The nitty gritty of watts and amps

jdhdiggs

The VC might take that many watts if there was a load on the woofer (enclosure). Without one, the speaker would throttle itsef mechanically. Also, ROT from the infinate baffle guys, if you don't have an enclosure, de rate your speaker 3X. to handle 900W, you would need at 2700W power handling. Beyond that, your math is right.

Also, FYI: ~742 Watts = 1 Horse power IIRC

Tour2ma

WAAAAAAAAAaaaaaaaaa...

My Sunfire is pushin' nearly 3 hp!!!

Ceruleance

Originally posted by Tour2ma
WAAAAAAAAAaaaaaaaaa...

My Sunfire is pushin' nearly 3 hp!!!

you could mow your lawn with that thing!

John K.

The form of Ohm's law defining power can be stated as PM did in his question: P=VxA(power in watts equals volts times current in amps). Although power from the wall socket is in the area of 120V, it should be kept in mind that the amplifier doesn't use it at this voltage. The transformer in the power supply section of the amp or receiver steps down the voltage typically by about 4 to 1, which then is rectified to DC and is cleaned up and briefly stored by the filter capacitors. The amp section itself then uses the power at a voltage of about 30-35V.

When playing at a comfortably loud listening level speakers use about 1-2 watts. On brief peaks 100 watts or more can be used. If for example some extreme peak called for 150 watts, the application of Ohm's law would be 150 watts equals say 30 volts times the required amperage, which would call for a 5 amp capability.

motocrossluke

all amps and recievers are not created equal even though some of the ratings are the same.Current is the main factor. here is a good example. if you take 8 AA batteries and hook them in series, you will have a total of 12 volts DC. 8 x 1.5 volts,the exact same voltage as your car battery . Does that mean that you could start your car with the 8 double AA batteries.? Of course not. These 8 batteries are not able to supply the current needed to turn such a heavy load that a car battery is able to supply. Wattage measurments alone do not tell us how much current an amplifier can produce

bknauss

But you know currenty from just about every power spec out there since they quote the resistance used. Now if you're reffering to how much current a power supply can deliver (which ultimately becomes the biggest factor in most AVRs and their total average power), that's a completely different story. A company might quote X watts per channel, but they're only measuring 1 or 2 channels. When you drive all channels, the average power out of each channel can drop dramatically based on the power supply.

AsSiMiLaTeD

Holy **** this thread is like 2 years old!!!

How do we keep digging up these old threads?

cheddar

People keep telling newbies to use the search function to dig up stuff on amp separates etc...

steveinaz

I think when you talk in terms of "more power" being better, we're talking about being able to reproduce music in lifelike "scale" without the amplifier peaking out/clipping. Scale is every bit as important as any other specification as it is related to the term "fidelity." You have to be able to recreate the musicians scale to realistic levels to be true to (fidelity) the performance in dynamics.

Of course all of this dependant on room size, speaker efficiency, etc...

sickicw

steveinaz wrote:

I think when you talk in terms of "more power" being better, we're talking about being able to reproduce music in lifelike "scale" without the amplifier peaking out/clipping. Scale is every bit as important as any other specification as it is related to the term "fidelity." You have to be able to recreate the musicians scale to realistic levels to be true to (fidelity) the performance in dynamics.

Of course all of this dependant on room size, speaker efficiency, etc...

yea, I think that is something that most people don't understand. I feel that dynamic range is by far the most important attibute that defines what makes music sound good. In fact, I would prefer listining to mp3s on a good system with plenty of power than i would a sacd on a system that limits the dynamic range. Of course, most people that have sacds also have a decent system, so this might not be a good example.

aaharvel

Wattage never sold me on my amp. It was headroom capability and +/-100 current. Put those specs on a 10 watt tube amp and my speakers would sing.

bikezappa

motocrossluke wrote:

all amps and recievers are not created equal even though some of the ratings are the same.Current is the main factor. here is a good example. if you take 8 AA batteries and hook them in series, you will have a total of 12 volts DC. 8 x 1.5 volts,the exact same voltage as your car battery . Does that mean that you could start your car with the 8 double AA batteries.? Of course not. These 8 batteries are not able to supply the current needed to turn such a heavy load that a car battery is able to supply. Wattage measurments alone do not tell us how much current an amplifier can produce

This topic was very interesting in my EE class 40 years ago. All batteries have an internal resistance. The D cell battery has more internal resistance than a car battery. That's why you can measure 1.5 volts from a battery that is dead and can't light the flash light. It's internal resistance has increased over time due to the nature of the chemical reaction going in the battery.

Let me expalin it another way. A D cell has 1.5 volts. It also has an internal reisitance called Ri. From ohms law Voltage = current X Resistance or

E=IRi
or
I=E/Ri

When you short out a D cell battery, that is connect a wire directly from the + terminal to the - the amount of current is 1.5 volts divided by the internal resistance. If the internal resistance is 0.1 ohms then you will have 15 amps going thought the wire. This makes for a very hot battery very quickly. It can explode. The internal resistance of the battery is not constant and will increase with time. Also the internal resistance will increase with heat which will reduce the current with time of a shorted out battery. By careful if you try this D cell battery experiment at home kiddies.

To summarize, the important factor in power supplies is the internal resistance, lower is better. They all have the same 30 or so volts as some else has already noted.

A side note, the most effecent power transfer from amplifier to speakers is when the internal resistance of the amplifier = the resistance of the speaker.

Does every one get it?

Ohms law, E=IR and watts W=IE are two very powerful tools used to understand all our electrical equipment. If you use a little algerbra and substitue you can also make W=IIE or W=EE/R. Have fun.

LuSh

You're wasting your time, as soon as Japan took over the consumer electronics industry and basically used numbers to market products it was game over. I find the HK number very funny indeed. Using ohm's law caculate what 45 "AMPS" of current would be outputted to an 8ohm speaker...you're looking at over 16,000 watts to the speaker...The HK spec is horse s***. As is much of the industry when it comes to numbers. Ever wonder why a data projector can be spec'd higher in contrast then a $10k Sim2?

I can't stress enough if you're passionate about Audio stick with local dealers and help support the industry. The math is there only to fool you trying to entice you to buy based on price and spec'd performance allowing large national retail chains to retail niche products.

bikezappa

Lush

That's correct, and your amp would need to output about 600 volts to make 35 amps into an 8 ohm speaker.

Proper math doesn't lie. It's the data that the manufactures give that is a lie.

Do the test yourself Polkies. Measure the RMS voltage to you speakers to determine the Watts to the speaker.

W=IE

or substituting in ohms law I=E/R, E being the voltage to the speakers and R being the speaker resistance and I being the amps to the speakers

we get

W=EE/R

Do the experiment to see how loud you can take it and then measure that voltage. You will be surprised how few Watts you are using.

dudeinaroom

120v is the power supplied to the transformer, not for amptilication. The transformer then puts out a certain amount of voltage. this is the rectified(changed to dc) and then smoothed by capacitors(which also hold reserve energy for peaks) so now you have some thing like +35, common/ground, and -35 volts which feeds the amplifier circuit. which might be what the haron would use. but in the cheap aiwa it might be a stetup with +/- 12 volts or even just 12 volt and ground. an amp cannot put out more voltage/current than it is fed and will waste energy as the by product know as heat. so thats part of the reason why they don't have the same out-put and antoher part of it is the design of the amlifer ciruit which is a whole other can of worms

Later polkies