Could someone explain the amps to ohm relation?
Was hoping some could help me understand how this works ? Why if I buy a combo, lets say a GK 212 combo rated at 500amps does it only put out 350amps to the 2x12 speakers but will put out all 500amps if a 8ohm cab is connected and will be cut in half if another 4ohm cab is hooked up or even puts out less if a 2ohm cab is connected. It`s got me all confused. LOL thanks
Not that simple. Do a search here. There are tons of threads on the subject and no need to rehash it all again.
It's Watts not amps.
Try to match the loads.
Higher speaker Ohm rating is not too bad. (8 ohm cab with a 4 ohm amp=OK)
Too low is bad (2 ohm cab into a 4 ohm amp= no no)
Other than that, do a search. There is a lot to read here.
Edit to add: It's a GK/amp question so B-string will be swinging by any time now to answer it using about 10 words and completely resolving the issue. Ha!
And a G-K MB212 combo will not take an extension cab, but will put out all 500 watts to the internal 2X12".
Amps is current - like how much water is flowing in a river
Volts - is the force or pressure - like how hard the water is pressing against you.
Ohms is resistance to the flow - like a dam with a hole in it. The smaller the hole, the higher the resistance
Watts is Power - the combination of all of the above that tells you how much water is coming out of that hole depending on the pressure behind it, the size of the hole and the amount of water available.
By using water as a simile, you can see how lowering the resistance by making the hole larger, or by making 2 holes will increase the output (power). If the 2 holes are the same size, it doubles the flow (power) as long as there is enough water behind the hole. This is the same as adding a second 8 Ohm cab, reducing the resistance to 4 ohms, doubling the output power.
If the hole gets too large, there is not enough water flow and there is no output. This is the output limit of the amp.
Very nice summary Bassamatic. :)
In mathematical terms :
Volts = Amps x Ohms
Watts = Volts x Amps
Watts = Amps squared x Ohms
Watts = Volts squared ÷ Ohms
Two more important things to note :
In a series circuit all components get the same current (amperage) but the voltage is divided.
In a parallel circuit all components get the same voltage but the current is divided.
Ok, here's an anology. Let's say you have a 20 foot long garden hose. You turn on the water spigot, and there is a certain amount of pressure at the spigot, at the entry to the hose. This pressure is analogous to voltage, measured in volts. Some of the pressure is used just forcing the water down to the end of the hose, and is sensed at the spigot as back pressure, analogous to resistance/impedance. Any remaining pressure beyond what it took to force the water down the hose actually ejects water out of the hose. The amount of water ejected per second is analogous to current, measured in amperes, often abbreviated as amps.
So watts, the amount of power, is voltage times current (volts x amps). Let's say you were washing dried on dirt off of your car. The higher the wattage, the more work you could do washing dirt off. A low pressure stream of a lot of water would do as much cleaning as a high pressure stream of less water. Same amount of wattage, same amount of work. But for given amount of water (current), a higher pressure (voltage) would clean more dirt, and would be a higher wattage. Same the other way, for a constant pressure, the more water you pump out at that pressure, the more dirt you clean.
Speaker cabinets connect to the amp in parallel. In our hose analogy, this would be like getting a Y adaptor at your spigot and running two 20 foot hoses side-by-side. Can you see how the back-pressure, analogous to resistance (or impedance) would be lowered by this configuration? It would be almost the same as making the one hose twice as big in cross-sectional area, so water would flow more easily. So for the same spigot pressure, you would get almost twice as much water to flow through the combination of both hoses, so you could do almost twice the amount of work- almost twice the wattage.
Unlike a spigot, though, as the back-pressure decreases in an electric amplifier, instead of the voltage (pressure) dropping, it stays the same and more and more current is output, and eventually the amp overheats and burns up, because it is not able to pump out current with 100% efficiency. The work lost to this inefficiency becomes heat. This is why the amp puts out more and more wattage as the impedance goes down until it goes too low and then the amp either shuts down to protect itself or burns up and dies.
Oh bassamatic, you beat me to it with the analogy!
Good Job HolmeBass. Nice to see that someone else understands how it really works! I think that the relationship to water flow and pressure makes it a lot easier for most to understand.
Thanks guys , the whole golf ball thru the garden hose thing does really help, sorry that was another forum but I do get it a little better now. Thanks..Let me ask anther question ? the cab I bought is a GK MB210 combo not a 212(sorry). So, the combos impedance is 8 ohms ?? and if you add another 8ohm cab it changes the total impedance to 4ohms allowing the 350w going to the two 10s increase to 500w and split evenly at 250 to each cab ? did i get it right ?
Yes you did. Not all that hard was it? :)
Problem with all the flow stuff, is its AC, so it doesn't work like that and everything is dependent on frequency, which is why watts end up not a very useful way of figuring how loud your amp is.
Just don't get hung up on the Watts. You cannot hear moderate changes in power. If you double the amp power, the SPL increased from the speakers is less than 1 dB difference - really! You just can't hear that.
You will get double the sound level (SPL) by adding a second cab because you double the speaker area a the same time as you increase the power.
The point I am making is that 3dB in amp electrical power does not create 3 dB in SPL increase. Amp power and acoustic power are way different animals. Too many people believe that they are the same and then end up surprised and disappointed when their 500 Watt amp doesn't sound any different than their 350 Watt amp when using the same speakers.
When the results don't match up to the mathematical defintions ther are sound reasons. Better to idscuss the reasons than try to restate the maths incorrectly.
Every time I think I got it these conversations always take a nose dive.
2a.) You can add an 8ohms (or 16 Ohms) extension cabinet.
2b1.) With an 8ohms extension cabinet the impedance will be 4 Ohms total, so the amp makes 500 Watts max.
2b2.) As both cabs (combo/extension) have 8 Ohms, both get the same power - here 250 watts max. ...
2c1.) With an 16ohms extension cabinet the impedance will be 5.333 Ohms total, so the amp makes about 450 Watts max.
2c2.) The impedance of the extension cab is the double of the combo impedance. So, the ext. cab will get half of the combo power. With 450 watts max. @ 5.333 Ohms, the combo gets 300 watts max, the cab gets 150 watts max.
2d). A 4 Ohms extension cab is not possible because the total impedance would be 2.667 Ohms - and this is below the minimum impedance of the amplifier.
Your nomenclature above is incorrect.
It is "350 watts" not "350 amps". My arc welder is only a 50-amp device... :D
Bass amplifiers are voltage devices.
Voltage is what moves the speakers and makes the noise.
Amperage is the limiting factor, being the amount of amps the hardware can deliver within its power rating.
The power rating is an indication of how much amps x volts can be delivered into a given load.
Your combo is rated at 500 watts into 8 ohms.
The electrical components are rated to produce 63.2 volts for this load with a current draw of 7.9 amps. (W = V*A)
This is the roundabout way of telling you the amp has a 7.9 amp current limitation.
Amperage is the limiting factor.
When you add a 2nd speaker and cut the load impedance to 4 ohms, the 7.9 amp limitation still applies.
You have to reduce the voltage to 44.7 volts to keep the current draw within limits.
This translates to 353 watts at 4 ohms. (353 = 44.7 x 7.9)
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