The tube with a narrow end is the perfect example! Let's say you have a pump putting out X pressure of water on the wide side and the water at the narrow end, reaches 1 m distance. If the narrow end increases its diameter, the water will now reach a shorter distance. So, in order to reach the same 1 m, you need a pump with more pressure, hence you need more power to achieve the same result.
Hi Paul, please don’t ever stop making videos on those topics and how all that plays a part on how amplifiers works, making amplifiers I realize nearly everyday that even if you already have a great sounding installation or know how they function you always end up learning new things and information that in the end play a lot in any setup ! You’re a mine of information thanks :)
Wattage rating indicates "capacity". For a fixed voltage, provided the amp can source the current (maintain the voltage at the output) meaning negligible internal resistance, a 4 ohm speaker will take more current and thus dissipate more power ( w = I* v = I*I*R). Giving information about how the amp delivers power to two different impedance speakers says something about the amp's ability to handle changes in a given speaker's impedance as music plays. Some manufacturers gave average wattage and peak wattage ratings to help with this, especially if they had internal "soft clipping" like the NADs.
The explanation that we are comparing the two potentials of the amp with the speaker in between instead of between the amp and the speaker just turned a huge light on. Thanks Paul!
This just appeared on my youtube feed and I watched it all the way through. Thanks for explaining the issue clearly. As a power tool product developer, I focused more on the mechanical and human factors aspects until I hired an electrical engineer to develop some unique speed controls. I learned a lot in a hurry about what you just said. In my job I also learned about how electricity flows out of your 120V wall outlet through a circuit and into an electric motor. In your example we're looking at a speaker instead of a motor. Your amp and speakers are doing a lot more work than you think!
Paul, I’m watching your videos for … I don’t know exactly, a year? Two? I can’t tell 😄 and im not completely green and yet I find your videos absolutely bonkers. I wish you many years of educating dudes of how their blocks work 👍
Example: In smaller ROADS the vehicles must go faster in a single direction, Large roads the vehicles can go slower but at both directions. At the end of the day the smaller road will need more EFFORT and attention to mantein and prevent accidents than the bigger road. Road = Resistance Effort = Wattages Thanks from Mozambique...
Paul, your oratory skill are exceptional. I don't think I've heard a professional speaker (and I've heard many), be it university lecturer, announcer, actor, news reader, et al, talk with such engaging clarity rendering difficult topics/concepts accessible for us non-experts. No ums or ahs or slips, no hypes, no agenda, no inconsequential tangents. You address each subject directly, stick with it until the end where you still have my undivided attention! If you ever come to London, please let me know as I'd love to invite you to give a talk at one of UCL's Chemical and Physical Society evenings, here in down town Bloomsbury. Martin
Thanks Paul. After reading the comments here, I like your description much better. When you "get it" you "get it", maybe. I like your way of using for instances, like things getting hot & batteries blowing up. Talking about air as an insulator; I like one of my descriptions. While being trained at the Electric Company we had to gear up before getting within 14 feet 6 inches of certain equipment. If you didn't you became a circuit part. If you could possibly run 35 feet away, not 14 feet 6 inches, it might let you go. You might get a clue of the danger beforehand, like your hair standing up, but you might not. Consider that little voice your guardian angel.
If you're riding a bike with gears (speaker resistance) and want to maintain 10mph (output wattage). In a small gear, you don't have to pedal as fast to maintain 10mph (amplifier doesn't have to output as much power to drive the speaker). Shift into a bigger gear, you have to pedal much faster to maintain 10mph (the amplifier has to give more power to maintain same speaker level).
Hey, I got it! I really got it! Hot dang Paul, that was a great explanation. Thanks for doing what you do. Your videos most always put a smile on my face. ;)
Thanks Paul, I think finally my lightbulb has gone on with enough resistance so that my brain does not shortcut 👍 I think I have grasped the concept. Now I also understand why the electromagnet my brother and I built when we were young, could not be powered by my model train transformer but worked with a battery. It got very hot but fortunately never exploded.
