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My understanding of amplifiers

One more quick point: the power rating on an amp is not related to the voltage it produces. Most home audio amps are, from what I gather, somewhere around 30dB gain. (I think THX standard is 29dB.) So, a higher powered amp will not produce a higher voltage at the output terminals. Rather, a higher powered amp can produce more current to keep the voltage at the desired level. Remember again Joule's law: P = I * V. Voltage (V) isn't changing from one amp to another (at least not that much) so higher power (P) means higher current (I).
 
Heeman likes analogies, and so do I. So here's one that I'm just making up now as I type this. (More of a DC analogy than AC... but hey.)

Think of a speaker like a big tank of water. The water is the current. The level of the water in the tank is the voltage. The tank has a hole at the bottom where water flows out to do some work: this is the current/energy used by the speaker to create sound. The word "impede" means "to block" - the higher the impedance, the more the speaker blocks the current - the more a plug in the water tank's hole blocks the water from flowing out. And vise versa: a low impedance plug in the hole will not block as much water from flowing out, the water flows faster.

The power supply is water flowing into the top of the tank from a supply hose. The goal of the supply hose is to keep the level of the water (voltage) at the desired height, while water is flowing out the bottom. If there is a low impedance plug (speaker) at the bottom, more water will flow out faster, meaning the supply hose has to work harder to refill the tank (more current is needed). If more water is flowing out the hole at the bottom than can be piped in by the supply hose, then it will be impossible to get the water to the desired level (the amp will clip because the current is inadequate to maintain the desired voltage).

How's that?
 
If a perfect power amplifier is rated at 200 watts into 8-Ohms, you connect that amplifier to a speaker with a nominal impedance rating of 4-Ohms, the perfect amplifier would then be capable of delivering 400 watts. What would happen to this rating if this same amplifier were connected to a speaker with a nominal impedance of 16-Ohms?

Rope
 
PaulyT said:
heeman said:
The load (speaker) impedance is directly related to the signal from the amp to the speaker. The speaker has no impedance if there is no signal. Yes, the speaker has a nominal Impedance Rating of 8 or 6 or 4 ohms, however that is just a basic rating at a specific frequency.

The signal from the amp to drive the speaker is a complex signal that correlates directly to how the load (speaker) impedence changes.

Yup, the impedance of a speaker is highly dependent on frequency and the spec rating is an approximate average. Some of the links in my posts above talk more about this, why speakers are like this, and what speaker designers do to compensate to some degree.

And of course, even to talk about a single static voltage at the output terminals is a simplification, even if there's only a single frequency; the nature of music and sound and hence electronic audio signals is that they are waves. An audio signal oscillates in a complex way between positive and negative voltage.

FYI that's why volt meters have "DC voltage" and "AC voltage" settings, they are not measured the same way, at least not in a time-average sense.


A DC Potential/Voltage has no relation to time and there for no frequency or sound.

An AC Potential/Voltage typically measured in Hertz (Hz) has a sound component. For instance an A (440Hz) note on an instrument will oscillate or have a frequency of 440 cycles per second. The domestic AC Voltage in our homes is (60 Hz); 60 cycles per second. Human hearing (I know this info is basic) is supposed to be 20 Hz - 20,000 Hz. Most of us can not hear the total spectrum.

So now think of an orchestra or jazz or rock band or a choir. Many notes (frequencies) all at once. Our amps and speakers must properly process (maybe not a good word) the signal and produce a sound that sounds just like the origional. Also, think of a movie on your HT and all the different frequincies there.........

Pretty complex to say the least?
 
PaulyT said:
Heeman likes analogies, and so do I. So here's one that I'm just making up now as I type this. (More of a DC analogy than AC... but hey.)

Think of a speaker like a big tank of water. The water is the current. The level of the water in the tank is the voltage. The tank has a hole at the bottom where water flows out to do some work: this is the current/energy used by the speaker to create sound. The word "impede" means "to block" - the higher the impedance, the more the speaker blocks the current - the more a plug in the water tank's hole blocks the water from flowing out. And vise versa: a low impedance plug in the hole will not block as much water from flowing out, the water flows faster.

The power supply is water flowing into the top of the tank from a supply hose. The goal of the supply hose is to keep the level of the water (voltage) at the desired height, while water is flowing out the bottom. If there is a low impedance plug (speaker) at the bottom, more water will flow out faster, meaning the supply hose has to work harder to refill the tank (more current is needed). If more water is flowing out the hole at the bottom than can be piped in by the supply hose, then it will be impossible to get the water to the desired level (the amp will clip because the current is inadequate to maintain the desired voltage).

