CloudedScoundrel Posted January 27, 2015 Posted January 27, 2015 Hello everyone. I'm new to the vape community and have been learning quite a bit in the past couple of weeks. With that said, I keep hearing the term "voltage drop". I'm not quite sure what exactly that means or how to tell if it is happening with my device. If this topic was already posted, I appologize. I spent a little while looking, but gave up pretty quickly.
Tug Posted January 27, 2015 Posted January 27, 2015 http://www.atmizoo.com/blog/voltage-drop/?lang=en CloudedScoundrel 1
Bebop Posted January 27, 2015 Posted January 27, 2015 In lay terms it is the loss of voltage that is absorbed by the components of your device before it reaches your atomizer. For example if your battery is putting out 4 volts but your atomizer or coil is seeing 3.8 volts then you have a voltage drop of .2 volts. You would have to use some equipment to measure the voltage drop accurately. Different devices have different amounts of voltage drop depending on their construction and the materials used. A switch for example with poor conductivity due to inferior materials would create more voltage drop than a high quality switch with premium materials. CloudedScoundrel 1
Compenstine Posted January 27, 2015 Posted January 27, 2015 Voltage Drop Test, How to do one. CloudedScoundrel 1
Bebop Posted January 27, 2015 Posted January 27, 2015 He doesn't say but looked like 3.75 on the coil?
Compenstine Posted January 27, 2015 Posted January 27, 2015 looked like a 3.78, but it is hard to read.
CloudedScoundrel Posted January 27, 2015 Author Posted January 27, 2015 Thanks so much guys. I really appreciate the responses.
vaping_jake Posted January 27, 2015 Posted January 27, 2015 when he measured the 4.7 it was power source (battery) unloaded. When he attached the atty to the battery it was now under load. His test is somewhat valid just need a few more steps to make it accurate. he should have but a white ceramic load on it so you get current draw.. then put the atty on and see what the total drop was. But then put in the tolerance of the meter used and it all gets lost in the wash.
vaping_jake Posted January 27, 2015 Posted January 27, 2015 In a very rough explanation; am sorry about the wording, English is my second language. If anyone can translate italian i would be happy to explain it that way or just look up KIRCHOFF'S laws. Basically one would need to know the voltage potential right before the load. take that potential and subtract it from the source potential. this will give voltage drop up to source. Then do it after load and at load. Voltage total = Voltage potential till load + Voltage drop on load(coil)+Voltage potential after load. The sum of the voltage potential before load and after load is what the voltage drop of the device would be. yes its over kill.and fluctuating magnetic fields aren't being considered ... and assumption that the electric charge (Q) remains constant..lol jeeze now i need to vape CloudedScoundrel 1
WillBlack Posted January 28, 2015 Posted January 28, 2015 (edited) Voltage Drop Test, How to do one. It wasn't really a test of the mod's voltage drop. Here's a pic of Field of Vapors tester and results. What he's pressing is one of Zen's mechs. The load is .5Ω. Assuming that the meter and load is accurate we see an interesting result. Using 3.51V and .5Ω in a calc we get 7.02A, but the thing has measured the amps at 6.6. There is a difference of .42A. Can we safely assume that the mod itself has actually sucked up nearly half an amp? Can we say that the mod itself has a parasitic resistance of .03Ω? I like this test, at least for the voltage drop part of the the actual mod itself. He messes up on the amps though he should be adding the 0.04 that his meter is showing at rest. Edited January 28, 2015 by WillBlack jasonculp and Compenstine 2
jasonculp Posted January 28, 2015 Posted January 28, 2015 It wasn't really a test of the mod's voltage drop. Here's a pic of Field of Vapors tester and results. What he's pressing is one of Zen's mechs. The load is .5Ω. Assuming that the meter and load is accurate we see an interesting result. Using 3.51V and .5Ω in a calc we get 7.02A, but the thing has measured the amps at 6.6. There is a difference of .42A. Can we safely assume that the mod itself has actually sucked up nearly half an amp? Can we say that the mod itself has a parasitic resistance of .03Ω? I like this test, at least for the voltage drop part of the the actual mod itself. He messes up on the amps though he should be adding the 0.04 that his meter is showing at rest. I would love to have a schematic of this. I don't really need it, but I would like to know how it works!
vaping_jake Posted January 28, 2015 Posted January 28, 2015 for some reason i can't seem to get the video, just going via picture. I agree many assumptions to consider: did he use precision (<2% tolerance), high wattage ceramic resistors Why did he add a switch which adds a load, instead of three individual paths of resistance Accuracy of meters and were all connectors calibrated out but then again do we need a grenade to take out an ant hill or one of my favorite sayings is measure with a micrometer, mark with a chalk, cut with ax
WillBlack Posted January 28, 2015 Posted January 28, 2015 (edited) You're right, seems this forum doesn't like start points. The testing starts at about 8min. Edited January 28, 2015 by WillBlack
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