The Boominator - another stab at the ultimate party machine

[uranhjort]

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?

[Excile]

Originally Posted by uranhjort

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?

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.

[skipper87]

Originally Posted by Saturnus

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.

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).

[craler]

I did my half boominator in stereo and with two ports. Hard to say if I would ever hear a difference between stereo and mono, maybe if I played chiptunes or something like that.

Check out pics of it here in the boominator facebook group.

[aaroe]

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

[skipper87]

Originally Posted by aaroe

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

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]

Originally Posted by skipper87

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.

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?

[skipper87]

Originally Posted by aaroe

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?

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.

[aaroe]

Originally Posted by skipper87

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.

Ah ok, I get it :-)
I actually thought that 1,5v (12v/8) was enough for the LED's to light up.

[Phaedras]

Originally Posted by aaroe

Ah ok, I get it :-)
I actually thought that 1,5v (12v/8) was enough for the LED's to light up.

Use this calculator for easy calculations on resistors and LEDs in parallel/series arrays

LED series parallel array wizard