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Gear / Technical Help => Remote Power => Topic started by: nic on February 09, 2004, 10:06:33 AM
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according to ADK, the power consumption is <4mA....is this the power it draws from phantom?
pardon the ignorant question, but trying to work out the power requirements I will need for a true portable rig
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Include the mic draw to the draw of the pre. Say the pre darws 500mAh....add 8mAh for the TLs.
It's really not much to worry about.
EDIT
This may be obvious but in case it isn't:
8mAh = .008 Ah
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ah, now see I would not have thought to ADD it to the preamps draw.
duh....
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I inquired with Sound Devices about MP2 run-time a while back and their tech responded with the equation he uses to calculate run-time. It's not simply a matter of adding the 4mA (x2) to the preamp's draw. The process of stepping up the voltage requires some juice, too, which adds more to the equation than the 4mA (x2). Dang, wish I could find the email, but I think I deleted it.
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I inquired with Sound Devices about MP2 run-time a while back and their tech responded with the equation he uses to calculate run-time. It's not simply a matter of adding the 4mA (x2) to the preamp's draw. The process of stepping up the voltage requires some juice, too, which adds more to the equation than the 4mA (x2). Dang, wish I could find the email, but I think I deleted it.
I'm not sure why stepping up the voltage would INCREASE the draw....if anything it should lower it (for the mics). As voltage increases current decreases. I also don't understand why amperage would be lost due to increasing the voltage.........I'm also not as familiar with DC systems as I am AC. I'm sure I'm missing something here though and I'm very curious......if no answer I'll ask at school wed night.
:P
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I emailed my teacher a sample of our question...here's my email to him:
The microphone preamp is powered by 6 volts dc and has a current draw of 500mAh (with phantom power off). The preamp also provides 48 volt phantom power to the microphone. The draw of the microphone is 4mAh.
What is the TOTAL draw of the preamp when suppling 48 volt phantom power to the microphone?
and here is his reply:
I would suggest that we take the power consumed by the preamp. 6 Volts X .5 Amperes = 3 Watts and add the power consumed by the mike, 48 Volts X .004 Amperes = .192 Watts and add the powers together which gives 3.192 Watts. Now divide the 3.192 Watts by 6 Volts and you will have .532 Amps or 532 MA.
Now since this calculation assumes 100% voltage conversion efficiency, which doesn't exist, I would expect to see slightly more total current. Probably something like 550 MA or so which would represent a 64% conversation efficiency which would seem likely.
So it seems that we can't just add the mics in...you're right mr skalinder!
:)
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Nice!! Thanks for the clarification, Brian - looks just like what the SD guy sent me in an email a ways back. Good to know! All this hashed out, at least it doesn't have a massive impact on run-time calculations - they're just rough numbers, anyway. ~50mA here or there shouldn't result in power failure in the field as we should all be carrying around a bit extra battery capacity anyway (if we know what's good for us).
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Right...I knew it wouldn't be much but it's nice to see it all spelled out.
I'll have to bring my teacher an apple tomorrow night. :P