Paul please understand this, I might get tired of it if you made about this topic about 1000th time …. But even then I’ll keep watching it. Love your passion, love your experience, love your calm demeanour. Keep it coming. Love from Pakistan 🇵🇰
Excellent video paul. You just accidently agreed with saying.any current limited or adjusting voltage rail voltage due to imp. Is wrong modern bs.. yes your correct. Any amp that cant follow ohms law cant faithfully deliver the source correctly.. YES OHMS LAW tells the truth. Ill go further. Any amp that the voltage rails move ANY..no matter what load is compromised.. GOOD JOB.. PAUL
The water pressure is the voltage, the amount of water is the amperage/current. Lower resistance/larger diameter tube can "draw" a higher volume of water than a tube with a smaller diameter opening. And the wattage is the water pressure (voltage) multiplied by the water volume (amperage/current). So the combination of more water pressure and water volume of a larger diameter tube is more wattage.
In college I remember my professor saying "It takes voltage to push amperage through resistance" (Ohms law E over I x R) It always made sense to me as the lower the resistance, the easier for the amps to flow. That's why power doubles (in reality its less) when speakers resistance is cut in half by wiring in parallel. If wired in series then its the opposite. Be careful when wiring in parallel as home systems cannot normally handle any speaker 4 ohms or >. Unlike car amps, which was made to run much lower (I had a US amp that could run @ 1/2 ohm) and those was called cheater amps. They allowed you to cheat in SPL contest, where you were limited to a 50 watt or 100 watt amp. Bur with a cheater, you could wire those subs in parallel and get extreme power out of an amp! I had a friend that used the leads of a similar amp weld with it!! Those HCCA's back then was amazing!
Amplifiers work at a maximum output voltage and the amperage changes depending on the speakers resistance. Remember these electrical math formulas. volts x amps = watts. volts ÷ ohms = amps For example an amp has amaximum voltage of 100 volts. At 1 ohm 100 amps would flow and that would give you 10,000 watts of power. At 2 ohms your power will cut in half to 50 amps or 5000 watts of power. At 4 ohms 25 amps of current or 2500 watts of power. At 8 ohms 12.5 amps of current or 1250 watts of power. And 16 ohms would further cut that in half to 6.25 amps or 625 watts. That's why when reducing the resistance the amplifier increases power. However your amplifier has a maximum amperage it can support also before it blows up or goes into clipping based on the internals and power being supplied to the amp.
a simple way to think of it is the speaker's resistance acts as a barrier that the amp's power must pass through to reach and drive the speaker. a speaker with high resistance will pose a greater barrier thus less overall wattage makes it past to drive the speaker. conversely a low resistance speaker puts up less of a barrier to the amp's power, enabling it to deliver more watts to the speaker.
The lightbulb went off, great explanation, very helpful and thank you for sharing! Now this would lead me to the question in which I think I know the answer. If you have a lower wattage amp, you probably would want higher impedance speakers. Of course it tells you this in the manual and I get that, but if you did not know, you would probably want to start out with higher impedance speakers.. right?
Hahaha .... I only watched this to see how well you're doing in explaining this topic .... you did well .. in the end 😂😂... maybe take 3 ... bit later ... I don't know why so many people have such difficulty understanding this simple subject .... I truly think more people should be introduced to Ohm's Law and the basic Power formulae.... I'll give you 8/10 for this lol 😝.... keep trying ... I love your sense of humour... so irreverent too ... 😝
Good explanation….. now people just need to realize that the ohm of a speaker is just a calculated average over industry standard frequency range. Most 8ohm speakers actually dip down to 4ohms or possibly lower in frequencies 100-300hz
Great Explanation!. I have a few questions remaining. If we have a music signal with content on all the music spectrum. What is the real impedance the amp actually "sees" from the speaker considering the speaker has an impedance and phase curve? why bass content requires more power? Is is because on lower frequency the impedance is lower too? If the impedance of some frequencies are greater (or lower) than others. Would those lower impedance frequencies sound louder and unbalance the sound? Why even if you probably don't use more than 10 watts (because he speaker sensitivity is high) at home a more powerful amplifier sounds more dynamic and engaging than a less powerful one with the same speakers? It is because the more powerful has faster slew rate?
