18V vs 9V

[Jonotallica]

Hi, I'm not sure the right section to post this but I'll give it a shot.

When you design or when you already have a pedal that's been designed for 18V.. can you run it at 9V and what would be the things that need changing in order to convert it one way or the other?

Would it simply be an issue of the voltage dividers feeding things like op amps and making sure the electrolytic caps have enough voltage to handle it?

The main advantage of 18V is extra headroom, so if it's been designed for 18V and it was running at 9V would would be the effects?  Lifeless sounding, more dull, more clipping or something else?

I have a wah that I'd like to convert from 18V to 9V, but it's also a general question in pedal building because I'd like to build some dirt boxes in the near future and would like to think about the option of optimising it from 18V vs 9V or having the option of both.

Is there any information online about this in terms of resources?  Has anyone else done this before?

Thanks

[Mark Hammer]

One needs to consider how much gain is being applied by the circuit in question.  If there is little danger of the signal exceeding the voltage swing that the circuit is capable of, then there is unlikely to be much advantage coming from operating it at a higher supply voltage.  Similarly, if the signal is physically constrained from exceeding some value by diodes, a higher supply voltage adds little value.

If, on the other hand, the circuit does add gain, such that there is a risk of the signal bumping up against the headroom limitations, then a higher supply value can be your friend.  And I suppose there will be those rare situations where everything else is operating with a higher supply voltage and it is an inconvenience for one solitary pedal to work from less.  That would be an instance where the higher supply is not necessary, and doesn't add any real value, but simplifies matters via using the same supply throughout.

[Jonotallica]

That helps, thanks.

[amptramp]

I would much prefer to see ±9V or ±12V supplies so you have enough headroom to add regulators.  This will enable you to separate the effects of bad power supplies and ground problems.  Our dependence on single 9V batteries and a positive shell for wall wart adapters is a holdover from the 1960's and it's about time we left that behind.  There are a number of circuits, both discrete and op amp, that behave differently as voltage declines because 9 volts is a rather low voltage.

Op amp circuits do not necessarily have the headroom we normally run into and a change to a different manufacturer may leave the designer with a circuit that no longer works.  Also, some op amps have the output go to the opposite rail when the input exceeds the input common mode range, so you can get horrendous noise or latchup (with CMOS op amps) that causes the circuit to fail.

Discrete circuitry may work better and if not, you can always regulate the voltage down to what you can use.  With a transistor amp, the output impedance of the transistor is much higher than the load resistor, so you have no power supply rejection.  Not usually a problem with a battery but a big issue with a wall wart.

[Pojo]

Here's something I've noticed regarding discrete common source jfet amplifiers and changing the supply. When they've been designed using a standard +9V DC supply. Increasing the supply to +18V doesn't simply mean there is just a proportional increase in headroom (although it is increased).

Let's say for example (using measurements from an actual fet in one of my pedals) you have a 2N5457 stage with a 2.7K at the drain and 470 ohms at the source. For this FET my DMM is showing 6V on the drain. With a 9V supply the Vdrain / Vsupply ratio is 6v/9v or 2/3 or 66.666666% of the supply. So what happens if the Vsupply is increased to 18v? Does the same 2/3 ratio apply here? If so, 18 x 2/3 would be 12V. Well lets swap the supply on the circuit and plug in an 18V adapter to find out. My DMM shows 14.8V. So it appears the Vsupply to Vdrain ratio has just changed to 18v/14.8V or 82.222222% of the supply. From my limited understanding of how these stages work, any deviation from Vdrain = Vsupply/2 results in asymmetrical clipping when the stage is overdriven. So in this case my guess (and in my experience from just listening) the increased voltage gives a little bit more headroom from the stage but also leads to further asymmetry when clipped unless the stage is rebiased according to the supply voltage change.

I know the original question asked about what would happen supply a pedal designed for 18v with 9v and I've, in my unscientific way, gave an example of the opposite. But it is one thing to consider and of course the devil is in the details.

[Mark Hammer]

It's also worth noting that there are plenty of effects where specific components set the potential headroom, independent of the power supply.  For example, 3080 OTAs used in compressors and other effects have serious headroom limitations that a higher supply will not overcome.  There may be some improvement in headroom for BBDs, but not enough to really justify going with a higher supply in all cases.  The JFETs used as variable resistors in many phasers also have headroom limitations that are not really addressed via increasing the supply voltage.

That is not a wholesale thumbs down on 18, 15, or 12V.  The utility will always depend on context.  All of which is to say that one should consider what any change in the supply will and won't do.  In the case of many 2nd generation BBDs, for example, there will often be a 5V regulator in the circuit, to provide a stable bias voltage to the input of the BBD.  You can crank the supply voltage to the overall pedal, but that BBD is still getting 5V to power it, and some fixed portion of that 5V to bias it.

[Frank_NH]

Here's something I've noticed regarding discrete common source jfet amplifiers and changing the supply. When they've been designed using a standard +9V DC supply. Increasing the supply to +18V doesn't simply mean there is just a proportional increase in headroom (although it is increased).

Let's say for example (using measurements from an actual fet in one of my pedals) you have a 2N5457 stage with a 2.7K at the drain and 470 ohms at the source. For this FET my DMM is showing 6V on the drain. With a 9V supply the Vdrain / Vsupply ratio is 6v/9v or 2/3 or 66.666666% of the supply. So what happens if the Vsupply is increased to 18v? Does the same 2/3 ratio apply here? If so, 18 x 2/3 would be 12V. Well lets swap the supply on the circuit and plug in an 18V adapter to find out. My DMM shows 14.8V. So it appears the Vsupply to Vdrain ratio has just changed to 18v/14.8V or 82.222222% of the supply. From my limited understanding of how these stages work, any deviation from Vdrain = Vsupply/2 results in asymmetrical clipping when the stage is overdriven. So in this case my guess (and in my experience from just listening) the increased voltage gives a little bit more headroom from the stage but also leads to further asymmetry when clipped unless the stage is rebiased according to the supply voltage change.

I know the original question asked about what would happen supply a pedal designed for 18v with 9v and I've, in my unscientific way, gave an example of the opposite. But it is one thing to consider and of course the devil is in the details.

I have a spreadsheet for calculating bias for a common source jfet stage.  I plugged your numbers for source/drain resistors and supply/drain voltages (also making some assumptions for Vgs and Idss) and was able to replicate your observations.  That is, if you keep the source and drain resistors fixed and go from a 9V to 18V source voltage, your drain voltage is indeed going to be higher than desired.  The fix is to increase the drain resistance, in your case from 2700 ohms to about 5400 ohms to achieve a drain voltage of 12V.  Anyhow, I think it's always a good idea to make the drain resistor a trim pot so you can dial in the drain voltage you want.  Just make sure the pot value is no more than twice the expected drain resistance (i.e. don't use 100K if you don't need it!  Makes adjustment less precise).

I have a developed a vero layout for the Runoffgroove Eighteen circuit which uses a charge pump to run at 18V (Eighteen at 18!  ), and so I will hopefully soon be able to study for myself the use of 18V source voltages for jfet amp sims.  Of course, the Supreaux Deux does this too, but I wanted to be different.