More Headroom?

[marnold]

OK, in descriptions of some guitar effects pedals and some bass active circuitry, they say that if you run it at 18v instead of 9v you will have more "headroom." In this context, what does this mean? I understand that if an amp has more "clean headroom" it will stay clean at louder volumes and take longer to overdrive. I don't get what it means here.

[Andy]

I think it means at the higher voltage the "clipping point" of the effect is higher
my ocd seems to run much cleaner and has a bit less dirt when running at 18v.

I'm sure the pedal gurus can give a better explanation tho

[markb]

I think it means at the higher voltage the "clipping point" of the effect is higher
my ocd seems to run much cleaner and has a bit less dirt when running at 18v.

I'm sure the pedal gurus can give a better explanation tho

That's exactly what I found with a Fulldrive. The higher voltage tightens up the tone and means you have to dig a bit harder for the same amount of clipping as using 9v. It makes the pedal feel quite different in its reaction to your attack. Sometimes though a weak battery is good too. I believe Duane Allman swore by almost dead batteries for his fuzz pedal.

[Ch0jin]

Yeah you've all got it right.

When people refer to headroom in pedals they mean exactly the same thing as when they say it about amps.

If you want I can dive into the tech aspects of why, but you seem to understand what it means in practice anyway. More voltage = Louder clean tone before breakup just like amps. When discussing amps you'd usually say more wattage = louder cleans, but your actually saying the same thing anyway because higher wattage amps (generally) run higher voltage too.

The battery deal Mark mentioned is also spot on. Both my germanium Fuzz box's (fuzzface and Fulltone '69) sound much more to my liking with a cheap 9V battery thats running around 8.8V under load. Given that all my Ge Fuzz's are positive ground I can't use my Boss 9V adapter anyway, so they are always on battery.

I only have one pedal that runs 18v and thats my delay and I run it at 18v specifically because I want it delaying, not distorting.

Does that help? If not post back and I'll have another crack at it.

[markb]

There's also an internet rumour (started on Harmony Central, I think) that to get the best out of an old RAT you should use a PSU rated for 10v rather than 9v. My feeling is that this is feeding into the fact that old zinc/carbon batteries had a higher actual voltage than alkalines but suffer a voltage drop in use more than the modern battery. My old RAT certainly sounded better with zinc/carbons (though I'm not claiming to be Eric Johnson here :) ). Again it was this impresion of (a little) more headroom and greater "feel" from the guitar. I remember Guitar Player running some kind of test of this theory with a selection of batteries some years ago.

[marnold]

Thanks, it makes sense to me. I don't know if I understand why you'd want more headroom in an overdrive pedal for example, but hey. I thought that's why they created clean boosts.

[Ch0jin]

Mark,
I hadn't stumbled across that one about the Rats but it makes sense though. Whilst 'mojo' may defy the laws of physics, fortunately I can guess at an explanation of this one with science :)

I need to make a couple of assumptions to get this to all work, but here goes..

Lets say the 9V adapter is rated to supply 9V @ 300mA. Using Ohms law that means that adapter can supply (9v x 0.3A) 2.7W of power.

Lets then say that the Rat draws >300mA when driven. For sake of example say it draws a hefty 400mA. As per the above example, that means the pedal needs 3.6W of power to work as intended.

So what happens? Well the first thing you're going to notice is your power supply is going to get warm.

Electronically though what I'd expect is the voltage is gong to drop. If (P)ower = (I)current X (V)oltage then V=P/I so if I have this right (and this is real rusty for me so I might not), then V= 2.7W (what the adapter can supply) divided by 0.4A (being the current required by the pedal) meaning you'd see 6.75V under full load instead of 9.

Big difference huh.

Based on the above you might have already worked out that even if the 10V adapter has the same current rating as the 9V one, it's power capabilities are higher. (10 X 0.3= 3W as opposed to the 2.8W of the 9V version) so under load the voltage drop isn't as severe.

It's also a pretty fair assumption that the 10V adapter has a higher current rating though, and that would increase the power rating and.....well I guess by now you see where I'm going....

@Marnold.

Why would ya? In practice, changing the supply voltage will change the overall tonal characteristics of any audio circuit meaning it's going to sound different at 10V than it does at 9V. Some people may dig that, some might not.

Remember that what we do to guitar amps breaks a LOT of the "rules" of amplifier design too. Talk to a HiFi amp guy about distortion and they'll be horrified. Amplifiers were never originally designed to distort the source signal. The fact that it can be musical when they do was a total fluke accident.

Anyway, I was going to do a bit about internal resistance of batteries too but I'd better do -some- work before lunch ;) If anyone cares, post back and I'll try and explain that too :)

[tot_Ou_tard]

I run my Timmy at 18 Volts, but that is because I use overdrive sparingly & am looking for a full, open, sound with as much articulation & dynamic range as possible. I should go back to 9V & see how it does.

I run my Diamond compressor at 24 Volts, but you don't want a compressor distorting.