Basic Electronics Info For Noobs
- Thread starter Mark Linehan
- Start date
- th_trl_thread_readers 0
-
- Tags
- madvapes
- Status
Really? I saw a post somewhere that someone had their PV vaping at only 1a. I'll probably never find it now.... ummm nope. I know I saw one, but you know far more than I do so I will accept your word over theirs. 
What is the minimum amperage for a standard 510 then, just out of curiosity since the topic has arisen.
What is the minimum amperage for a standard 510 then, just out of curiosity since the topic has arisen.
I measured voltage of a PT and it gives 4.2 V and, when connected to a wall charger that gives 1 A, it woks nice for a 510 atty. With a standard battery it is supposed to work at 1.5 A but I think it really works at 1.2 A, thus there is very little noticeable difference.
you are correct in Im making a 5v passthru and I know that I need at least 2a of power to it. I will also be using the Joye 510 attys 2.2-2.4Ohm
SO if Im understanding you right I don't need any resistor added to the wire (because of the resistance of the addy) and the passthru will only pull the amps it needs?
I really am great-full for all the help and direction you guys are giving me.
To me a 510 will always be a Joye 510, 2.1-2.2Ω. Just saying '510' means nothing. Which one? A Joye, Bauway, SLB, Boge? The current requirements of each is significantly different. For 5V operation it's from 1.6 to 2.4A.Really? I saw a post somewhere that someone had their PV vaping at only 1a. I'll probably never find it now.... ummm nope. I know I saw one, but you know far more than I do so I will accept your word over theirs.
What is the minimum amperage for a standard 510 then, just out of curiosity since the topic has arisen.
Use this calculator.
Online Conversion - Ohm's Law Calculator
Plug in 1A and 2.2Ω and check what the volts are.
No meaningful statements can be made without all details, ohms being the most important.
Yup, pretty much, except "and the passthru will only pull" it's the atty that determines the current draw.
And we always want vaping (loaded volts) when figuring things out or giving out voltages.
With my PT, a Joye 510 and this 2000mA adapter.
I get 4.17V under load (vaping). If I was to check the volts without an atty it will of course be ~5V, but that means nothing as far as vaping.
Here's an example of a 'reviewer' who says he's proving a vendors PT is in fact 5V. This man knows nothing, he has proven that USB ports output 5V, nothing else. He should get a clue before doing such a stupid and inane video that only misinforms. Without a load, his 'test' is a joke.
And what is a 'properly powered hub'? What are the specs? He obviously doesn't know.
Now here's how to test vaping, under load volts. We get important details, brands, ohms etc, a real test.
First off I am back. I will post later about the limits I have now. But to answerer the question:
Loaded vs. unloaded voltage:
The total voltage in a circuit it equal to the total of all the individual voltage drops in the circuit.
LOL… what heck does that mean?
The voltage in a circuit will divide itself across all the individual resistances in a circuit and add up to the total supply voltage.
Let’s look at a very simple circuit.
A 5 volt battery and a 5 ohm load.
What we have is 5 volts across a 5 ohm load so all the voltage would be “felt” across the single resistor (well not really but hang with me for a bit and I will explain). In this circuit we would have 1 amp flowing… right? I = E / R 1= 5/5 The voltage drop across the one resistor would be (E = I x R) 1 amp X 5 ohms = 5 volts The power, or heat from that one resistor would be 5 watts (P = I 2 x R) 1X1X5 =5.
Now let’s take that same circuit and add 2 resistors:
5 volt supply with 2 – 2.5 resistors in series:
Each resistor will “drop” 2.5 volts across it. So 2.5 volts plus 2.5 volts will still equal the 5 volt supply, and we would still have 1 amp flowing in the circuit. You can prove this with ohms law. 5 volts total with 5 ohms total still equals 1 amp total (I = E / R) 5/5=1. Now to see the voltage drop across each resistor you would use:
E = I x R 1 amp X 2.5 ohms = 2.5 volts. Each resistor will only “feel” 2.5 volts. When we add all the drops in the circuit we get 2.5 + 2.5 = 5 volts.
The power (heat) from each resistor would be (P = I 2 x R) 1 X 1 X 2.5 = 2.5 watts.
OK, one more circuit to make this clear.
5 volt supply with 2 resistors that are 1 ohm and 4 ohms in series.
We still have a 5 volt supply and still have 5 ohms total so we still have 1 amp flowing, right? BUT the voltage drops will be different. The one ohm resistor will drop 1 volt
(E = I x R) 1 = 1 X 1 and the 4 ohm will drop 4 volts. (E = I x R) 4 = 1 X 4. Total voltage drops equal 1+4=5.
So, as I said before. The total of the voltage in a circuit will always equal the sum of the individual voltage drops.
Sorry for the tech stuff above, but I had to do that to answer the loaded voltage question.
If we look at a simple mod circuit we have: 1-battery, 1- atty or carto, and one switch. Correct? And in a perfect world we could say, we have a voltage supply, a resistor (atty), and a switch, but we do not live in a perfect world. What we REALLY have is….. A voltage supply, a resistor (atty) , a switch, AND the internal resistance of the battery. I bet you thought that the battery was a voltage supple and not a resistor….It is both.
When you measure unloaded voltage your circuit would be …. Supply, supply resistance, switch and meter. The meter “looks” like a HUGE resistor to the circuit and therefore “sees” almost all the supply voltage. Remember in the above circuit the 4 ohm resistor saw much more voltage that the 1 ohm one. In this cast the meter is 10,000 or more ohms, and there is very little current flowing, so almost all the voltage in the circuit is seen buy the meter.
When you measure loaded voltage you are looking at the voltage across the 2-4 ohm atty minus the voltage across the batteries internal resistance.
The higher the battery resistance, the regulator chip, wire you are using, or even the switch resistance, you have the greater the difference in loaded vs. unloaded with be.
Last edited:
- Status
Similar threads
