P = I^2 x R
I'm doubling the current by using two supposedly identical channels in parallel to drive the loadNo, you're
Everyone has been very polite in pointing out Ohm's law and the limitations andIt's my amps.... it's my speakers.... I'm gonna get my resistors, just in case I will have a fire extinguisher.
In any event, I'm gonna get this thing cranked to 11.
Running our of power supply juice on just one channel? That's a non sequitur... come on. It's not like I'm gonna drive this to 12... only 11.
yet you insist it doesn't apply to you. You can violate the laws of man but not
physics. Good luck cranking to 11 but I predict you will only get to 9.
Get a TPA3255 amp and you'll get to 13.
G²
If you parallel the amplifiers, and you want to use twice the currrent, you need to half the load. If you use 8R speakers and you parallel two amplifiers, the power will be almost identical. If you want to increase the power of the parallel amp set up, you need to half the speaker load resistance AKA use two sets of identical speakers in parallel.
Put another way, whether I use a 12V 7.2AH battery or a 90AH 12V car battery, the light bulb will be the same damned brightness. If you want more light, you parallel more light bulbs on your larger batttery.
Current is drawn, not given.
Put another way, whether I use a 12V 7.2AH battery or a 90AH 12V car battery, the light bulb will be the same damned brightness. If you want more light, you parallel more light bulbs on your larger batttery.
Current is drawn, not given.
This is maybe the clearest way to see where your thinking is wrong:
Current is drawn, not given.
If you have the amps, and the speakers, and you don't care if it blows up, you can try:
Not knowing anything about the build of these amplifiers but knowing most amplifiers have floating outputs, the OP might be able to connect the two amps in series (-+-+) to get twice the voltage output. If the outputs don't float, one amp will short but most amplifiers have protection that will kick in.
Not knowing anything about the build of these amplifiers but knowing most amplifiers have floating outputs, the OP might be able to connect the two amps in series (-+-+) to get twice the voltage output. If the outputs don't float, one amp will short but most amplifiers have protection that will kick in.
What's the advantage over putting them in bridged mode?
Most amplifiers used to have a negative output terminal that was connected to the shield of the input. If you then connect the inputs in parallel and outputs in series, you short one of the outputs.
Most amplifiers used to have a negative output terminal that was connected to the shield of the input. If you then connect the inputs in parallel and outputs in series, you short one of the outputs.
I've seen both but more where the speaker output is differential floating - especially in car stereos.
If the amp can be set to bridged mode, there's no benefit I can see to series connection.
If the amp can be set to bridged mode, there's no benefit I can see to series connection.
Current is drawn, not given.
Now for a well needed diversion on the subject of Ohms Law.
I routinely test 9v batteries with my tongue. I can pick out feel the tingle of 7V, 8V and a fresh 9V battery. Now I can take a 9V power supply that has 100 amps behind it and I will draw no more amps thru my tongue than the 9v battery, as the resistance on my tongue does not change (with same distance apart as the battery terminals).
Now a video - not me, but shows the point (this guy acts like a bafoon, but actually an EE). My kids thought he was hilarious.
link
BTL amps quadruple the power output. You double the Voltage which doubles the current.What's the advantage over putting them in bridged mode?
Most amplifiers used to have a negative output terminal that was connected to the shield of the input. If you then connect the inputs in parallel and outputs in series, you short one of the outputs.
Abracadabra ! A 4 Ohm load with a 14V supply gets you about 5 Watts but in bridge mode,
20 WATTS! That was a big deal. I built a BTL amp in my car back in '79 before it was popular.
I worked in a stereo repair shop and snagged a Marantz power amp module from E-Waste
(didn't call it that back then), repaired it, set the gain to -1, connected the Marantz input to
the normal speaker output in the Pioneer KP-500 and connected the speaker '-' to the
Marantz output. I was using 8 Ohm speakers so the Pioneer saw a 4 Ohm load as did the
Marantz, 10 Watts into 8 ohms was a massive improvement. It's exactly the same as balanced
line audio in XLR connectors, just at the final output end. Nothing shorts if done correctly
BUT YOU NEED TO BE CAREFUL DOING MEASUREMENTS. BOTH the speaker outs, + & - are
active and your dummy loads must be isolated AND the scope inputs. Easiest is 4 channel scope
with 1 probe on each output and add the signals with scope math.
G²
Excellent analogy with the bulbs !If you parallel the amplifiers, and you want to use twice the currrent, you need to half the load. If you use 8R speakers and you parallel two amplifiers, the power will be almost identical. If you want to increase the power of the parallel amp set up, you need to half the speaker load resistance AKA use two sets of identical speakers in parallel.
Put another way, whether I use a 12V 7.2AH battery or a 90AH 12V car battery, the light bulb will be the same damned brightness. If you want more light, you parallel more light bulbs on your larger batttery.
Current is drawn, not given.
G²
Ya this guy is a riot 🙂Now for a well needed diversion on the subject of Ohms Law.
I routinely test 9v batteries with my tongue. I can pick out feel the tingle of 7V, 8V and a fresh 9V battery. Now I can take a 9V power supply that has 100 amps behind it and I will draw no more amps thru my tongue than the 9v battery, as the resistance on my tongue does not change (with same distance apart as the battery terminals).
Now a video - not me, but shows the point (this guy acts like a bafoon, but actually an EE). My kids thought he was hilarious.
link
A 4 Ohm load with a 14V supply gets you about 5 Watts
24
98but in bridge mode, 20 WATTS
You need to work on the math a bit. If you're lucky you'll get 13.5 Vp-p. Divivde by 224
98
for peak Volts and then the square root of 2 to get RMS Volts (4.77). Square that and
divide by 4 (E²/R) load to get 5.69 Watts. Quadrupling that is 22.78 Watts.
G²
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Actually the document shows both parallel and bridge mode.Here is how Quad do it with the 405
https://www.dadaelectronics.eu/uplo...Documents/Quad-405-Monoblock-Instructions.pdf
This allows a doubling of power but only if you half the impedance
The guy is an idiot BUT he got over 7 million views (NOT an idiot) so INow for a well needed diversion on the subject of Ohms Law.
I routinely test 9v batteries with my tongue. I can pick out feel the tingle of 7V, 8V and a fresh 9V battery. Now I can take a 9V power supply that has 100 amps behind it and I will draw no more amps thru my tongue than the 9v battery, as the resistance on my tongue does not change (with same distance apart as the battery terminals).
Now a video - not me, but shows the point (this guy acts like a bafoon, but actually an EE). My kids thought he was hilarious.
link
wonder how much he was paid. As a first time viewer I added to it.
G²
Has He don't want to understand Ohms Law, I would suggest He experiments and after both amps & speakers are burnt that should convince him.
To increase the current through the load you MUST either increase the voltage to the load or decrease the load impedance...merely making more current available without changing the speaker impedance will accomplish nothing, it will not increase the power output.OK... Two Amplifier Channels. Double The Current when in Parallel.
As opposed to Two Amplifier Channels. Double the Voltage when Bridged.
Either way I'm quadrupling the voltage... subject to the limits of the power supply and heat sink.
Is that so hard to understand? And it's late here.
Mike
Yep. I was assuming +/-14V rails. Shoulda noticed you were a car guy ;-)You need to work on the math a bit. If you're lucky you'll get 13.5 Vp-p. Divivde by 2
for peak Volts and then the square root of 2 to get RMS Volts (4.77). Square that and
divide by 4 (E²/R) load to get 5.69 Watts. Quadrupling that is 22.78 Watts.
G²
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