I think its important to watch what amplifier is deliver in lower impdence loads, no amp will struggle at 8 ohms, with 4 and 2 ohms loads its a different story, i run a Hegel h390 through b&w 702 s2 speakers, and it drives those speakers easily, its more than comfortable with 4 and 2 ohms loads as well
Speakers have a crazy dynamic resistance curve based on frequency. Current drive under duress is what's needed, this was proved by NAD over 40 years ago.
Very well explained Paul. Love your show. There used to be a formula for figuring out how many watts you need from a amp for a given size room. I can’t seem to find it anywhere. It went cubic feet of a room divided by ? Equal watts you need for that room. Anybody help me with this. Thanks
One way is begin with a nice powerful class D amp with an affordable preamp. Whatever speakers you use will work. Just don't melt or scorch anything turning it up too high. Next acquisition finetune speaker size if feasible.
The way I understand it is the lower the impedance, the lower the resistance. The lower resistance lets in more power. 4 ohm load draws more power than 8 ohm load.
Most people don't realise the watts are a consequence of load. They often think the watts are always there or increase with volume even in open circuit.
You jus ligheten up my "life" Paul!!!! Thank YOUR pacience explaining this again and again for we, people less experinced with the physics, especialy with the audio P.S. Question , it might be as stupid as the subject, or even worse, on the headsets its the same?
Ok, as resistance drops p/v=v/r the wattage relationship has to increase. Got it. But you said resistance is between the output and ground. Is ground considered the state of moving from a higher entropy to Negentropy? In other words back to a resting neutral state. Back to the horn analogy, the horn in its resting state would represent ground but as the air is manipulated in the horn sound/work is done but the each horn will have a different resistance or fight against the entropic state that is in direct relationship with its tubing diameter. So in essence ground is like the ultimate resistance setting the standard or coefficient for all other sudo grounds? I say sudo because at some point all potentials return to a resting orderly state then at some point back to disorder hence our existence 🙂
well yeah but why is it different with headphones and inears?? low impedance headphones they say are easy to drive and hi impedance ones are hard to drive so you need an headphone amp. ?? anyone? thanks
Hi, i understand the theory of wattage in solid state amplifier. But when the amp encounters higher impedance when it's supplying a fixed voltage, would the transistors produce higher voltage to mitigate the higher impedance it sees at the speakers at certain frequencies? Thanks
I’ll make this simple. Your power amplifier is a VOLTAGE amplifier. Example. A monoblock power amp has a fixed gain. For a gain of X100 then 100mV in becomes 10V out. Power is the voltage (here = 10V) squared divided by the speaker resistance. 10V output squared is always 100 for both. We then divide by the resistance for watts. 8 ohms will divide 10V by 8, 4 ohms divides 10V by only 4, So 4 ohms is double the watts. So then, what changed? The current did, 10V in 8 ohm is 1.25A, 10V in 4 ohm is 2.5A. The amplifier is required to have a power supply that can provide double the current in order for the power reach the 10V required into the 4 ohm load. Weak PSU = no doubling of power.
And… why if a bridge an amplifier to use it mono, one speaker… the ohms go from 8 to 4??? More resistance if I have one speaker instead of 2 (stereo) (Parasound amplifier manual)
It's in the words themselves: "Resistance", "Impedance". How about "Opposition"? Opposition to electron flow (current). The higher the impedance, the more opposition to current.
An audio amplifier is not a constant current or constant power device. The Volume control sets the voltage output. When you disconnect the 8 Ohm load for example, the output drive voltage should remain constant. Power, P = I^2R = V^2/R So if you increase the resistance, R, whilst keeping the current constant the power will increase. If you increase the resistance, R, whilst keeping the voltage constant, then the power will decrease. Keeping the volume control at the same setting will keep the voltage constant, so increasing the resistance will actually decrease the power.
