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Question about ohms and bridging...
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<blockquote data-quote="jsloan10" data-source="post: 929112" data-attributes="member: 560397"><p>3.5.3 What happens when an amp is bridged?</p><p></p><p>-------------------------------------------</p><p></p><p>Basically, one channel's signal is inverted, and then the two channels</p><p></p><p>are combined to form one channel with twice the voltage of either of</p><p></p><p>the original channels.</p><p></p><p>Ohm's Law for Alternating Current states that I = V/Z where I is</p><p></p><p>current, V is voltage, and Z is impedance. We also know that P = IV,</p><p></p><p>where P is power. If we use Ohm's Law and substitute into the power</p><p></p><p>equation, we get P = V(V/Z), which can be rewritten as P = (V^2)/Z.</p><p></p><p>Therefore, power is the square of voltage divided by impedance.</p><p></p><p>Now, why do we care about all that? Because it explains precisely what</p><p></p><p>happens when an amp is bridged. I'll give a practical example and</p><p></p><p>explain the theoretical basis of that example.</p><p></p><p>Imagine you have a two-channel amp that puts out 50 watts into each</p><p></p><p>channel when driven into a load of 4 ohms per channel. Since we know P</p><p></p><p>and Z, we can plug these numbers back into our power equation and find</p><p></p><p>V. 50 = V^2/4 -&gt; V = sqrt(200). So, we're seeing a voltage of 14.1</p><p></p><p>volts across each channel.</p><p></p><p>Now, imagine we bridge this amp, and use it to push just one of those 4</p><p></p><p>ohms loads. When the amp is bridged, the voltage is doubled. Since we</p><p></p><p>know the voltage (2*14.1 volts), and the impedance (4 ohms), we can</p><p></p><p>calculate power. Remember that P = V*V/Z. That means P = (28.2)^2/4,</p><p></p><p>which is 198.1 watts. It should be clear by now that the new power is</p><p></p><p>approximately 200 watts - quadruple the power of a single, unbridged</p><p></p><p>channel!</p><p></p><p>You can probably see that should be the case, especially if you look</p><p></p><p>back at the power equation. Since P = V*V/Z, if you double V, you</p><p></p><p>quadruple power, since V is squared in the power equation.</p><p></p><p>Now, all this assumes the amp is stable into 4 ohms mono. The mono</p><p></p><p>channel is putting out four times as much power as a single unbridged</p><p></p><p>channel, so it must be putting out twice as much as the two single</p><p></p><p>channels combined. Since the voltage on the supply side of the amp is</p><p></p><p>dependent on the car's electrical system, it doesn't change (OK, the</p><p></p><p>increased current might cause a voltage *drop*, but let's not worry</p><p></p><p>about that now). Looking at the first power equation, at the supply</p><p></p><p>side of the amp, we see P = IV. Now, when we bridged the amp, we</p><p></p><p>doubled the power, but the input voltage stayed the same. So, if we</p><p></p><p>hold V constant, the only way to double the power is to double the</p><p></p><p>current.</p><p></p><p>That means the amp is now drawing twice as much current when it's</p><p></p><p>running at a given impedance mono than it would be running two stereo</p><p></p><p>channels at the same impedance. There are only two ways the amp can do</p><p></p><p>that - it can simply draw more through it's circuits, and dissipate the</p><p></p><p>extra heat, or it can utilize a current limiter, to prevent the</p><p></p><p>increase in current. Of course, using the current limiter means you</p><p></p><p>don't get the power gains, either! So, if the amp can't handle the</p><p></p><p>extra current, and it doesn't limit the current in some way, kiss it</p><p></p><p>goodbye. For that reason, an amp is typically considered mono stable</p><p></p><p>into twice the impedance it is considered stereo stable.</p><p></p><p>3.5.4 Does bridging an amp would halve the impedance of the speakers?</p><p></p><p>----------------------------------------------------------------------</p><p></p><p>Impedance is a characteristic of the speakers. The speakers don't give</p><p></p><p>a flip how the amp is configured: they have a given impedance curve,</p><p></p><p>and that's that. It should be clear that when you bridge an amp, you</p><p></p><p>are changing *the amp*. The speaker's impedance is *not* a function of</p><p></p><p>the amp, but the amp's tolerance to a given impedance depends</p><p></p><p>completely on the way the amp is configured. If you'll remember from</p><p></p><p>section 4, an amp bridged into a given impedance draws twice as much</p><p></p><p>current as it would if it were driving two separate channels, each at</p><p></p><p>that impedance. So, a four ohm speaker stays a four ohm speaker, if</p><p></p><p>it's hooked to one channel, a bridged channel, a toaster, or the wall</p><p></p><p>socket. But, it is more stressful for the amp to drive any impedance</p><p></p><p>bridged than unbridged.</p><p></p><p>So, why do people talk about the impedance halving? Well, it's a</p><p></p><p>simple model that isn't correct but is easy to explain to people who</p><p></p><p>don't know what's really going on. It goes like this: When you bridge</p><p></p><p>the amp, each channel is "seeing" half the load presented to the amp.</p><p></p><p>So, if you bridge an amp to 4 ohms, each channel "sees" 2 ohms.</p><p></p><p>Therefore, each channel puts out twice as much power, and the combined</p><p></p><p>output is quadruple a single channel at 4 ohms.</p><p></p><p>Why is that still wrong? Because each channel isn't really used as a</p><p></p><p>single channel. You've used part of one channel, and an inverted part</p><p></p><p>of another channel to create a totally new channel, the bridged</p><p></p><p>channel. Also, there's no way for a channel to "see" only part of a</p><p></p><p>circuit. If it's "seeing" half the speaker, it's "seeing" it all.</p><p></p><p>Second, it makes it awkward if people believe that the impedance is</p><p></p><p>really, literally, changing. If you use that model, is it safe to run</p><p></p><p>a 4 ohm mono stable amp into a 4 ohm speaker? It should be, but we</p><p></p><p>just said the impedance halves, so that's now a 2 ohm speaker, and you</p><p></p><p>can't use it. That's wrong, and confusing, and it makes people think</p><p></p><p>they can't do things they really can.</p></blockquote><p></p>
