I have a RANDALL RPA-2 that is rated at 4 ohms (marked by each 1/4" jack). Since this will be my first p.a. set-up I'm a little bit lost. All the loudspeakers I've been looking at say they are 8 ohms. Is the 8 ohms rating the total between the 2 speakers which translates to 4 ohms per sie? I know that the Randall is an older 70's amp, but that is all I got to work with at this time ... so any suggestions pointing towards me replacing my amp is totally out of the question... Thank you in advance for any and all constructive help that is given....
The 4 ohm rating on the amp is most probably the minimum rating for the amp and your speakers are 8 ohm (higher than 4) so you will be fine. If you use two separate 8 ohm boxes per channel wired in parallel you will then be at the 4 ohm minimum. According to the amp manufacture that should also be fine, IE always stay above 4.
That's a so-called "4-channel" PA system, those were almost always nominally rated 4-ohm because they had an internal pair1+pair2 parallel configuration for convenience. Cool vintage box 🙂
Magnetar, is correct, the 4-ohm rating on the back of you amp is the minimum. Standard loads are between 4-ohms and 16-ohms. Generally speaking anything outside that range is impractical. Lower than 4 ohms draws too much current, and loads higher than 16-ohms receive too little power.
So, that is your working range.
Many professional PA/guitar type power amps are pretty forgiving. I've worked with a few garage bands, and they just hook up whatever they have, and somehow survive.
It's best to know in advance how to hook your PA speakers up, so you aren't trying to figure it out on the spot. By using various combinations of series and parallel, you can keep the total load in the acceptable working range.
As an example, if you have seen Marshall Stacks with four 12" speaker each, they are arranged to so the total impedance is the same as a single speaker.
Each pair of two are wired in parallel, that give you two gangs of two with a 4 ohm impedance each, then those two gangs are wired to each other in series creating (4 + 4) a cabinet impedance of 8 ohm. Most guitar players use a double stack, two 8-ohm cabinets that are plugged straight into the Amp for a combined impedance of 4 ohms.
If you eventually have a lot of speakers, you might want to make some Series/Parallel boxes. Boxes with gangs of 1/4" phone plugs that either place speaker in series or parallel. That way, you can easily create the proper combination.
This is probably more than you want to know, but here is how you combine the impedance of several speakers -
1.) When you are combining two speaker of equal size, they simply divide in parallel and add in series.
- two 8 ohm speakers per channel, wired normally (parallel)
Rt = 8/2 = 4 ohms
Actually, this hold true for any number of equal sized (impedance) speakers in parallel. Divide the value of a single speaker by the total number of speakers.
three 8 ohm speakers in parallel-
Rt = 8/3 = 2.667 ohms
four 8 ohm speakers in parallel-
Rt = 8/4 = 2 ohms
- wired in series or in a chain, they add
Rt = 8 + 8 = 16 ohms
Actually, in series, they just add
Rt = R1 + R2 + R3 + R4 + R...
a 6 ohm, an 8 ohm, and a 16 ohm in series-
Rt = 6 + 8 + 16 = 30 ohms
2.) Two speakers of unequal impedance in parallel.
Rt = (R1 X R2) / (R1 + R2)
Example: 6 ohm with 8 ohm
Rt = (6 X 8) / (6 + 8) = (48) / (14) = 3.43 Ohms
3.) Multiple speaker of unequal value in parallel
Rt = 1 / [(1/R1) + (1/R2) + (1/R3) + ...]
If you have a common calculator this is easy. Most cheap calculators have a reciprocal key [1/x] meaning ONE divided by 'X' where 'X' is the number.
So, using a calculator with the key strokes shown in square brackets [] and numbers in round () brackets, here is the sequence for 6 ohm with 8 ohms with 16 ohms -
Rt = 1/(1/6 + 1/8 + 1/16)
(6) [1/x] [+] (8) [1/x] [+] (16) [1/x] [=] [1/x]
Here is what you should see -
6 [1/x] = 0.166666666
[+]
8 [1/x] = 0.125
[+]
16 [1/x] = 0.0625
[=]
= 0.3541666666
[1/x]
= 2.824 ohms
Sorry if that last bit was boring for you.
Steve/BlueWizard
So, that is your working range.
Many professional PA/guitar type power amps are pretty forgiving. I've worked with a few garage bands, and they just hook up whatever they have, and somehow survive.
