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Impedance and Resistance
Impedance and Resistance
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Old 15th January 2004, 12:19 PM   #1
newbie1 is offline newbie1  United States
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Default Impedance and Resistance

I haven't been able to get a clear understanding of the real differences in these two. I understand that resistance is a measure of DC resistance and impedance applies to AC but does not necessarily equate to resistive heat dissapation.

So what am trying to understand is this. If a speaker is rated at 4 ohms impedance, how much of the current is being dissapated into heat, and how much is actually being used to drive the speaker? Even though there is an AC signal coming from the amplifier output terminals there must be some heat dissapation.

Once more question. What is the correct way to measure the amount of output power of an amplifier? I have a car amp hooked up to a 4 ohm speaker with the gain cranked but I only measure about 10VAC? The amp is rated at 60W/channel. I have to admit I don't know the rail voltage only the input voltage (12V). But would it be right to assume that the rail voltage would be 15V buy using P=V^2/R and solving for V? Also how would I check the rail voltage in the amp anyway....I would imagine you would do that somewhere after the bridge rectifier?

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Old 15th January 2004, 01:00 PM   #2
SY is offline SY  United States
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Impedance and Resistance
One at a time:

Impedance is a general term. Mathematically, it's a complex number, that is, it can be expressed as a sum of a real part and an imaginary part (you remember imaginary numbers from high school algebra). The real part is "resistance." The imaginary part is "reactance."

The ratio of the amount of power used to create sound versus what creates heat is referred to as "efficiency." A typical driver efficiency is (depressingly) on the order of 1% or so.

The nominal impedance is a work of fiction. Driver impedances vary wildly with frequency.

To measure amp power, it's best to use a dummy load like a noninductive power resistor. You absolutely need to put an oscilloscope (or computer equivalent) across the load to monitor for clipping and to give an accurate idea of the actual AC voltage. Increase the signal in until the clipping point, back off until the clipping goes away, then look at the peak AC voltage. Average power is (Vp^2)/2R.
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Old 15th January 2004, 01:13 PM   #3
Richard C is offline Richard C  United Kingdom
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The proportion of power dissipated as heat in a loudspeaker is related to the efficiency of the speaker.

The power output of your amp is as you state: P=V^2/R, your meter should give a reasonably accurate rms AC reading so your amp is in fact 25W rms, don't be mislead by the manufacturers claim.

I guess from the output power that your amp is bridged and the rail voltage is the DC input voltage of 12V. There is no bridge rectifier.
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Old 15th January 2004, 05:03 PM   #4
soundguymark is offline soundguymark  England
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Default ohms

what ohms would a 4 and 8 ohm speaker be connected in parralell
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Old 15th January 2004, 09:43 PM   #5
Sch3mat1c is offline Sch3mat1c  United States
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Assuming they act the same (which they don't), it would go by the equation:
     R1 * R2
R = ---------
Or the reciporical of the sum of reciporicals, if you want to parallel more than two resistors of dissimilar value. (1/R = 1/R1 + 1/R2 + 1/Rn...)

Note that the 8-ohm speaker will recieve half as much power, and given the same sensitivity, will be half as loud as the other speaker.

Say you parallel a resistance with an impedance, given the impedance remains constant (it often changes with frequency or even signal strength, hence why I say "they don't" about the speakers above), the resulting impedance will obey the parallel (or series, as the case may be) rule for resistances.

Also, since inductances are frequency-dependant impedances, inductors obey the same law when paralleled: L = L1L2 / (L1+L2).

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