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Why a low impedance speakers harder to drive?

TSX

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Hi,

I am wondering why low impedance speakers are harder to drive than high impedance speakers? If it takes, let's say, 50 Watt to drive a 8Ω speaker, why does it require more power to drive 4Ω speakers? It sounds counterintuitive

Thanks!
 
It's not 'harder' as such, but it has different requirements of the amp. Depending on what the design decisions were for the amp it may or may not be a problem. For a given output voltage the 4Ω speakers will draw twice the current of the the 8Ω speakers and receive twice the power. If the amp is capable of delivering current into the lower load it will 'double down' as the load decreases, so with your 50W into 8Ω example it would deliver 100W into 4Ω and 200W into 2Ω. Not many amps can do this - it costs more, and 2Ω is a very unusual load, so the designer works to a lower cost, lower current design. With lower impedance loads the amp will now be current limited rather than voltage limited. You will see this calculation discussed in threads about amps like the Aiyima A07 and recent Fosi amps as people decide which power supply to go for.
 
Because of Ohm's law. They draw more current.
 
High impedance loads mean the amplifier can run out of rail voltage when pushed to its limits.

Low impedance means the amplifier needs to sustain higher currents which can strain the power supply or put the amplifier beyond its heat extraction capability when pushed to its limits.
 
High impedance loads mean the amplifier can run out of rail voltage when pushed to its limits.

Low impedance means the amplifier needs to sustain higher currents which can strain the power supply or put the amplifier beyond its heat extraction capability when pushed to its limits.
I understand the second statement, but not the first. Can you please elaborate?
The amp tries to give a certain Voltage out,a high impedance makes it easier/less current ..the voltage is the same…
 
Amplifiers are mostly voltage-gain devices. The load is what "decides" the current demand. P=V^2/R or I^2xR. So for a fixed Power, as the resistance goes up so must the voltage (1st formula); or as the resistance comes down, so must the current go up (2nd formula).
 
@somebodyelse gave you the best answer:

It's not 'harder' as such, but it has different requirements of the amp. Depending on what the design decisions were for the amp it may or may not be a problem. For a given output voltage the 4Ω speakers will draw twice the current of the the 8Ω speakers and receive twice the power. If the amp is capable of delivering current into the lower load it will 'double down' as the load decreases, so with your 50W into 8Ω example it would deliver 100W into 4Ω and 200W into 2Ω. Not many amps can do this - it costs more, and 2Ω is a very unusual load, so the designer works to a lower cost, lower current design. With lower impedance loads the amp will now be current limited rather than voltage limited. You will see this calculation discussed in threads about amps like the Aiyima A07 and recent Fosi amps as people decide which power supply to go for.

In an ideal world, there would be no problem. Amplifiers could be made to deliver more-than-enough current to all speakers. However, we don't live in an ideal world. We live in a world where manufacturing costs are very important.

Not only that, but the amp design needs to take into account not only impedance, but also reactance (both inductive reactance and capacitative reactance). Reactance is the monster hiding under the bed. It's described here:


So ... if an amp is a samurai, it's a samurai who is fighting an easy foe right in front of him (impedance) as well as two hardened and cunning foes trying to stab him in the back (capacitative and inductive reactance), and he has to fight all three of them at the same time.

Some samurai fight this battle with expensive, fine-quality swords (power supplies), and they do just fine. Unfortunately, some samurai have to fight this battle with cheaper, lower-quality swords. Therefore, their fight is a great deal more difficult.

Just listening to an amp won't tell you diddly-squat. The correct tests and measurements will tell you what you need to know.


Jim
 
Hi,

I am wondering why low impedance speakers are harder to drive than high impedance speakers? If it takes, let's say, 50 Watt to drive a 8Ω speaker, why does it require more power to drive 4Ω speakers? It sounds counterintuitive

Thanks!
More current is needed to drive the lower impedance, and that makes extra demand on electronic components, mostly due to heating. The more current a wire, resistor or other electronic device carries, the warmer it gets.
 
To add some historical perspective. Historically, speaker drivers have been made to suit the amplifiers available at the time. 32 ohms was common in early vacuum tube days, or 16 ohms. Transistors are better driving lower impedances, like 8 or 4 ohms. Most amps even today struggle with 2 ohms and below.
 
To add some historical perspective. Historically, speaker drivers have been made to suit the amplifiers available at the time. 32 ohms was common in early vacuum tube days, or 16 ohms. Transistors are better driving lower impedances, like 8 or 4 ohms. Most amps even today struggle with 2 ohms and below.
And some tube amps used a workaround - multiple taps on the output transformers to match them to a range of speaker impedances. From the the amp (primary) side the 4 ohm load on the 4 ohm tap and the 16 ohm load on the 16 ohm tap look the same.
 
I guess short circuits (i.e., ca. zero ohms) are either very easy or very hard to drive -- depending upon one's perspective.

A low impedance load isn't necessarily hard to drive. E.g., consider a (nearly) purely resistive low impedance load consisting of a sensitive driver (or drivers) and benign crossover. Not necessarily hard to drive a priori.

