Voltage drop

[tomkp]

How do I find out what the voltage drop of my mech is?

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[KenD]

Check the voltage of your battery with a multimeter or an inline voltmeter (one that screws in between the mod and the atty, for checking the battery charge fire wirhout the atty attached). This is your battery charge level. Insert the battery into the mod, screw on an rda or an rta deck without the topcap, fire the mod and measure the voltage at the positive and negative posts of the atty (with an inline voltmeter you simply screw the it in between the mod and atty, fire the mod, and you'll see the voltage under load). This is your voltage under load. Subtract the voltage under load from the battery charge level and you'll know the voltage drop. Note however that the main factor in voltage drop is the internal resistance of the battery. The internal resistance rises as the resistance drops, and therefore you get more of a voltage drop at lower resistances. Measuring the voltage drop caused by the mod itself (the mod's resistance actually) is difficult.

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[Robert Cromwell]

Check the voltage of your battery with a multimeter or an inline voltmeter (one that screws in between the mod and the atty, for checking the battery charge fire wirhout the atty attached). This is your battery charge level. Insert the battery into the mod, screw on an rda or an rta deck without the topcap, fire the mod and measure the voltage at the positive and negative posts of the atty (with an inline voltmeter you simply screw the it in between the mod and atty, fire the mod, and you'll see the voltage under load). This is your voltage under load. Subtract the voltage under load from the battery charge level and you'll know the voltage drop. Note however that the main factor in voltage drop is the internal resistance of the battery. The internal resistance rises as the resistance drops, and therefore you get more of a voltage drop at lower resistances. Measuring the voltage drop caused by the mod itself (the mod's resistance actually) is difficult.

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good explanation. Most of the reviewers I have seen are actually measuring battery voltage sag under load. All batteries output voltage drops some under load. Some drop more than others.

 

[Robert Cromwell]

According to Mooch's graph a fully charged 25R will only put out 3.8 volts at 20 amps. And drops down from there as it discharges.

 

[tomkp]

Thank you everyone for the replies. My dumbass forgot about the inline voltmeter lol

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[mhertz]

If I could please ask a question also about this, please...

Yeah, an e.g. 0.18 build runs usually around 3.7-8v under load with good batteries e.g. samsung 25r's...

As stated this mostly is battery sag and not mod voltage-drop, and the battery internal resistance rises upon higher loads(lower ohms).

Now when calculating amp draw for a specific load, e.g. 0.18, then is is then correct that the mod resistance i.e. voltage-drop should of course be included, but not the battery sag, as the internal resistance of the battery hinders the amp drawn out from it?

So, just as an example, if a 0.18 build on a fresh 4.2v battery on a mod with e.g. a 0.1v voltage drop, and then under load the output is meassured at 3.7v. Then is it correct that the amps drawn at the beginning would not be calculated upon 3.7v, but of 3.8v, as the sag should be omitted, but not the 0.1v mod voltage-drop, correct?

So the amps drawn in above example is 3.8/0.18 = 21.11 amps. Of course its easier to just go by 4.2v for a quick calc, but I just wanted to learn if the above was correct to calculate the actual amp draw and not a "fictional" one...

To sumerize; should battery internal resistance be omitted or included if wanting to calculate the actual amp draw from the battery? I know for sure the mod-drop should be included, but am not 100% sure about the battery internal resistance should be omitted?

Thanks in advance!

EDIT: Sorry, of course it is correct...

 

[tomkp]

When using an inline voltmeter, without Atty, the voltage shown would be the voltage used in the calc as that is the voltage that hits the atty

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[Robert Cromwell]

If I could please ask a question also about this, please...

Yeah, an e.g. 0.18 build runs usually around 3.7-8v under load with good batteries e.g. samsung 25r's...

As stated this mostly is battery sag and not mod voltage-drop, and the battery internal resistance rises upon higher loads(lower ohms).

Now when calculating amp draw for a specific load, e.g. 0.18, then is is then correct that the mod resistance i.e. voltage-drop should of course be included, but not the battery sag, as the internal resistance of the battery hinders the amp drawn out from it?

So, just as an example, if a 0.18 build on a fresh 4.2v battery on a mod with e.g. a 0.1v voltage drop, and then under load the output is meassured at 3.7v. Then is it correct that the amps drawn at the beginning would not be calculated upon 3.7v, but of 3.8v, as the sag should be omitted, but not the 0.1v mod voltage-drop, correct?

So the amps drawn in above example is 3.8/0.18 = 21.11 amps. Of course its easier to just go by 4.2v for a quick calc, but I just wanted to learn if the above was correct to calculate the actual amp draw and not a "fictional" one...

To sumerize; should battery internal resistance be omitted or included if wanting to calculate the actual amp draw from the battery? I know for sure the mod-drop should be included, but am not 100% sure about the battery internal resistance should be omitted?

Thanks in advance!

The actual internal resistance of the battery should be included in the overall resistive load. the actual voltage sag/internal reisistance will vary a bit from battery to battery for a really accurate amp measurement an amp meter should be used.
Using full batery voltage for you calc is fine for figuring safety limits as you do not run on the very ragged edge for reasons of safety.

 

[mhertz]

Thanks both of you, I appreciate it! I didn't think this completely through I can see, sorry! I've never measured v-drop/sag or anything myself, but was just interested in the theory of it, but I can see that it means "back to school"(net-reading) for me then Of course I know that for safety it should just be easily calc'ed without any of these considerations i.e. based always on just 4.2v regardless...

Thanks again!