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Quick noob question.
- Thread starter H4X0R46
- Start date
Yes, and here is how:
(Wattage used/single battery V cutoff/number of batteries)/mod efficiency %
(Wattage used/single battery V cutoff/number of batteries)/mod efficiency %
Thanks! I might get me a new mod just to have another on hand, I'm used to my RX200S, just wanted to make sure there's nothing drastically different if I choose to go with a parallel mod! Thanks!
Hey nightshard, quick question. So I run my RX200S at 50w most of the time, and the amp draw that shows on the screen when hitting it shows an amp draw of about 9amps. Well....... Doing the math, at 3.0v it would come out to about 5.5amps per cell, where is the board getting this 9amp figure from?
Since the coil is not directly connected to the battery and passes through a regulator, an easy way to visualize it is to think of two separate circuits, first one which includes the chip and the battery and the second one which includes the chip and the coil.
The formula above lets you calculate the current for the first one and ohm's law for calculating the second one.
So based on ohm's law 50W @ 9A means you're using a coil of around 0.6 ohm.
The formula above lets you calculate the current for the first one and ohm's law for calculating the second one.
So based on ohm's law 50W @ 9A means you're using a coil of around 0.6 ohm.
Alright, so this would mean I'm correct in saying that 5.5amps is what's being pulled from each battery cell, and the 9amp figure I see is what the regulator chip is feeding the coil?
I am using a 0.5ohm coil, so that sounds about right as well.
I don't think so, 50W/3V/(2 batteries) = 8.3A. If you're at 90-95% efficiency then that comes out to about 9A if it's smart enough to know its own efficiency. Remember this is only when the battery is empty though, when it has more charge than 3V the current will be lower.
Using an RX200S so it has 3 cells. Thanks for the reply!
Thanks for the reality check! So yes you're right, 5.5A per cell if you're not taking efficiency into account.
adding efficiency would be dividing the amp draw by the decimal correct? Example: 50(watts)/3(batteries)/3(worst case scenario)/.90(efficiency assuming it's 90%) right?
You got it. So then for 90% you're looking at 6.2A or so.
Thanks Jon! I should take this opportunity to give kudos to liionwholesale, got my batteries from you guys and they're legit!