Solid State Guitar Amp Forum | DIY Guitar Amplifiers

Solid State Amplifiers => The Newcomer's Forum => Topic started by: Bulb on January 16, 2019, 09:18:13 PM

Title: Fender bass cab questions
Post by: Bulb on January 16, 2019, 09:18:13 PM
Hi all, Don't know if I'm posting in the right place but here goes. I just got this SWR bass amp and with it a Fender BXR Spectrum cabinet. 2X10 (Celestions) and 1X18 (EmB). All the speakers are 8 ohm. I've read that it's impossible to wire three 8-ohm speakers and get an 8-ohm cabinet. But, when I check the cab at the jack (with the old multimeter on the free end of the cable) I keep getting 6.5-6.6 readings. Wouldn't that make it an 8-ohm cab? You can tell I'm new at this. But loving it. I'll try and attach my drawing of the wiring. Appreciate any help, comments. Peace
Title: Re: Fender bass cab questions
Post by: phatt on January 17, 2019, 06:37:59 AM
There are many ways to wire up speakers.
If the big one was 16 Ohm and the 2 smaller ones were 8 Ohm then yes that gives 8 Ohm load. After removing the wires Check the big speaker at the terminals you may find it is 16 Ohm.
Phil.
Title: Re: Fender bass cab questions
Post by: Jazz P Bass on January 17, 2019, 11:25:32 AM
If the are all 8 ohms, then the equation works out to 5.33 ohms.
Title: Re: Fender bass cab questions
Post by: Bulb on January 17, 2019, 12:54:33 PM
What exactly is the equation? The 18" is also 8-ohm. I had read that the cab should be 5.33 ohms. Am I getting the higher reading because measuring the speaker cable is an inaccurate (ie. dumb) way to determine the cab's ohms? Is it because of the wiring (splits at both terminals of the jack before going to speakers)?  Cool learning experience for me. Thanks
Title: Re: Fender bass cab questions
Post by: g1 on January 17, 2019, 02:31:39 PM
The two 10's are wired in series.  So that part is simple addition, if they are 8 ohms each then the two of them in series is 16ohms.
Now you have 16ohms wired in parallel with 8ohms (the 18).
The equation for parallel is product over sum.  So (16x8) divided by (16+8)

Maybe your meter is reading high, disconnect one of the speakers and measure it by itself to get an idea of what your meter reads for an 8 ohm speaker.  Also, DC resistance readings on a meter are usually lower than the impedance rating.  So for a calculated 5.33ohm impedance, we expect the meter to read a little less.
Also note what your meter reads with the probes shorted together.  This should be zero, if it is 1 or 2 ohms or something, you must subtract that from your measurements.