How's that?


Please forgive ( I guess I owe you a bottle of bourbon) :happy-smileygiantred:

I would think of the water as the voltage and the amount of flow the current. The speakers impedance is always changing therefore the hole in the bottom is constantly changing (opening and closing at different rates).

I found this to explain most of the analogies with respect to Voltage (DC) and Water.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html
 
Rope said:
If a perfect power amplifier is rated at 200 watts into 8-Ohms, you connect that amplifier to a speaker with a nominal impedance rating of 4-Ohms, the perfect amplifier would then be capable of delivering 400 watts. What would happen to this rating if this same amplifier were connected to a speaker with a nominal impedance of 16-Ohms?

Rope

I get the feeling that this is a trick question and you're looking for something other than "100 Watts" as the answer?

The amp rating doesn't change when you plug in different speakers. Ideally what the ratings tell you is, for example, whether the amp is able to produce enough current to double the power if the impedance is halved. E.g. if it can do 100W @ 8ohms, can it do a full 200W @ 4ohms? 400W @ 2ohms? For an ideal amp the answer is yes. But for a real amp, as the impedance drops, at some point the current limitations prevent it from reaching the ideal power output.

I've never seen a speaker rating claiming anything about going the opposite direction, to higher impedance. Presumably, if it's capable of 100W @ 8ohm, it will be capable of 50W @ 16ohm because the current requirement is less.
 
heeman said:
PaulyT said:
Heeman likes analogies, and so do I. So here's one that I'm just making up now as I type this. (More of a DC analogy than AC... but hey.)

Think of a speaker like a big tank of water. The water is the current. The level of the water in the tank is the voltage. The tank has a hole at the bottom where water flows out to do some work: this is the current/energy used by the speaker to create sound. The word "impede" means "to block" - the higher the impedance, the more the speaker blocks the current - the more a plug in the water tank's hole blocks the water from flowing out. And vise versa: a low impedance plug in the hole will not block as much water from flowing out, the water flows faster.

The power supply is water flowing into the top of the tank from a supply hose. The goal of the supply hose is to keep the level of the water (voltage) at the desired height, while water is flowing out the bottom. If there is a low impedance plug (speaker) at the bottom, more water will flow out faster, meaning the supply hose has to work harder to refill the tank (more current is needed). If more water is flowing out the hole at the bottom than can be piped in by the supply hose, then it will be impossible to get the water to the desired level (the amp will clip because the current is inadequate to maintain the desired voltage).

How's that?



I would think of the water as the voltage and the amount of flow the current.

Umm.... yes, that's pretty much exactly what I said, isn't it? (highlighted in red)


The speakers impedance is always changing therefore the hole in the bottom is constantly changing (opening and closing at different rates).

Yes the frequency dependence on impedance is hard to fit into my analogy, that's why I said it's more of a DC than an AC analogy.
 
I took a quick look at Parasounds' website to see which amplifiers are actuallly rated at 2 x the 8 ohm rating at 4 ohms.

Only the JC 1 has this rating:

Continuous power output, 20Hz - 20 kHz:
400 watts RMS x 1, 8 Ω
800 watts RMS x 1, 4 Ω
1200 watts RMS x 1, 2 Ω

and current rating of this:

Current capacity:
135 amperes peak


The others fall short of the 2 X power rating from 8 - 4 ohms.

I do not see any spec's for 16 Ohm. Pauls explaination should theoretically be correct.
 
Holy shit, that's a beast!

jc1_angle_792.jpg


Net weight: 64 lb. (for one channel!!!!)

Input sensitivity:
1 V for 28.28 V, THX Reference Level (ok that agrees with what I said earlier about most amps having similar gain)
 
PaulyT said:
heeman said:
PaulyT said:
Heeman likes analogies, and so do I. So here's one that I'm just making up now as I type this. (More of a DC analogy than AC... but hey.)

Think of a speaker like a big tank of water. The water is the current. The level of the water in the tank is the voltage. The tank has a hole at the bottom where water flows out to do some work: this is the current/energy used by the speaker to create sound. The word "impede" means "to block" - the higher the impedance, the more the speaker blocks the current - the more a plug in the water tank's hole blocks the water from flowing out. And vise versa: a low impedance plug in the hole will not block as much water from flowing out, the water flows faster.