As an engineering idiot and sticking with the tube analogy it seems that if blowing into the tube is the amplifier, then it would take more power to get the same water pressure from the larger (lower impedance) tube. Is that what he said? (Either he)
If an amp 'doubles' then the marketing dept has had its fingers in the equation. Nothing wrong with accurate specs like 100 and 165 but the marketing dept will claim 80 and 160. A recent amp I saw ($7000 USD) was 150/300 but the reviewer came up with 190/320.
Okay. But with high impedance headphones the opposite usually happens. High impedance headphones usually need amplifiers with many watts of output to drive them well. Why does this happen with headphones and not with speakers? I have some Sennheisers with 300 ohm impedance and without a powerful amplifier, it is very difficult to get the best out of them.
That's sort of true but not the way you think. Higher impedance headphones need more voltage, not current. Most headphone amps have limited voltage because of their lower overall wattage ratings. So getting a headphone amplifier with more wattage means it can provide more voltage (and current) when needed for higher impedance headphones. In actual use, the higher voltage into higher impedance does not use more wattage. Good to remember that watts is amps times volts. If you have high voltage but low amps you still have low wattage.
You don't need many watts to drive those 300 ohm headphones, you need many volts to push some current through that high impedance load. A high wattage rated amplifier also happens to supply the high voltage you need so that's what you buy. If you hook up that high wattage amplifier to your headphones and actually measured the current flowing to the headphone you would find it to be very low (0.08 amps) at around 25 volts giving you 2 watts to your headphones. That same 25 volts into an 8ohm speaker would give you (3.1 amps) and that equated to ~ 80 watts of power going to the speaker. You don't need the high wattage capacity of the amplifier but you do need the high voltage that a high wattage amplifier can provide.
It's worth pointing out that power DOES NOT double with impedance is halved. Power is halved when impedance doubles. If you make an amplifier that has 100 watts of power at 8 ohms, that means you only need an amp that can deliver 28 ish volts and about 3.6 amps. If you plugged in a 4 ohm load to this amp, we do not get 200 watts out. We actually do indeed get 50 watts out before you overdrive the amp. In order to get 200 watts out, you need to supply double the current. Well guess what? If you were to double the current, you no longer have a 100 watts 8 ohm amp, you have made a 200 watt 4 ohm amp. And if you double the current again, you have made a 400 watt 2 ohm amp. You don't plug in a 4 ohm load into a 100 watt 8 ohm amp and magically double your power and I think this is where people get confused. I think 99% of the confusion would go away if we changed our language to say that we started with a 200 watt 4 ohm amp and the power is halved every time impedance is doubled.
These videos are all awesome. Power = Voltage x Voltage / Impedance. Twice the voltage means 4 times the power. Twice the impedance means half the power.
no twice the voltage doesn't mean necessarily four times the power it's not voltage that determines it it's amps, depends on the amps, if your power supply is not capable or the amplifier is not capable producing the amps you won't get anywhere near four times or even twice as powerful
An audio amplifier is not a constant current or constant power device. The Volume control sets the voltage output. When you disconnect the 8 Ohm load for example, the output drive voltage should remain constant. Power, P = I^2R = V^2/R So if you increase the resistance, R, whilst keeping the current constant the power will increase. If you increase the resistance, R, whilst keeping the voltage constant, then the power will decrease. Keeping the volume control at the same setting will keep the voltage constant, so increasing the resistance will actually decrease the power.
@@hoobsgroove Well, generally it works like that and a power amplifier is mostly a voltage amplifier. The internal power rails of the amp defines the clipping level and the max RMS sine wave power level where the peaks of the sine waves start hitting max voltage. Of course at any given time, actual power is lower than peak power due to volume setting and music content. Actual peak RMS power is close to the calculated power from peak RMS voltage that is derived from the power rail voltage divided by square root of 2. Any decent amplifier can maintain the power rail voltage near constant at peak power. Resistance is not the right term here as we are dealing with AC signals.
