Well, this seems to come up an awful lot so I figured I'd provide the best explanation I could. I'll do my very best to simplify this and put things in laymen's terms. Here goes. Bridging - this is when a load is connected across two channels of a stereo power amplifier. The term bridging itself refers to the act of connecting across the channels like a bridge connecting two points. Consider the following: Each channel (or half) of the power amplifier drives half of the load. Therefore, when connecting an 8 Ohm load, each channel of the amplifier sees a 4 Ohm load. Power is doubled when bridging to a given load - IE, an amplifier that can make 50 wpc into an 8 Ohm load will make 100 w into 8 Ohms when bridged. [This does not apply to all makes and models of amplifiers - explanation to follow.] Now, some amplifiers require you to throw a switch on the rear panel when bridging and others do not. This switch may have several functions, depending on the design and make of the power amplifier. The net result is that one channel of the amplifier will drive the positive half of the waveform and the other channel of the amplifier will drive the negative half of the waveform. McIntosh power amplifiers are somewhat different in their fundamental design from many other power amplifiers in that the channels have Common terminals. These Common terminals are always negative. In addition, the positive terminals are always HOT. Other amplifier designs invert one of the amplifier channels - RED is HOT on one channel (typically LEFT) and BLACK is HOT on the other channel (typically RIGHT). The signal is inverted at the output so that a bridged load can be connected across these terminals which are obviously both HOT without having to have a switch to do so. Because the one channel is inverted at the output, the signal is also inverted as it comes into the input of the amplifier on the same channel. This allows both channels to operate in phase with one another when using the amplifier in stereo and allows both channels to operate out of phase with one another when bridging the amplifier - all without having to flip any kind of switch. There are surely other methods, but these would be the two most common. For the following computations, consider: P = Power I = Current in Amperes E = Voltage in Volts R = Resistance in Ohms Now let's apply Ohm's Law (I = E / R), the Power formula (P = I x E), and combinations thereof to a few examples to see what's really going on when we bridge a power amplifier. Now keep in mind that when we talk "R" here, what we're really talking is a load resistor on a test bench - a constant, which is required to allow us to apply Ohm's Law and the Power formula in an effort to make comparisons. A 50 wpc power amplifier will make 20 Volts per channel into an 8 Ohm load. Let's first prove this: P = E^2 / R P = 20^2 / 8 P = 400 / 8 P = 50 Watts Bridging Example 1 - The power amplifier doubles the power of a single channel when bridging into an 8 Ohm load: P = E^2 / R P = 28.28^2 / 8 P = 800 / 8 P = 100 Watts Note that the voltage increases accordingly, but it is not doubled. Bridging Example 2 - The power amplifier doubles the voltage of a single channel when bridging. Therefore, it will quadruple the power of a single channel - (2^2 = 4) - when bridged into an 8 Ohm load: P = E^2 / R P = 40^2 / 8 P = 1,600 / 8 P = 200 Watts McIntosh amplifiers with autoformers follow the first example while some other brands of power amplifiers follow the second example (specifically those that permit bridging but do not have switches to do so). It is important to note that the designers may choose to current limit a power amplifier which can show up in the amplifiers specifications - IE: 200 wpc stereo into 8 Ohms, 300 wpc stereo in to 4 Ohms, 600 wpc bridged into 8 Ohms. The easiest way to limit current is to limit voltage to the output devices to prevent the amplifier from doubling power when impedance is halved. This can prolong the life of the output devices or allow an amplifier to provide the desired play time given the mass of heat sink the bean counters approve in the design phase (this is referred to as thermal time - time to thermal shut down). Don't let this throw you off when applying