It's best to know in advance how to hook your PA speakers up, so you aren't trying to figure it out on the spot. By using various combinations of series and parallel, you can keep the total load in the acceptable working range.
As an example, if you have seen Marshall Stacks with four 12" speaker each, they are arranged to so the total impedance is the same as a single speaker.
Each pair of two are wired in parallel, that give you two gangs of two with a 4 ohm impedance each, then those two gangs are wired to each other in series creating (4 + 4) a cabinet impedance of 8 ohm. Most guitar players use a double stack, two 8-ohm cabinets that are plugged straight into the Amp for a combined impedance of 4 ohms.
If you eventually have a lot of speakers, you might want to make some Series/Parallel boxes. Boxes with gangs of 1/4" phone plugs that either place speaker in series or parallel. That way, you can easily create the proper combination.
This is probably more than you want to know, but here is how you combine the impedance of several speakers -
1.) When you are combining two speaker of equal size, they simply divide in parallel and add in series.
- two 8 ohm speakers per channel, wired normally (parallel)
Rt = 8/2 = 4 ohms
Actually, this hold true for any number of equal sized (impedance) speakers in parallel. Divide the value of a single speaker by the total number of speakers.
three 8 ohm speakers in parallel-
Rt = 8/3 = 2.667 ohms
four 8 ohm speakers in parallel-
Rt = 8/4 = 2 ohms
- wired in series or in a chain, they add
Rt = 8 + 8 = 16 ohms
Actually, in series, they just add
Rt = R1 + R2 + R3 + R4 + R...
a 6 ohm, an 8 ohm, and a 16 ohm in series-
Rt = 6 + 8 + 16 = 30 ohms
2.) Two speakers of unequal impedance in parallel.
Rt = (R1 X R2) / (R1 + R2)
Example: 6 ohm with 8 ohm
Rt = (6 X 8) / (6 + 8) = (48) / (14) = 3.43 Ohms
3.) Multiple speaker of unequal value in parallel
Rt = 1 / [(1/R1) + (1/R2) + (1/R3) + ...]
If you have a common calculator this is easy. Most cheap calculators have a reciprocal key [1/x] meaning ONE divided by 'X' where 'X' is the number.
So, using a calculator with the key strokes shown in square brackets [] and numbers in round () brackets, here is the sequence for 6 ohm with 8 ohms with 16 ohms -
Rt = 1/(1/6 + 1/8 + 1/16)
(6) [1/x] [+] (8) [1/x] [+] (16) [1/x] [=] [1/x]
Here is what you should see -
6 [1/x] = 0.166666666
[+]
8 [1/x] = 0.125
[+]
16 [1/x] = 0.0625
[=]
= 0.3541666666
[1/x]
= 2.824 ohms
Sorry if that last bit was boring for you.
Steve/BlueWizard
All of the replies have been very helpful ... even the added info that was offered up. I'm thankful for your help and have saved the added info for future reference.
After reading your question a second time, I just thought I'd point out that the impedance figures you have been quoted apply to woofers and other full range type drivers. If you have a crossover and you wire an 8 ohm woofer with an 8 ohm tweeter, the impedance is still 8 ohms. The reason is that they are each looking after their own part of the spectrum and do not overlap, therefore there is no dividing by two when in parallel.
Just to avaoid any confusion.
Just to avaoid any confusion.
With reference to what Cal Weldon was saying, and I am assuming he is responding to your statement -
All the loudspeakers I've been looking at say they are 8 ohms. Is the 8 ohms rating the total between the 2 speakers which translates to 4 ohms per sie?
This is a little confusing as you don't make a distinction between speaker cabinets, that is cabinets with one or MORE speakers in them, and just plain raw speakers
If I understand correctly, you don't need to be concerned with what is in a speaker cabinet. Just the manufacturer's rating for the cabinet as a whole.
If the cabinet has twelve 10" speakers and the manufacturer rates the total as a nominal 8 ohms, then you have to assume the manufacture did what ever he had to do inside the cabinet to make that happen.
So, as a user, you only need to consider cabinet ratings. As a builder of cabinets, you need to be concerned about the properties of the individual raw speakers.
The same is true in Cal Weldon example with a crossover network. As a user, you have to assume that whatever is inside the cabinet will take care of itself. As a user, you only need to consider the rating of the cabinet as a whole.
