Re: Questions on impedance matching
Ted Rook
Yes I agree, with 600 ohms attached the terminal voltage is the nominal value, with open
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circuit it is double. In audio we NEVER match source and load impedances, not with small signals nor with power amplifiers and loudspeakers. The universal system comprises low source impedances and high load impedances, ratios of 1:10 to 1:1000 are common. Things are happening in audio at such a slow speed that reflected energy is irrelevant. Reflections don't cause "distortion" in audio. Distortion in audio occurs due to nonlinearity in the amplification and from abuse of amplification by the connection of load impedances below the rated value. DECIBELS ratio of two voltages expressed in dB is 20 times the log base10 of the ratio in round numbers: double is +6dB half is 6dB 3x is +10dB 10x is +20dB 100x is +40dB ratio of two powers expressed in dB is 10 times the log base 10 of the ratio in round numbers: double is +3dB half is 3dB 10x is +10dB 100x is +20dB Hope this helps Ted
On 12 Feb 2018 at 6:34, David Berlind wrote:
thank you @tedR: when you say "if it (the 600 ohms) is absent," you actually mean if there's no load at all. Because, according to the math, other loads (above or below 600 ohms) should not necessarily yield double the voltage. It will be somewhere in between. I'm trying to understand the use cases for where maximum power transfer (watts) trumps maximum voltage (amplitude) and vice versa. My assumption is that as you move off the maximum power transfer point (where impedances match), you get signal distortion due to refection and at that point, maximizing voltage means you're just maximizing a distorted signal which is not helpful in radio or cable communications, but might be desirable in audio. The relationship to dB is a new dimension for me. What's the math behind any dB calculations and what is optimal? What dB am I shooting for?