@@PetraKann Since an audio signal is AC, the correct measure is impedance rather than resistance. A power amplifier is actually a voltage amplifier but with adequate max current capability to drive speaker loads. Of course any amplifier will have a limit of power that might be due to the transformer not able to output adequate current. To save money, many amps can deliver peak power by using capacitors to maintain constant power rail voltage during the peaks only.
Yes but where it gets difficult and needs an explanation why a given speaker has 8 or 4 Ohm. How is it build and why is it exactly 8 or 4 Ohm? Why not 6,435 Ohm?? or any Ohm you design your speaker? does it have to do about building your woofer into certain materials or amount of materials depending how loud you would like to have it? This is where probably people never get to understand this over and over again. Your answers are always the same: to short in this case. Why not try to explane how speakers are build making them 4 or 8 Ohm. Like to hear this from you. 🙂 Friendly greetings from the Netherlands, Onno Nugteren Utrecht.
The sorry state of elementary physics and mathematica teaching in western secondary education becomes very clear when most people do not know or are not able to use the simplest of simple formulas which is Ohm's law.
This is not about knowing formulas, but about understanding how it works in our daily use of our equipment. Just punching numbers in a formula don't explain anything, although they could be correct.
@@robinr5787 The ability to mentally extrapolate from a formula to 'understanding' is a basic skill you obviously lack.. You need to be explained something in inaccurate words while all is told in the equation. Enjoy your amp meal.. I recommend some chilly sauce on the side to spice it up.
@@razisn I don't think you understand, as long as educators think like that they keep failing to educate. Formulas are not real, just numbers. You can't eat them, can't hear them, can't feel them. They can be usefull of course. I'm just greatfull people like Paul also exist to teach based on real life. End of discussion for me, have a nice day and enjoy the music.
@@robinr5787 lol ok. All science is based on equations, equations are one of the pillars of our civilisation btw, but never mind.. Enjoy your amp meal, I recommend a dash of chilli sauce on the side to spice it up.
@@johnmoser1162 Firstly, amplifiers are supposed to maintain constant voltage, regardless of load, so your 100W amp, should double the power output for every halving of impedance. Secondly, actual measurements show relation between distortion and load. Thirdly, mathematics. Another point worth mentioning, is th3 discrepancy between nominal impedance and that "seen" by the output stage, when dealing with an actual speaker system. Stereophile's EPDR represents this well.
@@johnmoser1162 Indeed, human hearing is so poor at resolving low frequency range, that you can pretty much drive it with a spade. Distortion figures of between 2 and 20% are tolerated by listeners and achieved in practice by majority of commercial speakers.
My 2-ohm speaker power recommendation is 150-600 watts. I feed them ~250, and I'm sure they coast along well under that. Amp manufacturer claims -100/150/250. I've never seen a testbench of this model but their other models all beat specs.
DC -Power = I xV .Then there comes ac .now that's another demon on its own .True power and apparent power and capacitance and inductance and resistance and frequency .AC theory is now burning my brain cells
Paul, lately ( as of last year) you seem to be really bothered and preoccupied with bad comments by people when they either don’t agree with you or they say it in a way that doesn’t rub you the right way. I’ve watched dozens and dozens of these videos you put out and I always go to the comment section afterwards and what I’ve seen is maybe 1 out of 30 or 40 comments is what you would call nasty. That’s a pretty good batting average where I come from. I just don’t see all of these “nasty” comments you are complaining about. Overwhelming the comments have been positive and pleasant both towards you as well as what you are telling people. I just don’t see why you are so bothered by such a small fraction of people who disagree. Maybe you need a vacation.
I feel pain when you try to explain this. How about using the age old water analogy raising a water tank as an analogue for voltage? No need to talk about complex loads before the basics are understood.
Instead of saying “more energy” into the amplifier why not say more current. If things are done properly you won’t have a “short circuit” from a battery or an amplifier You’re getting away from a simple engineering class where the explanation would take about five minutes on the blackboard so everyone can understand. You’re making things complicated. Some of the few reasons why you would want to have a short circuit is when you discharge a capacitor or turn on a light switch. That’s only two examples. If you show it out of speaker then you destroy it more than likely.