[QUOTE="jsloan10, post: 929112, member: 560397"] 3.5.3 What happens when an amp is bridged? ------------------------------------------- Basically, one channel's signal is inverted, and then the two channels are combined to form one channel with twice the voltage of either of the original channels. Ohm's Law for Alternating Current states that I = V/Z where I is current, V is voltage, and Z is impedance. We also know that P = IV, where P is power. If we use Ohm's Law and substitute into the power equation, we get P = V(V/Z), which can be rewritten as P = (V^2)/Z. Therefore, power is the square of voltage divided by impedance. Now, why do we care about all that? Because it explains precisely what happens when an amp is bridged. I'll give a practical example and explain the theoretical basis of that example. Imagine you have a two-channel amp that puts out 50 watts into each channel when driven into a load of 4 ohms per channel. Since we know P and Z, we can plug these numbers back into our power equation and find V. 50 = V^2/4 -> V = sqrt(200). So, we're seeing a voltage of 14.1 volts across each channel. Now, imagine we bridge this amp, and use it to push just one of those 4 ohms loads. When the amp is bridged, the voltage is doubled. Since we know the voltage (2*14.1 volts), and the impedance (4 ohms), we can calculate power. Remember that P = V*V/Z. That means P = (28.2)^2/4, which is 198.1 watts. It should be clear by now that the new power is approximately 200 watts - quadruple the power of a single, unbridged channel! You can probably see that should be the case, especially if you look back at the power equation. Since P = V*V/Z, if you double V, you quadruple power, since V is squared in the power equation. Now, all this assumes the amp is stable into 4 ohms mono. The mono channel is putting out four times as much power as a single unbridged channel, so it must be putting out twice as much as the two single channels combined. Since the voltage on the supply side of the amp is dependent on the car's electrical system, it doesn't change (OK, the increased current might cause a voltage *drop*, but let's not worry about that now). Looking at the first power equation, at the supply side of the amp, we see P = IV. Now, when we bridged the amp, we doubled the power, but the input voltage stayed the same. So, if we hold V constant, the only way to double the power is to double the current. That means the amp is now drawing twice as much current when it's running at a given impedance mono than it would be running two stereo channels at the same impedance. There are only two ways the amp can do that - it can simply draw more through it's circuits, and dissipate the extra heat, or it can utilize a current limiter, to prevent the increase in current. Of course, using the current limiter means you don't get the power gains, either! So, if the amp can't handle the extra current, and it doesn't limit the current in some way, kiss it goodbye. For that reason, an amp is typically considered mono stable into twice the impedance it is considered stereo stable. 3.5.4 Does bridging an amp would halve the impedance of the speakers? ---------------------------------------------------------------------- Impedance is a characteristic of the speakers. The speakers don't give a flip how the amp is configured: they have a given impedance curve, and that's that. It should be clear that when you bridge an amp, you are changing *the amp*. The speaker's impedance is *not* a function of the amp, but the amp's tolerance to a given impedance depends completely on the way the amp is configured. If you'll remember from section 4, an amp bridged into a given impedance draws twice as much current as it would if it were driving two separate channels, each at that impedance. So, a four ohm speaker stays a four ohm speaker, if it's hooked to one channel, a bridged channel, a toaster, or the wall socket. But, it is more stressful for the amp to drive any impedance bridged than unbridged. So, why do people talk about the impedance halving? Well, it's a simple model that isn't correct but is easy to explain to people who don't know what's really going on. It goes like this: When you bridge the amp, each channel is "seeing" half the load presented to the amp. So, if you bridge an amp to 4 ohms, each channel "sees" 2 ohms. Therefore, each channel puts out twice as much power, and the combined output is quadruple a single channel at 4 ohms. Why is that still wrong? Because each channel isn't really used as a single channel. You've used part of one channel, and an inverted part of another channel to create a totally new channel, the bridged channel. Also, there's no way for a channel to "see" only part of a circuit. If it's "seeing" half the speaker, it's "seeing" it all. Second, it makes it awkward if people believe that the impedance is really, literally, changing. If you use that model, is it safe to run a 4 ohm mono stable amp into a 4 ohm speaker? It should be, but we just said the impedance halves, so that's now a 2 ohm speaker, and you can't use it. That's wrong, and confusing, and it makes people think they can't do things they really can. [/QUOTE]
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