#1
What wattage resitors will I need to build this? I plan to use it to bring down the headroom of a clean boosted, slightly modded Epi VJ with a 12AU7 in the preamp. I'm thinking of a 5-10 db reduction.

Thanks!
#4
How much did you have to drink today?
None are more hopelessly enslaved than those who falsely believe they are free.
#5
Dead clean and sober, I swear.

Any thoughts on the required wattage for the resistors?
#8
Yes, but they differ in wattage rating. Most of the ones in stompboxes never see more than 9, 12, or 18 volts, they are usually 1/4 watt, whereas some of the ones in tube amps requring over 450 volts need to be WAY higher.
I know the ohms I need, but the article says nothing about the needed wattage.
#11
Wow, idiot alert. I just realized I never posted any real info or link explaining the attenuator I want to build.
Sorry!
I would edit my original post, but I need the bump.


A bridged-T attenuator is another purely resistive design, but one which reportedly has superior sonic characteristics to the L-Pad.
FM Systems, Inc. has a page about the bridged-T. I found the link on the Ampage BBS. In this circuit, the values of R3 and R4 are always equal to the characteristic impedance of the whole thing. I.e. if you’re building one for an 8 ohm load, R3 and R4 are both 8 ohms.
FM Systems’ page describes the math for figuring R1 and R2. I’ve included the same table below for the Bridged-T as above for the L-Pad.


dB Z R1 (ohms) R2 (ohms)
-3 4 9.70 1.65
-6 4 4.02 3.98
-9 4 2.20 7.27
-12 4 1.34 11.92
-15 4 0.87 18.49
-3 8 19.39 3.30
-6 8 8.04 7.96
-9 8 4.40 14.55
-12 8 2.68 23.85
-15 8 1.73 36.99


Edit: The link:

http://www.fmsystems-inc.com/index.cfm?tdc=dsp&page=engineers_detail&pid=7
#13
I would think at least rated for 10 watts, since the VJ will exceed 5 watts, when cranked.

EDIT: By that I mean, the total attenuator circuit should be rated for 10 watts, not each resistor. I don't know how a Bridged-T attenuator is wired, but here's an example of how to figure it out in an L-Pad application:
http://www.diyaudioandvideo.com/Electronics/RCL/
Last edited by forsaknazrael at Dec 11, 2008,
#17
Ah, so the bridged-T is different than the normal T-Pad. Got it.
The L-Pad design and the Bridged T aren't that different, actually.

I'm going to use that same equation I posted....

So!
For an L-Pad application...


R1 here is basically the same thing as the R2, R3, and R4 in a Bridged T. Just work it out.
So, we're going to refer to that total resistance of R2, R3, and R4 as RT. R3 and R4 are the same as the characteristic impedance...in this case, an 8 ohm speaker. Add them up, since they're in series, so it's (R3 and R4) in parallel with R2. R2 = 162.17 ohms.
so now we have:
1/R-L = 1/16 + 1/162
162 ohms is the number i got from the chart for a 10dB attenuation circuit. I rounded.

So R-L = 14.562.

So going back to the wattage rating for an L-Pad...

W1 = (14.562 * P)/(14.562+ 35)
W2 = (35 * P)/(14.562 + 35)

W1 is the rating of R-L. Now, R-L is a circuit of resistors, so since it's pretty much just additive, divide whatever W1 is by 3.
W2 is the rating of R1
P is the wattage rating of your speaker.


I think that's how you would do it. I think.
Last edited by forsaknazrael at Dec 12, 2008,
#20
Wait, wouldn't it be the amp's output, not the speaker power rating?
I apologize for my lack of math/elec chops, my speaker is 50 watts, amp is 5.
#22
What page?
And my speaker is 50 RMS like I said in my last post.
#23
Right, I saw your speaker rating.

http://www.diyaudioandvideo.com/Electronics/RCL/
And this page. They want the speaker's rating there.

If you do the math, W1 = 14.7. So divide that by 3, and we get about 5W per resistor.
Then for W2, you get 10.3 W.
So, in the diagram you posted, R1 needs to be rated for about 10.3 W, though I imagine 10W would be fine. For R2, R3, and R4, you would need resistors rated for 5W each.

I would PM SomeoneYouKnew, though, and ask him to check this out. I'm not 100% certain of my answers, and would feel a lot more comfortable if he weighed in.