Trying to understand impedance

[bigguyca]

It is exactly proportional. 300 to 16 is a current increase of 300/16 = 18.75 times greater. 8 to 4 is 2 times greater.

Ohm's Law is Z (impedance) = V (voltage)/I (current). The math is pretty simple.

This of course is nonsense. Ohm's Law in this form is R = V / I, where "R" is resistance.

Impedance includes both resistance and reactance and varies with frequency.

As a "Technical Expert" with tens of thousands of posts, you really should understand these very basic concepts and use them correctly in your posts. Others who lack a basic technical background may read your posts and internalize incorrect information and concepts. Two people already "liked" your post which is sad.

 

[JustAnAudioLover]

This of course is nonsense. Ohm's Law in this form is R = V / I, where "R" is resistance.

Impedance includes both resistance and reactance and varies with frequency.

As a "Technical Expert" with tens of thousands of posts, you really should understand these very basic concepts and use them correctly in your posts. Others who lack a basic technical background may read your posts and internalize incorrect information and concepts. Two people already "liked" your post which is sad.

I don't think it's that important, given that "Z" seems to appear in some other forms of Ohm's laws, like V = Z * I

(from Wikipedia: We can now write

where V and I are the complex scalars in the voltage and current respectively and Z is the complex impedance.)

Also letters don't matter that much, it's better to use standards but the formula is still correct especially given the explanation of all letters was given in the comment

 

[Cbdb2]

Well technically shouldn't speaker amplifiers be able to drive any headphone? Just put the amplifier's master volume at super low levels and it'll work fine! Right?

Or a dozen of any HPs. This is what many recording studios use to use to drive large numbers of HPs.

 

[SIY]

This of course is nonsense.

I'm sorry you're confused by generalizing Ohm's Law to AC. Nonetheless, it still holds.

I'm afraid to mention the most general version, J=σE, with bolding indicating vector quantities. Nonetheless, if you're interested in improving your understanding but aren't ready to be tortured by Jackson, I highly recommend David Griffiths's "Introduction to Electrodynamics."

 

[Cbdb2]

This of course is nonsense. Ohm's Law in this form is R = V / I, where "R" is resistance.

Impedance includes both resistance and reactance and varies with frequency.

As a "Technical Expert" with tens of thousands of posts, you really should understand these very basic concepts and use them correctly in your posts. Others who lack a basic technical background may read your posts and internalize incorrect information and concepts. Two people already "liked" your post which is sad.

Pretty clear here. Seen it written that simply in many EE texts. V and I can be complex or not, he did not actually have to write V=Acos(wt+Phase), which might confuse newbees more than clarify.

 

[antcollinet]

Ok so my first impression was right, but then does that mean most headphone amplifiers can just accommodate a 18x current increase going from 300 to 16 Ohms? And if so, why can't speaker amplifiers do the same?
Sorry if these seem like dumb questions but I have a hard time wrapping my head aroudn it...

I suspect:

Headphone amps might be voltage limited. You can put more power into a low impedance with the same voltage

P = V^2/R
P is also = I^2 * R

So with a 300ohm headphone you need 4.33 times more volts to get the same power compared to 16 ohm.

Speaker amps on the other hand are normally current limited - especially at 4ohm load. You need 1.41 times as much current into 4ohm for the same power.

 

[solar_sustem_s_user]

The KZ Castor Bass (Harman Target with Improved Bass version) is listed as 16–20Ω impedance (AC) on the manufacturer's site.

However, when I measure them with a multimeter (DC), I get resistance values of 110Ω, 130Ω, and 160Ω, depending on the switch position.

Strangely, they still sound loud and clear from a smartphone 3.5 mm jack, no signs of underpowering. The same happens with KZ Castor Silver — identical resistance readings.

For comparison: other hybrid KZ IEMs like ZS10 Pro X or BA10 show resistance close to what’s stated — ~16–24Ω.

Why is Castor’s DC resistance so high?
Could this be a circuit design with internal capacitors or switching elements that “bypass” the resistance at audio frequencies (AC)?

Would appreciate if someone could explain this behavior.

 

[eddantes]

This picture from the video actually shows a bad analogy.
View attachment 442390

The electronic-hydraulic analogy equates voltage to pressure, flow resistance to electrical resistance, and volumetric flow rate to electrical current. However, an amplifier magnifies voltage according to its gain, which means output voltage = input voltage × gain. Therefore, it is the voltage that the amplifier works to maintain.

Using the hydraulic equivalent, the pump (amplifier) works to maintain the target pressure (voltage). If the flow resistance (electrical resistance) is low, the pump has to work very hard to maintain the target pressure (voltage) because that will require a high flow (current). That's why it is harder for a voltage amplifier to drive into a low impedance load, and is easier for it to drive into a high impedance load.

If the amplifier is a current amplifier (that tries to deliver a target current, which is NOT the case in standard audio implementations), the situation reverses.

Great explanation!

 

[DVDdoug]

is listed as 16–20Ω impedance (AC) on the manufacturer's site.

However, when I measure them with a multimeter (DC), I get resistance values of 110Ω, 130Ω, and 160Ω, depending on the switch position.

That's unusual. A speaker or driver will usually measure lower with DC than with a signal.

Have you seen an impedance vs. frequency curve? MAYBE it's not even true...

Could this be a circuit design with internal capacitors or switching elements that “bypass” the resistance at audio frequencies (AC)?

I'm not making any assumptions about how the thing is built, but you could make something like that with a high-impedance bass driver and a low-impedance high-frequency driver/tweeter. The high frequency driver would have a capacitor in series as part of the crossover and capacitors have infinite resistance at DC so it won't be measured with a multimeter.

...Not THAT extreme but some cheap 2-way speakers simply put a capacitor in series with the tweeter to act as a crossover (with no filtering to the woofer). If the woofer and tweeter are both 8-Ohms, the speaker impedance is 8-Ohms at low frequencies and 4-Ohms at higher frequencies where both are getting a signal. (That's ignoring the inductance of the woofer's voice coil which raises the impedance a bit.) (And it's usually sold as an "8-Ohm" speaker.)

 

[solar_sustem_s_user]

Have you seen an impedance vs. frequency curve?

Thanks for your insights!
Yes, I don’t have the impedance vs frequency curve, unfortunately. I based my info on manufacturer specs and my DC resistance measurements with a multimeter.