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What about two amps on one set of preouts?
Quote:
Yes. When two loads are connected in parallel (such as with a Y-cable) they get the same voltage as each other. They do NOT get the same voltage as if only one load was connected because the head-unit has an internal resistance (typically around 600 ohms). So, given that the amp has a typical input impedance of around 10k ohms then we get something like this: Code:
----------------------------- ---------------------------- HEAD UNIT ________ | | AMP | ______| |_________Vamp___________ | | | R(head)| | | | | _ | __|__ |________| | | __|___ |__| - _ | / \ | | | | | -___|__ | Vi | | | |R(amp)| | _- | \_____/ | | |______| __| _- | |_______________________________|________| - | | | | ----------------------------- ---------------------------- So, assuming a typical head unit and single amp the voltage seen at the amp (Vamp) is given by (Ohms law/Kirkov's law/1st year EE/high school electronics technology class/etc.): Code:
R(amp) Vamp1 = Vi * ------------------ R(amp) + R(head) Vamp1 = Vi * 0.94 Code:
10000/2 Vamp2 = Vi * --------------------- 10000/2 + 600 Vamp2 = Vi * 0.89 If you had a more typical 1V preout you would get Vamp1 = 0.95V and Vamp2 = .89V, also not a noticeable drop. This is also why this is slightly more susceptible to noise than a direct one-to-one connection. If the noise level inserted due to cabling was 0.1V per cable then the noise level in the signal reaching each of the two amps would be a slightly higher percent of the signal level but not doubled. (this is also why the 4V head unit is favored over the 1V unit for noise immunity: 0.1V noise / 3.76V or 3% is much less than 0.1V noise / 0.95V or 10% even in a one to one connection). |
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