltage will get past D1 to be shunted to ground by D2... So D2 can be removed regardless of D3...
Reply 9 years ago on Introduction
Ive been thinking about why this is distinctly different than the analysis I had done for this.... That schematic I posted is incorrect. It is a bridge rectifier in function- that ground between D2 and D4 should be the other wire from the pickups- the pickup is wired to the D1/D3 node and D2/D4 node. The D3/D4 node is grounded to the ground from the cable and the D1/D2 node is the output
8 years ago on Introduction
What I want to do is add both versions of this with an on-off-on switch. But I'm having trouble figuring out how to make that happen.
The 2-diode version is easy enough - "tap" the signal from the volume pot, bleed to ground. The 4-diode version needs to be added back to the signal path, and that's where I'm running into confusion.
Here's where I'm at with it. Any help would be appreciated!
8 years ago on Introduction
This doesn't sound like a useful improvement over the traditional circuit.
The Black Ice (and its predecessors) attempt to model amp clipping by bleeding signal peaks to earth.
By putting the diodes into the signal path, you're doing things the other way round -- blocking the signal below the peaks, and only letting the peaks through. You'll get a distorted sound, right enough, but it will be a very different beast from typical guitar overdrive/distortion.
When the diodes bias and debias, you get a very sudden change in signal, which by the time it reaches you speaker is a "high-energy transient". You're asking your drivers to jump suddenly from one position to another, accelerating and braking far beyond their rated levels, this is very bad for your speakers or headphones.
Speaking of things that are bad for your speakers, why did you want a rectifier in the first place? It'll emphasise the second harmonic, but a rectified signal stays one side of 0V, and only uses half the travel of the driver coils. This won't do your speakers any favours either.
8 years ago on Introduction
8 years ago on Introduction
Cool!