New Adcom 👀585 SE. anybody heard this monster

Hi Auxinput,

No, you were right the first time, and if Erik disagreed he was uncharacteristically incorrect.

From the perspective of each amplifier channel a given output voltage will result in twice as much current being drawn from that channel in bridged mode than in stereo mode. That is because in bridged mode a voltage that is equal in magnitude but opposite in polarity will appear on the other side of the load, driven by the other channel, resulting in twice as much voltage across the load than each channel is outputting individually.

Twice as much voltage across a given load means twice as much current, of course, per Ohm’s Law (stated for this purpose as I = E/R; if E is doubled and R remains constant I is doubled). And from the perspective of each channel having to deliver twice as much current for a given output voltage corresponds to each channel "seeing" the load impedance divided by two, again per Ohm’s Law (stated for this purpose as R = E/I; if I is doubled and E remains constant R is halved).

The factor of 4 relates to power, not to impedance, as doubling the current into a given impedance corresponds to a 4x increase in power. As I’m sure you realize, for a resistive load Power = (Current squared) x Resistance.

So for an ideal design a bridged amp can deliver 4x as much power as each channel can deliver by itself, in stereo mode. In practice, though, current capability and thermal limitations usually reduce the factor of 4 significantly.

Best regards,
-- Al

Aragon Palladium is BTL of two 8008!

Auxinput 7-1-2018

If you look at Parasound A21 specs:

250 watts x 2 into 8 ohm

400 watts x 2 into 4 ohm

750 watts x 1 into 8 ohm (bridged)

If each channel of the amp only saw half the load in bridged mode, it would want to push 400 watts, not 750 watts.

Hi Auxinput,

My previous response, while accurate, did not specifically address this comment.

As I’m sure you realize, for a resistive load P = E x I, where P is power, E is voltage, and I is current.

The increase in the maximum power rating of solid state amps which occurs when load impedance is reduced from 8 ohms to 4 ohms reflects the fact that nearly all solid state amps act as voltage sources. Meaning that they will maintain an essentially constant output voltage as load impedance varies, as long as they are not asked to exceed their maximum voltage, current, power, and thermal capabilities.

So the increase from 250 watts to 400 watts that you referred to reflects I (current) increasing substantially as a result of R decreasing, ***but reflects E remaining the same.*** And as I’m sure you realize, some amps can provide up to 2x as much power into 4 ohms as into 8 ohms, reflecting a 2x increase in current due to R being cut in half while E remains constant.

In bridged mode, on the other hand, E across the load and I through the load ***BOTH*** double (compared to stereo mode, and provided that the amp can supply the required current), while R remains the same. Therefore, since P = E x I, ideally P can increase up to 4x, as I said in my previous post.

Best regards,
-- Al