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-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of Anthony Lapinski
Sent: Tuesday, April 26, 2011 9:31 AM
To: phys-l@carnot.physics.buffalo.edu; tap-l@lists.ncsu.edu
Subject: [Phys-l] speakers
I have an old Fisher speaker system. Remove the grill and it shows a
woofer, midrange, and a tweeter. Red/black connectors in back. Works
well
with a Radio Shack amp.
On the back is a sticker with information about this unit:
impedance = 8 ohms
input power = 5 - 50 watts[Bill Nettles] The low number usually refers to the minimum power required to overcome "frictional" forces in the cone and produce an audible signal. The upper is usually the maximum continuous power the speaker coils can handle without overheating, or without causing the speaker cones to mechanically be destroyed by excessive movement.
crossover = 1.5/6 kHz[Bill Nettles] Crossovers can range from very simple RC or RL networks, to very complicated multiply connected R's L's and C's. The frequency quoted is generally the frequency at which half the speaker power is feed to the "lower" element and half goes to the "upper". So at 1.5 kHz, half the power goes to the woofer and half the power goes to the mid/tweeter, most of it to the mid. At 6 kHz, half the power goes to the tweeter and half to the mid/woofer, most of it to the mid. BTW, half power points are also 3dB decreases from full. Typical humans barely notice changes of 1 dB, so although half-power sounds like a lot, the actual volume/loudness change, while noticeable, isn't huge.
I have a few questions.
1. Does the impedance of a speaker affect the sound quality/loudness?
So[Bill Nettles] You can still get the power (subject to amp limitations), but you might have to change the volume knob. Again, the impendance is for matching the output loading design of the amplifier for optimum power transfer and amplifier protection.
for a given power, you need a certain "ideal" impedance (resistance).
And
if you use a speaker with a higher impedance, the power output will
decrease (P = V2/R)?
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