The power supply is water flowing into the top of the tank from a supply hose. The goal of the supply hose is to keep the level of the water (voltage) at the desired height, while water is flowing out the bottom. If there is a low impedance plug (speaker) at the bottom, more water will flow out faster, meaning the supply hose has to work harder to refill the tank (more current is needed). If more water is flowing out the hole at the bottom than can be piped in by the supply hose, then it will be impossible to get the water to the desired level (the amp will clip because the current is inadequate to maintain the desired voltage).

How's that?



I would think of the water as the voltage and the amount of flow the current.

Umm.... yes, that's pretty much exactly what I said, isn't it? (highlighted in red)


The speakers impedance is always changing therefore the hole in the bottom is constantly changing (opening and closing at different rates).

Yes the frequency dependence on impedance is hard to fit into my analogy, that's why I said it's more of a DC than an AC analogy.


2 statements confused me "the water is the current" and "the level of the water in the tank is the voltage"

I learned that the water is the Voltage and the flow of water is the Current.

Think of measuring the Voltage at one of your outlets in your home. The voltage is there and is approx. 120V. There is no current flow, until you connect a load. You basically connect the circuit and get Flow, which is the current.

If I go outside and go to my water spicket, there is alway water available, but no flow unless I open the valve (flow). If I connect a sprinkler I now have impeaded flow and pressure (current).

Maybe we are saying the same thing?????

This has been fun :eusa-clap:
 
Yes, we are saying the same thing. I should have just said "the water flow is the current." Right, putting it in writing is the hard part; it all makes sense in my head, but that doesn't mean others will get what I'm saying.
 
I wasn't going to do this however.............

I looked a Krells' website and found that many of there amps double the output power from 8-4 ohms, however the output voltage is much higher and the peak current is lower (but that makes sense).

We have all learned that it is all about the Voltage, right????

So what is more expensive to make a higher Voltage Amplifier or a higher Current Amplifier?

What is better in the Audio and HT application or does it not matter??

We also learned that the higher the Voltage, the less current we need for the same load.

Hummm............ :think:

The Parasound JC1 is $3825

I was unable to find a price for the Krell Evloution One Mono Block amp rated at 450W and 900W; 8-4 ohm.
 
PaulyT said:
Rope said:
If a perfect power amplifier is rated at 200 watts into 8-Ohms, you connect that amplifier to a speaker with a nominal impedance rating of 4-Ohms, the perfect amplifier would then be capable of delivering 400 watts. What would happen to this rating if this same amplifier were connected to a speaker with a nominal impedance of 16-Ohms?

Rope

I get the feeling that this is a trick question and you're looking for something other than "100 Watts" as the answer?

The amp rating doesn't change when you plug in different speakers. Ideally what the ratings tell you is, for example, whether the amp is able to produce enough current to double the power if the impedance is halved. E.g. if it can do 100W @ 8ohms, can it do a full 200W @ 4ohms? 400W @ 2ohms? For an ideal amp the answer is yes. But for a real amp, as the impedance drops, at some point the current limitations prevent it from reaching the ideal power output.

I've never seen a speaker rating claiming anything about going the opposite direction, to higher impedance. Presumably, if it's capable of 100W @ 8ohm, it will be capable of 50W @ 16ohm because the current requirement is less.

No, not a trick question, and your answer (100 watts) is exactly what I expected.

Now, lets assume the amplifier is rated @ 1 watt into 32-Ohms and connect a speaker that's nominal impedance rating is 600-Ohms.

Rope
 
heeman said:
I wasn't going to do this however.............

I looked a Krells' website and found that many of there amps double the output power from 8-4 ohms, however the output voltage is much higher and the peak current is lower (but that makes sense).

That's confusing. I don't know what they mean by "output voltage"... note that the gain on their amps is still in the 25-30dB range.

Ah, I think I see: using the same formula I've posted before, P = V^2 / R, note that for the "One" model, 450 = 60^2 / 8 - in other words, their RMS voltage (60V) is the V in that equation, and it's simply a restatement of the power specs (450W @ 8ohm). Same with the 900e: 85^2 / 8 = 903 (close to the 900W @ 8ohm spec). But it sounds cooler. ;)

Note that V(rms) = 0.7 * V(peak). So taking again the "One" model, 60 Vrms ~= 0.7 * (170/2). (where Vpeak is the highest + or - voltage, which is half the peak-to-peak voltage of 170V in their spec)

:text-link:
 
Rope said:
Now, lets assume the amplifier is rated @ 1 watt into 32-Ohms and connect a speaker that's nominal impedance rating is 600-Ohms.

Ok. So if the (ideal) amp is putting out the full power it's rated for, it'll do ... .5W/64ohm, .25W/128ohm, .125W/256ohm, 0.0625W/512ohm... so something on the order of 0.06W into 600ohm.

