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Thread: Opus mod

  1. #1
    Variable Bitrate Moneyfink's Avatar
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    Opus mod

    This is from the opus 150 owners manual:
    "The normal full load operating voltage range of the power supply is between 7V to 18VDC. The micro controller monitors the battery voltage so the battery will not be deep discharged. When the ignition switch is in the aux position or delayed turn off enabled the computer can be powered up. Under this condition the engine is not running, so the battery is not being charged. When the battery voltage dips below 11V for more than 10 seconds continually the SFX- DC-DC power supply will enter automatically into a shutdown sequence. This will protect the battery from deep discharge, and the battery is saved for its main intended purpose, to start the vehicle."

    What i want to do is put a resistor infront of the opus so that the opus sees a lower voltage than what the battery actually has. so for example, if the battery has 11.5 charge, the opus might only see 11.0, so that the opus would shut down. this resistor would just drop the voltage a bit. will this work? i've been having a lot of dead batteries b/c i go days at a time without driving, and make short trips. what kind of drain does an epia m10k with 2 usb devices have when the computer is off.
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    Maximum Bitrate binary.h4x's Avatar
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    you are actually going to draw more power. By adding a resistor you are making it harder for the voltage to flow to the opus, which means it will be draining more from your battery than without the resistor.

    You could always install a switch along the remote line and manually turn off the carputer when you know you are going to make a short trip.
    2007 Honda Fit Sport 1.5L SOHC-VTEC

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    Binary: But that's exactly what the resistor's for. The only additional battery drain would be in the excess heat wasted, though that should be minimal. Just make sure to choose a resistor that can conduct the full amperage that the Opus will pull. I tried to work it out, but the math it all way to complicated for me, highschool algebra being about 11 years ago.

    I agree that a switch would be a better way to go, but what I would do is cut all power to the opus/carputer by means of a relay. It's a bad idea to leave your battery half drained for days at a time, and does it all really need to have power when you're not using it? My system connects the coil on the relay to the ACC lead-- When the ignition is off, or the car is cranking, the audio components and laptop are all disconnected from the starter battery.

    A 20a Bosch Relay is $20 +shipping on ebay... You'll likely even be able to find one in a local auto parts store.

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    Why don't you hibernate it? There's very little power draw in hibernation, although there is the 5v drain from the USB devices.

    If you're really worried, get a delay timer that you can set to two days or so and operate a relay with it. Cut the power off after whatever time you'd like.
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    Maximum Bitrate binary.h4x's Avatar
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    Quote Originally Posted by raucous
    Binary: But that's exactly what the resistor's for. The only additional battery drain would be in the excess heat wasted, though that should be minimal. Just make sure to choose a resistor that can conduct the full amperage that the Opus will pull. I tried to work it out, but the math it all way to complicated for me, highschool algebra being about 11 years ago.

    I agree that a switch would be a better way to go, but what I would do is cut all power to the opus/carputer by means of a relay. It's a bad idea to leave your battery half drained for days at a time, and does it all really need to have power when you're not using it? My system connects the coil on the relay to the ACC lead-- When the ignition is off, or the car is cranking, the audio components and laptop are all disconnected from the starter battery.

    A 20a Bosch Relay is $20 +shipping on ebay... You'll likely even be able to find one in a local auto parts store.

    Right. If the Opus is drawing 150watts @ 12VDC then there will be 12.5A flowing to the Opus. Putting a resistor inline to drop 1VDC, it will need to disipate 12.5 Watts as heat. Summing these will give you a total power draw of 162.5 Watts.

    So infact by adding a resistor you will drain your battery faster, not to mention a 15Watt resistor isn't gonna be cheap.

    EDIT: Yup I screwed it up, Opus @ 150W and 11VDC uses 13.6A. So infact, you will be using 163.6Watts.
    2007 Honda Fit Sport 1.5L SOHC-VTEC

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    Quote Originally Posted by binary.h4x
    Right. If the Opus is drawing 150watts @ 12VDC then there will be 12.5A flowing to the Opus. Putting a resistor inline to drop 1VDC, it will need to disipate 12.5 Watts as heat. Summing these will give you a total power draw of 162.5 Watts.

    So infact by adding a resistor you will drain your battery faster, not to mention a 15Watt resistor isn't gonna be cheap.
    Correct me if I'm wrong, because I may well be. What I'm thinking is:

    Reversing the math, we take the 12.5A draw and multiply it by the reduced voltage (now 11V), we get 137.5 watts. Adding the 12.5 watts being dissipated by the resister we've now got, again, 150watts being used between the Opus and the resistor.

    Were you accounting for that 12.5 watts twice, or am I missing something?

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    Maximum Bitrate binary.h4x's Avatar
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    You should start with Power, because as long as the components of the carputer stay the same, they will draw the same amount of power. Its an Opus 150, nothing you do can change that.

    Power = Volts x Amps >> Amps = Power / Volts

    So if the Opus gets 11volts, Amps = 150W / 11VDC = 13.6A
    2007 Honda Fit Sport 1.5L SOHC-VTEC

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    Maximum Bitrate mushin's Avatar
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    If you assume a constant amp draw (and thus resistance) from the opus, wouldn't adding an resistor inline reduce the power draw (due to higher resitance overall) rather than increase it?

    Anyway, the simple equations are all moot because the opus is decidedly not a constant load. Amperage is going to be varying greatly with system state, and if you reduce the voltage the opus sees I'm sure it'll pull more amps to maintain regulated power.

    I think, anyway (IANAEE)

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    Maximum Bitrate binary.h4x's Avatar
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    Maybe this formula will help you understand.

    Assume a constant current draw:

    V=IR AND P=VI , substitution yields P = (I^2) x R


    Assume a constant current draw of 10 Amps, So:

    Power= 100A x R

    Increasing resistance will increase power.


    Think about it, if you are pushing 2 boxes, both are the same size, but one is light, and the other is heavy.
    Which one is gonna require more power to move?
    Why? Because the heavier box has more RESISTANCE to movement.(Inertia)
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  10. #10
    Maximum Bitrate mushin's Avatar
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    Binary, your conclusion there of increased power implies increased voltage, which isn't going to happen. When voltage is constant and you increase resistance, current, and thus power, will go down.

    I think what actually would happen if you were to try this, is that the Opus would adjust it's internal resistance down so that it can maintain pulling the required amps. Efficiency will likely go down, so it will need more amps, with the end result being a slightly lower total (opus + resistor) resistance and a slightly higher current and power draw.

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