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^ ownz you all ^
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One of the most basic laws in physics is ohm's law, it simply states that the current drawn by an electrical device is equal to the voltage applied across that device divided by its resistance (I=V/R). Preouts are extremely low current devices, they have a high output resistance (100-1000ohm normally) and are meant to simply feed a voltage to amplifers that have input resistances on the order of 10,000ohm or higher. This means that when they're sending a singal to these amps, the amount of current that is flowing through that output circuit is equal to the output voltage divided by the sum of the preout's internal resistance and the amp's input resistance, so in this case I=5/10100 = .5mA, or .0005 amps. We also know that due to the voltage division rule for a series circuit, the amp is getting 5*10000/10100 = 4.95 volts of that 5 volts that the preout is generating, the other .05 volts is being consumed by the preout itself due to its output resistance.

Now if you were to go in and attach a 4ohm speaker to these preouts instead of an amp, the current that would result would be I=5/104 = 48mA, or .048 amps. We also know that again, due to the voltage division rule, the sub would be receiving 5*4/104 = .2 volts of the 5 volts that the preout is generating, the other 4.8 volts is being consumed in the preout's output resistance.

This tells us 2 things
1 - as soon as you attached the speaker to the preout, the voltage would drop from 5V to close to .2V, meaning the speaker would be receiving practically no voltage at all. Since we also know that power is equal to the voltage squared divided by the resistance (P=V^2/R), the speaker would be receiving a whopping P=.2^2/4 = .01 watts.

2 - 48mA is WAY WAY too much for the preout stage in the headunit to handle. All of that current would fry the circuitry in a matter of seconds.
 
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