Now, typical impedance ratings are Nominal, meaning the speaker will present a more or less 8 ohm load. If you looked at some of the impedance graphs I reference earlier, or maybe that was another thread ...
Parts Express Project List-
http://www.partsexpress.com/projectshowcase/homeaudio.cfm
Select any speaker on the page, the from the speaker page, select 'Crossover Design', usually there will be a frequency response and impedance graph at the bottom. Select that graph for a larger view. The impedance graph is the lower one on the chart.
Notice that even with a nominal 8 ohm speaker it is possible for the impedance to drop below 8 ohm. So, if you have TWO 8 ohm speakers in parallel for a resulting 4 ohms. It is possible that the resulting impedance might dip to 3 ohms briefly. There is a bit of flex in the amps.
But if you take TWO 6 ohms speaker, one would assume the amp could handle the resulting 3 ohms; it did in the first example. But now, it is possible for the actual impedance to dip as low as 2 ohms, and that represents a huge current requirement, and the results may or may not be disastrous, but they will never be good.
So, my advice, arrange your speakers so the result is between 4 ohms and 16 ohms.
Willitwork points out that -
That's a so-called "4-channel" PA system,...
If the PA system REALLY has four output amps, then you don't need to combine (impedance-wise) the output of individual amps. The load on each individual amp can be 4 ohms.
Hope that helps.
Steve/BlueWizard
All the loudspeakers I've been looking at say they are 8 ohms. Is the 8 ohms rating the total between the 2 speakers which translates to 4 ohms per sie?
This is a little confusing as you don't make a distinction between speaker cabinets, that is cabinets with one or MORE speakers in them, and just plain raw speakers
If I understand correctly, you don't need to be concerned with what is in a speaker cabinet. Just the manufacturer's rating for the cabinet as a whole.
If the cabinet has twelve 10" speakers and the manufacturer rates the total as a nominal 8 ohms, then you have to assume the manufacture did what ever he had to do inside the cabinet to make that happen.
So, as a user, you only need to consider cabinet ratings. As a builder of cabinets, you need to be concerned about the properties of the individual raw speakers.
The same is true in Cal Weldon example with a crossover network. As a user, you have to assume that whatever is inside the cabinet will take care of itself. As a user, you only need to consider the rating of the cabinet as a whole.
Now, typical impedance ratings are Nominal, meaning the speaker will present a more or less 8 ohm load. If you looked at some of the impedance graphs I reference earlier, or maybe that was another thread ...
Parts Express Project List-
http://www.partsexpress.com/projectshowcase/homeaudio.cfm
Select any speaker on the page, the from the speaker page, select 'Crossover Design', usually there will be a frequency response and impedance graph at the bottom. Select that graph for a larger view. The impedance graph is the lower one on the chart.
Notice that even with a nominal 8 ohm speaker it is possible for the impedance to drop below 8 ohm. So, if you have TWO 8 ohm speakers in parallel for a resulting 4 ohms. It is possible that the resulting impedance might dip to 3 ohms briefly. There is a bit of flex in the amps.
But if you take TWO 6 ohms speaker, one would assume the amp could handle the resulting 3 ohms; it did in the first example. But now, it is possible for the actual impedance to dip as low as 2 ohms, and that represents a huge current requirement, and the results may or may not be disastrous, but they will never be good.
So, my advice, arrange your speakers so the result is between 4 ohms and 16 ohms.
Willitwork points out that -
That's a so-called "4-channel" PA system,...
If the PA system REALLY has four output amps, then you don't need to combine (impedance-wise) the output of individual amps. The load on each individual amp can be 4 ohms.
Hope that helps.
Steve/BlueWizard
Hi Steve,
They sometimes make mid cabinets and high driver cabinets for pro use. Each box has a bit of crossover built in. This is what Cal was probably referring to.
Hi mulliganmaster,
It's always well advised to research what you have and decide accordingly. Don't guess as this can lead to smoke and a trip to someone like me. I've repaired enough low budget band stuff over the years. Always check.
-Chris
They sometimes make mid cabinets and high driver cabinets for pro use. Each box has a bit of crossover built in. This is what Cal was probably referring to.
Hi mulliganmaster,
It's always well advised to research what you have and decide accordingly. Don't guess as this can lead to smoke and a trip to someone like me. I've repaired enough low budget band stuff over the years. Always check.
-Chris
- Status
- Not open for further replies.