Your point?
 
Sorry, that was rude, and I apologize. What I should say is, ok you're clearly moving towards some example in the headphone world. I'm listening... what's next?
 
Just a quick note on the Peak Current Ratings of the Parasound and the Krell Single Output Amps.

Krell = 57 Amps Peak

Parasound = 135 Amps Peak
 
I'm getting Dizzy! :confusion-seeingstars:


Which is pretty much normal! :shock:



Dennie
 
Following where I think Rope may be going, I found this on some random googled site, but it says exactly the same thing I mentioned before: that with high impedance headphones of similar sensitivity, you need higher voltage - to turn the amp up - to get to a decent SPL level. And that some amps may not be able to produce that voltage, so your headphones would never play loud enough. But the current through the high-impedance headphone is less. It is not a current limitation that prevents a lesser amp from driving them, but a voltage limitation.

That is the opposite from a low-impedance headphone, where current is more likely to be the limiting factor.


:text-link:

The UR20s also, by the way, have only 32 ohm impedance, which is important when you want to run headphones from equipment with weedy headphone outputs. The crummy headphone sockets in most home and portable stereo gear can't deliver a whole lot of power (by headphone standards - most headphones need a fraction of one per cent of the power of a loudspeaker), and often have serious voltage and current limits, too.

A strong headphone amp, such as you'll get in some quality hi-fi gear, and in pretty much any dedicated headphone amplifier, can drive all but the very strangest headphones as loud as anyone needs. If you've got headphones with very high impedance, though, you'll need a lot more output voltage to drive them to reasonable volume. More than a few high-end headphones have 600 ohm impedance, for instance; most headphone sockets can't get much volume out of those at all.

Voltage equals current times resistance (impedance is the more complex AC incarnation of resistance, but the simple DC version will do for now, and reduce the number of readers' heads that go round and round), and power equals voltage times current. So to drive a 32 ohm headphone transducer to, say, five milliwatts (mW; 0.005 watts), you need 0.4 volts. That'll result in 12.5 milliamps (mA; 0.0125 amps) of current flowing (as long, once again, as you ignore the Nasty Math that's involved in real world multi-frequency AC power calculations, and also as long as you remember your algebra. Otherwise, the numbers may turn out quite differently).

Step up to 600 ohms, and now five milliwatts requires 1.73 volts, with less than three milliamps of current flowing.

If the 600 ohm headphones are much more efficient than the 32 ohm ones (efficiency is the amount of noise you get per milliwatt; the UR20s are rated at a better-than-OK 97dB from one milliwatt input) then you'll be able to play 'em pretty loud even if your headphone output only goes up to 1.5 volts, but if the efficiency figures are much the same (as they often are, these days) then the 32 ohm 'phones will play considerably louder from a voltage-restricted headphone output.

12.5 milliamps isn't a great deal to ask for from practically any headphone socket, but more than 1.5 volts may well be; small battery powered devices, in particular, commonly don't go that high, at least not without nasty distortion. For them, low impedance headphones are definitely the way to go, particularly if you're buying at the bargain end of the market and so even a cheap add-on headphone amplifier is totally not an option.
 
PaulyT said:
Yes, we are saying the same thing. I should have just said "the water flow is the current." Right, putting it in writing is the hard part; it all makes sense in my head, but that doesn't mean others will get what I'm saying.

Ok, I often have tried to explain to my buddies that have inquired before making purchases and I am simple folk so I explain it in terms that I can relate to.

I explain it as water flowing through a fire hose. The pressure in the hose is equivalent to voltage (even most fire fighters think pressure is most important, but they are wrong) more to follow. The flow measured in gallons per minute (gpm) in the hose is equivalent to current. And the friction loss in the hose is similar to impedance.

Pressure is important so the fire stream can reach the seat of the fire, but GPM is what puts out the fire (you have to put enough GPM on the fire to overcome the BTU's of heat energy being produced). Friction loss is the loss of potential energy when moving the water through the hose.

So, kinda sorta:

Voltage= Pressure
Current= Flow or GPM
Impedance= Friction Loss

And, Pauly's explanation about trying to keep the water tank full visits water supply, which is similar to the recent "what do you feed your system" thread. ATI used to make an amp that required a dedicated 30 amp circuit.


But, I really prefer the even simpler approach and now look at it like this. Dude from Parasound said wight is good so, at nearly 75 lbs, my ATI 1505 is good. :handgestures-thumbup: :handgestures-thumbup: :handgestures-thumbup:
 
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