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04-26-2010, 01:34 PM
| | Registered User Endorsing Artist: J.C. Basses | | Join Date: Aug 2005 Location: Phoenix, Arizona 85029 | | | Aguilar OBP-3 - Mid push/pull pop?
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The thread title really says it all. Anybody have any suggestion for eliminating that damn pop when the mid push/pull is used? I suspect it's something to do with the two unused lugs.
__________________ Quote:
Originally Posted by McThumpenstein I don't think the wife would buy the "I need to take off this knob and put a whole new bass under it" story. | | 
04-26-2010, 01:40 PM
| | Registered User | | Join Date: Feb 2005 Location: S. Carolina | | | See the Aguilar site wiring diagrams, very helpful. has to do about resistors in the circuit. | 
04-26-2010, 01:58 PM
|  | Supporting Member | | Join Date: Jun 2008 Location: Close to Los Angeles, CA | | | Yep, resistors across the switch terminals should do it. | 
04-26-2010, 03:08 PM
| | Registered User Endorsing Artist: J.C. Basses | | Join Date: Aug 2005 Location: Phoenix, Arizona 85029 | | | I've found some 1M resistors in 1/4W, 1/2W, and 1W - Which should I use?
If my physics knowledge is correct, running the preamp at 9V will need at least the 1/4W and the 18V will need at least the 1/2W?
__________________ Quote:
Originally Posted by McThumpenstein I don't think the wife would buy the "I need to take off this knob and put a whole new bass under it" story. | | 
04-26-2010, 03:17 PM
|  | Supporting Member | | Join Date: Jun 2008 Location: Close to Los Angeles, CA | | Quote:
Originally Posted by FunkMetalBass I've found some 1M resistors in 1/4W, 1/2W, and 1W - Which should I use?
If my physics knowledge is correct, running the preamp at 9V will need at least the 1/4W and the 18V will need at least the 1/2W? | It doesn't matter what you wattage you use, they should all be sufficient.
Your entire preamp circuit draws about 0.009 to 0.018 watts at 9V, or 0.018 to 0.036 watts at 18V. Well below the 0.25 watt limit of your lowest wattage resistors.
Last edited by line6man : 04-26-2010 at 03:20 PM.
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04-26-2010, 03:19 PM
| | Registered User Endorsing Artist: J.C. Basses | | Join Date: Aug 2005 Location: Phoenix, Arizona 85029 | | Quote:
Originally Posted by line6man It doesn't matter what you wattage you use, they should all be sufficient.
First off, your entire preamp circuit draws about 0.09 to 0.18 watts at 9V, or 0.18 to 0.36 watts at 18V.
The current flowing through the leads for the EQ is probably quite low, significantly less current that the draw of the preamp as a whole. | Ah, okay, that's perfect then. RadioShack has a 5-pack of resistors for $0.99 and I've got two basses that need this mod. Thanks, L6M.
__________________ Quote:
Originally Posted by McThumpenstein I don't think the wife would buy the "I need to take off this knob and put a whole new bass under it" story. | | 
04-26-2010, 03:21 PM
|  | Supporting Member | | Join Date: Jun 2008 Location: Close to Los Angeles, CA | | Quote:
Originally Posted by FunkMetalBass Ah, okay, that's perfect then. RadioShack has a 5-pack of resistors for $0.99 and I've got two basses that need this mod. Thanks, L6M. | I misplaced the decimal point in the part you quoted, read the post again. | 
04-26-2010, 03:29 PM
| | Registered User Endorsing Artist: J.C. Basses | | Join Date: Aug 2005 Location: Phoenix, Arizona 85029 | | Can I ask how you calculated those? Either I'm estimating incorrectly, or I suck at crunching numbers (go figure that I'll be receiving a math degree w/ a physics minor soon...) 
__________________ Quote:
Originally Posted by McThumpenstein I don't think the wife would buy the "I need to take off this knob and put a whole new bass under it" story. | | 
04-26-2010, 03:40 PM
|  | Supporting Member | | Join Date: Jun 2008 Location: Close to Los Angeles, CA | | Quote:
Originally Posted by FunkMetalBass Can I ask how you calculated those? Either I'm estimating incorrectly, or I suck at crunching numbers (go figure that I'll be receiving a math degree w/ a physics minor soon...)  | It's a very simple formula:
Voltage * Current = Wattage
V= 9 or 18.
I= 0.001 to 0.002. | 
04-26-2010, 04:08 PM
| | Registered User Endorsing Artist: J.C. Basses | | Join Date: Aug 2005 Location: Phoenix, Arizona 85029 | | | The current is where you've lost me. I was just using P = V^2 / R.
__________________ Quote:
Originally Posted by McThumpenstein I don't think the wife would buy the "I need to take off this knob and put a whole new bass under it" story. | | 
04-26-2010, 04:36 PM
| | | | V=IR, so I = V/R
So,
P = V * I
is the same as
P = V * V/R
or
P = V^2/R
No reason to get lost. You two said the same thing.
I am interested in where the numbers for the OBP-3's current draw came from, though. | 
04-26-2010, 04:47 PM
|  | Supporting Member | | Join Date: Jun 2008 Location: Close to Los Angeles, CA | | Quote:
Originally Posted by FunkMetalBass The current is where you've lost me. I was just using P = V^2 / R. | See, you're overcomplicating things.
The way I saw it was that if I'm looking for a measure of wattage, I don't know how much current is being passed through the resistors in that part of the circuit, so I just looked at how much current is drawn by the preamp as a whole, which is a couple of milliamps at most.
The preamp is running on 9V or 18V, so I can just multiply by the current, and there is my maximum wattage.
FWIW, these resistors are just there to quiet the switch pop, they aren't really a part of the EQ circuit. | | Thread Tools | Search this Thread | | | |
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