Many Magneplanars manage to be very benign from an impedance perspective, although they are not very sensitive (nor very electrically efficient).

1703535109999.jpeg

source: https://www.stereophile.com/content/magnepan-lrs-loudspeaker-measurements

Taps are far from a workaround -- they improve the match between amplifier output and load -- which is always always a good thing. Thus does McIntosh still use outpur transformers (well... autoformers, which are functionally quite similar, although not using inductive coupling between primary and secondary windings) in some of their (solid state) products.
 
W=R*I^2 -> I=SQRT(W/R) <-> current=SQRT(power/resistance) -> power needed being equal, the amplifier must provide higher current on a lower resistance. In numbers, If an amplifier supplies 1 Ampere to an 8Ω speaker, in order to have the same volume with a 4Ω speaker of the same efficiency the amp must provide 1,414 Ampere. Higher currents require beefier power supplies.
 
It doesn't take more power to drive per se since the power sensitivity are usually comparable.

The answer gets long but here's a shortcut answer: Consider the Redbook 2Vrms output voltage coming out of your CD player or computer or whatever. This voltage is sufficient to drive most speakers to 80dB, which is plenty loud. Many preamps can output higher than 2Vrms too. So why aren't we directly connecting speakers straight to CD players or preamps? Because these equipment have a low current output limit, so the moment you connect the speakers and go above like 0.1V the current limit gets reached and the signal clips.

So speaker amplifiers, aside from generating around 10x to 100x the voltage, they also need to provide the current which is really their main thing, compared to preamps which only provides higher voltages. Thus, things that require higher currents out from speaker amps are usually called "harder to drive", since a speaker amp has no problem driving loads with higher impedances.

On the flipside, in the headphone realm there are also people who call high impedance headphones as "harder to drive", because typically headphone amps don't run into current limits while the voltage provided is usually plenty, however some headphones have high impedances which require higher voltages that typical headphone amps cannot output.

So in short, whether we call something "harder to drive" is based on whether it presents more work for the amplifier with respect to the amplifier's main job, all other things being equal.
 
A key thing to consider with loads that are hard to drive is speaker's overall impedance curve. Loudspeakers with significant reactive elements increase the strain because the current and voltage are out of phase, requiring more current and/or voltage at some frequencies than would be expected from a purely resistive load.
 
This may give a general idea of how many factors are at play, this is just one of the many calculators available:


It will also depend on the program material, listening distance and desired SPL level. That is, how many actual volts do you need for peaks and where in the frequency band they are.
 
50 Watt to drive a 8Ω speaker, why does it require more power to drive 4Ω speakers?
There are very small differences.

Look at datasheet of integrated amplifiers. For example LM3886 dissipates about 30W for 50W output for both 4 and 8Ohm, but you need different supply voltage to get the same efficiency. 4Ohm seems to look a bit more efficient.

Characteristics for TPA3255 show a small advantage efficiency for 8Ohm, like 3W supply more for 4 Ohm at 50W output.
 
Because of Ohm's law. They draw more current.
OT. It made me think of this old thread. Interesting, I think::)

 
There are very small differences.

Look at datasheet of integrated amplifiers. For example LM3886 dissipates about 30W for 50W output for both 4 and 8Ohm, but you need different supply voltage to get the same efficiency. 4Ohm seems to look a bit more efficient.

Characteristics for TPA3255 show a small advantage efficiency for 8Ohm, like 3W supply more for 4 Ohm at 50W output.
Yes. 3 dB (nominally... Ohm's law and all that). "Barely audible" in classic audio reckoning. :)
 
In order not to create a new thread, I'm resurrecting this one.
I'm in the process of developing a 2-way bookshelf and I'm faced with a dilemma: the most stable impedance for the system gives me a non-optimized frequency response (there's a natural hump in the woofer's response that I haven't been able to get rid of), while two possibilities for optimizing the response leave me with a very low minimum impedance.
Is this impedance below 2 ohms really something to worry about, considering that the amplifier used will be 4ohms, but will hardly be pushed to its limit?

Not so good FR
Captura de tela 2024-07-01 111844.png


Good FR, bad impedance (#1)
Captura de tela 2024-07-01 111909.png


Good FR, bad impedance (#2)
Captura de tela 2024-07-01 111937.png
 
Is this impedance below 2 ohms really something to worry about, considering that the amplifier used will be 4ohms, but will hardly be pushed to its limit?
That's impossible to answer with certainty. Speaker is a "nominal" specification and it's not unusual to drop below the spec.

You'll probably be OK, especially if you aren't pushing the amplifier.

You'll only occasionally be hitting that worst-case frequency and you might not be using all of the amplifier's power at that frequency.

With constant full-power test tones at the lowest-impedance frequency you'd be stressing the amplifier. It might simply distort, it might go into thermal protection mode, or it might burn-up. But you can also fry speakers with full-power test-tones.
 
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