When Bi-amping is there change in sensitivity

Almarg,
"Keep in mind that power equals voltage times current (or less, if the load is not purely resistive). The crossover circuit that is in the mid/hi section of the speaker prevents low frequency currents from being supplied by the mid/hi amp and flowing into that section of the speaker. The near zero current means that the amplifier is delivering near zero power at low frequencies, even though its output voltage corresponds to the full-range signal.

Likewise, the crossover circuit in the low frequency section of the speaker prevents mid/hi frequency currents from having to be supplied by the low frequency amp, resulting in near zero power being supplied by that amp at mid/hi frequencies."

I dont quite understand how the crossover circuit prevents the low frequency currents from being supplied by the mid/hi frequency amplifier since the amp is being fed the full range and only after it reaches the speaker does it get split off by the coil used in the crossover. How does that coil draw off the low frequencies if they never get there in the first place? That is why I dont believe that biamping with the same amplifiers changes the sound other than perhaps some additional headroom for the mid/hi amp.

Since the amp is being fed the entire spectrum and amplifying it to the speaker where it is diverted by the crossover, where does that diverted energy go?

Manitunc, I'm not sure I understand what you are not understanding. Energy is proportional to power (factored by time). Power is proportional to voltage times current. As you undoubtedly realize, in a passive biamp configuration there is no connection between the mid/hi amp and the low frequency section of the speaker, and there is no connection between the low frequency amp and the mid/high frequency section of the speaker. As a result of the high impedance that is presented by each section of the speaker at frequencies that it is not intended to reproduce, there will be little or no current flow at those frequencies, hence little or no power will be generated or delivered at those frequencies, hence there will be little or no energy to be diverted, absorbed, dissipated, or consumed at those frequencies.

I would draw an analogy with turning on a light fixture via a switch on the wall. When the switch is in the off position it presents a high (essentially infinite) impedance to the AC that is supplied through the house wiring. Therefore the light fixture draws no current and consumes no power or energy. Similarly, the high impedance of the mid/hi crossover at low frequencies prevents any current, power, or energy from being drawn from the mid/hi amp in response to the low frequency content (i.e., the low frequency spectral components) of the output voltage of that amp. Similarly, the high impedance that the low frequency crossover has at high frequencies prevents any current, power, or energy from being drawn from the low frequency amp in response to the mid/high frequency content of the output voltage of that amp.

Think of the output voltage of each amp as being a summation of many different frequencies. The amount of current that is drawn from the amp at each of those frequencies depends on the impedance of the speaker at each of those frequencies.

Regards,
-- Al

03-20-12: Ngjockey
This is why I don't turn the lights off. Don't want the electricity to build up in the wire like water behind a dam.:)

LOL!!
sheesh! don't confuse Manitunc any more......

>03-20-12: Manitunc
>I dont quite understand how the crossover circuit prevents the low frequency currents from being supplied by the mid/hi frequency amplifier since the amp is being fed the full range and only after it reaches the speaker does it get split off by the coil used in the crossover. How does that coil draw off the low frequencies if they never get there in the first place? That is why I dont believe that biamping with the same amplifiers changes the sound other than perhaps some additional headroom for the mid/hi amp.

You're confusing voltage (potential, like the 80 PSI or whatever that your pipes are at, or the voltage across the amplifier output terminals) and current (what's flowing - crack the faucet so there's high resistance to flow and only a trickle comes out. Open it all the way creating a low impedance and you get wet. Little energy flows into a high impedance load and lots goes into a low impedance).

Reactive components have inductive (coils) and capacitive (capacitors) impedance that varies with frequency.

A capacitor's impedance magnitude is 1 / 2 pi f C with f in Hz and capacitance C in farads.

As f approaches 0 impedance becomes infinite. With current flow voltage divided by impedance current and therefore power (V * I) approach 0 as you get down to DC.

An inductor's impedance magnitude is 2 pi f L with f in Hz and inductance L in Henries.

Impedance is proportional to frequency; so as frequency goes up, less current flows, and less power is delivered.

The voltage dropped across impedances in series is proportional to them. IOW, put 3 volts into 1 and 2 Ohms in series and you'll have 1A flowing with 1V dropped across the first resistor and 2V across the second resistor.

The simplest possible cross-over circuit (first order electrical) puts a capacitor in series with the tweeter and inductor in series with the woofer.

At the highest frequencies the capacitive impedance becomes negligible, current flow is as high as it can get due to driver impedance, and all the voltage goes into the tweeter. The inductive impedance is large, essentially no current flows, and the voltage drop is almost entirely across the inductor in the low frequency circuit.

At the lowest frequencies the inductive impedance becomes negligible, current flow is as high as it can get due to driver impedance, and all the voltage goes into the woofer. The capacitive impedance is large, essentially no current flows, and the voltage drop is almost entirely across the capacitor in the high frequency circuit.

Assuming a purely resistive text book driver load (which only exists where the capacitance of the moving mass equals the inductance from the suspension compliance and voice coil) with resistance R at the cross-over frequency of 1 / 2 pi R C about 29% of the voltage is being dropped across inductor and capacitor and the two drivers are each seeing about 71% of the total.