A simple question (or two) on G. Forrest Cook Guitar Reverb 2

[EAsch]

This will be my first build, and one I'm excited about as I have wanted a spring reverb for some time, and have never built a complex circuit before. As such, please excuse the nature of my question if it is painfully obvious... but, with regards to G. Forrest Cook's Guitar Reverb 2 available @ http://www.solorb.com/elect/musiccirc/reverb2/

I have made my parts list, but I noticed that none of the resistors have wattage rating listed, and while looking at parts did some basic research that suggested I might burn up resistors if this is value too low for a given circuit. Particularly for the power section I was concerned, but high wattage resistors are not inexpensive, so I thought I would ask rather than wasting money.

Will 1/4W resistors be sufficient for the majority of the circuit? It seemed that for audio only sections of the circuit, dealing with 1-2 Volts, 1/4W should be fine, but what about for the power supply filter section? Will I need higher watt rated resistors? Also, are there any other sections that 1/4W resistors will be insufficient?

Thank you all in advance for your help!

Also, minor secondary question: I assume that the pair of 20K, 20T pots for biasing are trimpots, but what does that 20T value refer to? Does he just mean 20 Turn trimpots? Or does he mean 20% taper?

[antonis]

It actually is a multi-turn pot (trimmer) of 20 x 360^o turns wiper travel but you can manage same results with a deca-pot (10 turns)..

You can also use any simple trimmer of single turn (300^o) but you have to be very lucky for precice BJT collecor bias... 

[amptramp]

There is a quick way to get an upper bound on power requirements: assume the supply voltage is dropped across the resistor and determine the dissipation.

P = V²/R

In this case, V is 12 so a 144 ohm resistor would dissipate 1 watt, a 288 ohm resistor would dissipate 1/2 watt, a 576 ohm resistor would dissipate 1/4 watt, an 1152 ohm resistor would dissipate 1/8 watt and a 1440 ohm resistor would dissipate 1/10 watt.  Anything above 1440 ohms can be a 1/10 watt resistor.  It is normal to allow for a derating factor of 0.5 for resistors to obtain long life and reliable service, so you can double the wattages above in this case and use 1/10 watt resistors for 2880 ohms and above.

You can prove that much of the circuitry sees less than 12 volts and you can determine the voltage and current through the resistors individually and the product of that is power.

[EAsch]

Ok, thank you both so much, that is very helpful.

Thanks amptramp, I knew the formula, but was spending to much time trying to figure out the specific values across each resistor based on its position, other resistors in series with it, ect. and it seemed like an absurd task to actually do that for every resistor, so thanks, that is a much more logical way of finding a general rule of thumb for the circuit.

Really appreciate the prompt help! You guys rock!