the above math but this is reality versus theory. Strapping - This is when a load is connected to both the left and right channels simultaneously. This is often referred to as Parallel-Mono as the stereo channels are wired in parallel with one another - Hot to Hot and Common to Common - and the load is then tied to them both. The term strapping itself refers to using short straps of wire to connect the channels together. The idea here is that when both channels are doing identical work and the signals each channel is amplifying are identical one can tie them together. [NEVER try this on an amplifier that was not specifically designed to do this! Especially when the Red on one channel is the HOT and the Black on the other channel is the HOT as described above.] When it comes to HiFi power amplifiers, McIntosh is the only company I'm aware of that permits this and this applies only to McIntosh amplifiers with autoformers. In the pro industry they all do it - Crown being I believe the earliest adopter - and they're all direct coupled designs. Many McIntosh and Crown power amplifiers allow either Bridged-Mono or Parallel-Mono wiring schemes. This makes a given power amplifier incredibly versatile. Now, the common perception is that when you strap channels you double the current. Is it true? Consider the following which I've copied directly from the McIntosh MC2300 owners' manual - a McIntosh SS stereo power amplifier with autoformers that can only be strapped to mono and cannot be bridged: Power Output / Stereo: 49.0 Volts across 8 Ohms (49^2 / 8 = 300) Power Output / Mono: 69.3 Volts across 8 Ohms (69.3^2 / 8 = 600) Recall, it's specifications are 300 wpc into 8 Ohms stereo and 600 Watts into 8 Ohms mono. The math is just as when bridging - note the difference in voltage. Hmmm . . . But is current doubled? Stereo: I = SQRT(P/R) I = SQRT (300/8) I = SQRT (37.5) I = 6.12A Mono: I = SQRT(P/R) I = SQRT (600/8) I = SQRT (75) I = 8.66A It isn't. The reason for this is because power isn't quadrupled when connecting the amp in mono with the McIntosh design. If it were, then yes current would be doubled. Plug 1200 Watts into the above formula and do the math . . . Let's look at a Crown design. This is copied directly from the Crown Macro Tech amplifier brochure: Bridge-Mono mode provides double the output voltage and Parallel-Mono mode provides double output current from a single channel. OK, so it's easy to see where the perception comes from. It's printed right there in the manual. Is it true for the Crown MacroTech 2400? Let's find out. Here are the printed specs on the Crown MacroTech 2400: 800 Watts into 4 Ohms per channel - let's look at both voltage and current: I = SQRT(P/R) I = SQRT (800/4) I = SQRT (200) I = 14.14A E = SQRT(RP) E = SQRT(4 x 800) E = SQRT(3,200) E = 56.57 Volts [Checking our math, P = I * E, P = 14.14 * 56.57, P = 800W] 2,070 Watts into 4 Ohms bridged-mono E = SQRT(RP) E = SQRT(4 x 2,070) E = SQRT (8,280) E= 90.99 Volts This is a case where the designers implement current limiting for some reason (prolong the life, only so much heat sink mass to dissipate heat, etc.). If the amp was not current limited and it did actually double the output voltage, it would quadruple power at 3,200 Watts into 4 Ohms bridged mono. [This is an incredibly powerful amplifier nonetheless.] 1605 Watts into 2 Ohms parallel-mono I = SQRT(P/R) I = SQRT (1,605/2) I = SQRT (802.5) I = 28.32A Interesting. The amplifier does double it's current BUT this results in only a doubling of output power as impedance is halved as well. Note that if we do a comparison in 4 Ohms parallel-mono, this is not comparing apples to apples as the amplifier only makes 1,035 watts when operated into 4 Ohms parallel-mono. [This amounts to only 16.1A at 4 Ohms in parallel-mono.] We really need to look at its 1 Ohm parallel-mono ratings to compare apples to apples, which is 2,080 Watts: I = SQRT(P/R) I = SQRT (2,080/2) I = SQRT (1,040) I = 32.24A So, there are no hard fast rules in regards to doubling of voltage when bridging or doubling of current when strapping that apply to all amplifiers. As you can see, this is greatly dependent on the design. I hope this helps. I'm sure some Crown fans will chime in on strapping.