More resistance is less load??


Hi, can someone explain, in "ohms for dummies" language, why a 4 ohm speaker, which has half the resistance of an 8 ohm speaker, is said to be more demanding on the amp? And the other way, why a 16 ohm speaker, with twice the resistance, is less demanding?
128x128jimspov

Hi, can someone explain, in "ohms for dummies" language, why a 4 ohm speaker, which has half the resistance of an 8 ohm speaker, is said to be more demanding on the amp? And the other way, why a 16 ohm speaker, with twice the resistance, is less demanding? jimspov


Simply, the lower the resistance of a speaker, the closer it is to a short (zero ohms).

EG: Like a car going up hill, the steeper the hill harder the engine has to work, if almost vertical it could blow if not built for it. In speakers v amp this means that amp has to supply current more and more the lower the speakers impedance (resistance).


Cheers George

Good responses by the others above, except that the reference to E/IR should be E = IR (meaning E equals I multiplied by R). (E is commonly used in the context of Ohm’s Law to denote volts, and means the same thing in that context as V).

So when Ghosthouse referred to V/I = R, he was correct. By simple algebra V/I = R is equivalent to V (or E) = I x R = IR.

Regarding the voltages that are typically provided to speakers, for a resistive load:

Power = (I squared) x R = (E squared)/R
where, for example, P is expressed in watts, I in amps, R in ohms, and E in volts.

Therefore E = (Square root (P x R)).

So for example 100 watts into an 8 ohm resistive load corresponds to:

Square root (100 x 8) = 28.28 volts.

Regards,
-- Al

In other words, think of the flow of electrons like a water hose as someone already mentioned. The less resistance, the more flow and therefore the more draw on the amplifier to push those electrons through the wires. The more draw, the harder the amplifier has to work. It's rather counter intuitive unless you think about it in those terms. 
Thanks this is very helpful. I agree falconquest it is counter-intuitive, but the equations and explanations are clear & helpful. I have some clarity now.

almarg has brought up the notion of "power".

Used to be, when I paid attention to these things, 40 watts per channel was 40 watts "RMS".

How is RMS (root mean square I believe) to be understood?
Yes, RMS = root mean square. As you probably realize, the signal provided to a speaker consists of various frequency components each of which is AC (alternating current). Amplifier power capability is defined based on the simplified assumption that the signal consists of a pure sine wave at a single frequency, with that single frequency being anywhere within some range of frequencies, such as 20 Hz to 20 KHz. The RMS value of a sine wave equals its peak (maximum) instantaneous value divided by the square root of 2, or approximately 0.707 x the peak value.

In audio voltages and currents are usually defined on an RMS basis, in part because the amount of power supplied to a resistive load that can be calculated based on RMS voltage and current numbers (even for waveforms that are not sine waves) equals the amount of power that would result from a DC (direct current) voltage and current having the same values, that amount of power in turn being proportional to the amount of heat that is produced when supplied to ("dissipated in") a resistive load.

By the way, one thing that often causes confusion in this context is that the word "peak" can be used to mean two different things. It can refer to the peak (maximum) value of a sine wave or other signal at any instant of time during each of its cycles (corresponding for a sine wave to the RMS value divided by 0.707), or it can refer to the peak (maximum) RMS value that can be reached by that sine wave or other signal during normal (or other) operating conditions.

Kudos for your interest in these matters. Regards,
-- Al