First impression: Herron VTPH-2A phono preamp
Extremely nice words and I agree the listening community will be at a loss. |
@jw944ts Al’s math is right. A doubling of output power to your speakers corresponds *approximately* (not exactly) to +3 decibels of increased SPL. But the gain from a phono stage is not amplifying power - it is amplifying the source signal’s voltage. When you double the voltage signal input to your power amplifier (e.g. from 0.5 V to 1.0 V), a linear amp responds by outputting twice the voltage AND necessarily twice (approximately) the current - because the speaker load (represented by R in ohms) remains constant and the law is: V = I * R. Since power is V * I (voltage times current), you end you with 4 times the output power from the doubling of signal voltage! Therefore a doubling of gain in your phono stage corresponds to (approximately) a +6 dB increase in SPL. This is also why bridging a stereo amplifier to mono nets you up to 4 times the output power, not just 2 times (assuming the power supply and heatsinks and output stage are up to task). It’s not "magic" or free power - the amp is working all that much harder to push the extra current (and be very wary of hooking a bridged amp into 4 ohm and less speakers)! This means 60 dB of phono gain is approximately equivalent to an amplification factor of 2^10 (a 6 dB voltage doubling, 10 times) or 1,024. More exactly, 60 dB of gain is exactly equal to 10 ^ 3 = 1,000 - since 20 dB corresponds exactly (not roughly) to a voltage multiplier of 10x. The 3dB / 6 dB doubling rules-of-hand are a (close) approximation to make the math easier by tossing out some nasty decimal digits. In short, a doubling of voltage (in most applications) results in a SPL increase of approximately +6dB; a 10x voltage amplification factor is exactly +20dB. You can mix-and-match these two shortcuts to approximate a wide range of gains. E.g. 32dB = 20dB + 6dB + 6dB = 10 * 2 * 2 = a voltage amplification factor of 40 times (decibels are added as amplification factors are multiplied). A doubling of *power* is an SPL increase of approximately +3dB; a 10x power amplification factor is exactlyequal to +10dB. Use these shorthand rules to impress your friends with Rainman-like quick calculations :) Also, the VTPH-2A is a very nice sounding phono stage :) |
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@mulveling , thanks for providing the comprehensive response to the question posed by @jw944ts. In doing so you saved me a lot of time, and I of course agree with what you wrote. JW’s calculation reflects a commonly held misconception that a "voltage db" and a "power db" both correspond to 10x the logarithm of the ratio between two voltages or two power levels, respectively. But they don’t, at least in terms of generally accepted usage, if not etymology as well. 10x is the proper multiplier to use when computing the difference between two power levels in db, but 20x is the multiplier that should be used in computing the difference between two voltage levels in db. A db is a db. It is not either a voltage db or a power db. The numerical result will be the same regardless of whether the number of db is calculated from voltages or from power levels, assuming impedances are the same in the two cases. That is a consequence of the fact that if impedances are the same power is proportional to voltage squared, as mulveling indicated. And therefore squaring a 2x increase in the voltage provided into a given impedance corresponds to supplying 4x the power, not 2x. (In the interests of simplicity I’m putting aside effects that occur when the load impedance is not purely resistive). And assuming the speakers and the rest of the system are operated within limits that allow them to perform in an essentially linear manner, a gain of 66 db corresponds to a voltage multiplication of about 1995x, that commonly being rounded off to 2000x. (20 x log(1995) = 66 db, where "log" is the base-10 logarithm). And if everything else is equal that results in approximately 4,000,000x (2000 squared) as much power being delivered to the speakers, and correspondingly to an approximate SPL increase of the same 66 db. (10 x log(1995 x 1995) = 66 db). Regards, -- Al |
