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And in terms of signal voltage, we're trying to distribute voltage these days, not power, so impedance mismatches are less significant. For transferring voltage, you just want the load to have a lot higher impedance than than the source. Besides, the 4073 has voltage to spare. I don't disagree that this is not best practice, but I think the odds are in Kevin's favor. |
Phantom POWER is POWER, not just voltage. Power = Current x Voltage. We already have existing examples of distortion due to low milliamp capability of some Phantom Power supplies.
I'm bemused by the "how to get by with less" attitude exhibited here. I once had a client suggest that he reduce his ad budget by 10% to see if his sales went down 10% or hopefully less (there's at least one in every crowd.) I suggested he shoot one toe off and see if it had any effect on his abilty to walk properly. If not, shoot another! If you're doing this for yourself, it really doesn't matter. If you are getting paid as a professional, it does. Thanks, Ty |
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drop with a 20dB pad would be the square root of 20 i.e. 4.5% of the mV/Pa? |
No, -20 dB is a drop to 10% of the reference level. dB is 20 log (new level / reference level). Going the other way, the ratio of new level to reference level is 10 raised to the power (dB / 20).
So for the earlier example of -10dB that was 10 raised to the power of -10/20 or 10 raised to the power of -1/2, which is 0.316 (not actually the square root of 10. +10dB is the ratiothat happens to be equal to the square root of 10). For the case of -20dB it's 10 raised to the power of -20/20, or 10 to the -1 power, or 1/10 |
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***edit*** why did the forum server make 2 of these?
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But yes Doug cleared it up, they are a great company making some great products with very good support. |
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The voltage factor for 10 dB is sqrt(10). In this case as you are padding (reducing) level you divide by sqrt(10) = 3.1622... Thats equivalent to multiplying by 1/sqrt(10) = sqrt(10)/10 = 0.31622.... I wasn't very clear on that.
There is no code requirement for any of the commercial radiotelephone licences. The commercial radiotelegrah licences did have a requirement. I think it was 25 wpm random groups of 5 for the first class. Impedance matching isn't too critical at audio frequencies as is clear by the fact that the sources (microphones) tend to have low output impedances and sinks (mixers, cameras, recorders) tend to have high input impedances. Thus it's clear the mic is intended to serve as a voltage source. So if, for example, the mic has an impedance of 100 ohms and a mixer 5900 (to make the math easy) 5900/6000 = 98.3% of the microphone's open circuit output voltage would appear across the mixer's input. If two mixer channels are paralleled (with a Y cable) the mic is loaded by 2950 Ohms so that 2950/3050 = 96.7% of the mic voltage appears across the input. This is a very small loss (0.14 dB) and all should be fine assuming that the Y cable is properly wired. |
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