The Boominator - another stab at the ultimate party machine

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Phaedras

Phaedras

Member
Joined 2010
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uranhjort

Member
Joined 2011
if i were to build a half boominator, it would be with the two hp-10W's in parallel on 1 channel in order to get them down to 4Ohm, or would you run 1 on each channel and just have them at 8Ohm?
also, if you were to run them in parallel on 1 channel, i guess, it would work best if the signal coming into the amplifier is mono. how would you do that?
 
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Excile

Member
Joined 2010
uranhjort said:
if i were to build a half boominator, it would be with the two hp-10W's in parallel on 1 channel in order to get them down to 4Ohm, or would you run 1 on each channel and just have them at 8Ohm?
also, if you were to run them in parallel on 1 channel, i guess, it would work best if the signal coming into the amplifier is mono. how would you do that?
Click to expand...
If you read the thread, your questions will be answered :)

Saturnus has earlier recommended recommended doing it with 2 channels. Theres both advantages and disadvantages. If you search this thread for "half", im sure you get a lot of good info.
 
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skipper87

Member
Joined 2011
Saturnus said:
That how (most of) those that have built the half version have done it. You should just be aware you shouldn't have holes in the centerbracing, and you should have a port in each side if you do it like that. I keep promising myself do make a drawing of it but I never seem to get around to it.
Click to expand...

Would it be a bad solution to do it like rubennn did with his Halfinator? This solution has holes in the center bracing and just one port. Can't see how it can be a problem when it works with a full size Boominator (maybe if i'm going stereo, but I think that mono will do).
 
aaroe

aaroe

Member
Joined 2011
Question regarding LED's...
I think I've read somewhere in this thread, that one of those small 2,1V 3mm LED's will consume 20mA.

So if I connect 6 of these in series to a 12V SLA - these 6 LED's will consume 120mA, right?

But what if I connect 8 in series? - I don't expect the power consumption to be 160mA (8x20mA) - because each of the 8 LED's will output a little less light, as the average current is now 12/8 = 1,5V - am I right about this?

Can someone clarify this?

Btw, sorry for going off-topic - but its regarding Boominator bling bling :-D
 
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skipper87

Member
Joined 2011
aaroe said:
Question regarding LED's...
I think I've read somewhere in this thread, that one of those small 2,1V 3mm LED's will consume 20mA.

So if I connect 6 of these in series to a 12V SLA - these 6 LED's will consume 120mA, right?

But what if I connect 8 in series? - I don't expect the power consumption to be 160mA (8x20mA) - because each of the 8 LED's will output a little less light, as the average current is now 12/8 = 1,5V - am I right about this?

Can someone clarify this?

Btw, sorry for going off-topic - but its regarding Boominator bling bling :-D
Click to expand...

If you connect a 2,1 V LED to 12 V, then you will need to have a current limiting resistor in series. Ohms law states that 20 mA will run through the LED if you have U/I=R => (12 V-2,1 V)/20 mA = 495 Ohm resistance.

If you want to have 8 LEDs in series then the voltage drops will exceed the supply voltage. A better idea is to make 2 series of 4 LEDs, and then put these 2 series in parallel. Total current consumption will be 40 mA if you use 180 Ohm in each of the branches.
 
aaroe

aaroe

Member
Joined 2011
skipper87 said:
If you want to have 8 LEDs in series then the voltage drops will exceed the supply voltage. A better idea is to make 2 series of 4 LEDs, and then put these 2 series in parallel. Total current consumption will be 40 mA if you use 180 Ohm in each of the branches.
Click to expand...

Ok, I get the calculation.
What you are saying is, that 2 x 4 LED's in series with a 180 ohm resistor on each branch, will consume less power than 8 LED's i series?
 
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skipper87

Member
Joined 2011
aaroe said:
Ok, I get the calculation.
What you are saying is, that 2 x 4 LED's in series with a 180 ohm resistor on each branch, will consume less power than 8 LED's i series?
Click to expand...

No, just that the total voltage drop will be too high (8 x 2,1 V=16,8 V ) and the LEDs will not light up. It is possible to find LEDs with lower voltage drops, but still too high for 8 in series with 12 V.
 
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