Will this work? (passthrough)

[modeezy]

I recently made a USB passthrough but every USB device I have feels like it doesn't provide nearly as much power as my 901 battery does. (multiple pc's, comcast dvr, even went as far as splicing a molex and trying to connect that way) I get 5v+ out of every device I test with a multimeter, but I think maybe the amps is what causes it to not provide much.

ht tp://w ww.radioshack.com/product/index.jsp?productId=2049707
(remove spaces between tt and www)
4.5V/1600mA AC-to-DC Power Adapter


Will this work good or is there too much mA?

 

[Kewtsquirrel]

I recently made a USB passthrough but every USB device I have feels like it doesn't provide nearly as much power as my 901 battery does. (multiple pc's, comcast dvr, even went as far as splicing a molex and trying to connect that way) I get 5v+ out of every device I test with a multimeter, but I think maybe the amps is what causes it to not provide much.

ht tp://w ww.radioshack.com/product/index.jsp?productId=2049707
(remove spaces between tt and www)
4.5V/1600mA AC-to-DC Power Adapter


Will this work good or is there too much mA?

Thats 7.2watts, which is a bit high, throw a couple diodes or resistors in there and try to get it a bit lower.

 

[modeezy]

Thats 7.2watts, which is a bit high, throw a couple diodes or resistors in there and try to get it a bit lower.

Thanks for the reply! Any recommendation of what OHM of resistor I would need?

 

[Laredo7mm]

Thats 7.2watts, which is a bit high, throw a couple diodes or resistors in there and try to get it a bit lower.

How does that work? Even a 3.3 ohm atty at 4.5 volts will only pull 1.36 amps. Most atty's seem to be around 3.7 ohms which will draw even less current. Just because the power supply is capable of putting out 1.6 amps doesn't mean that it will run the atty at 1.6 amps. Your logic is like saying that if you turn on your 100W light bulb that it is going blow your circuit breaker because it is going to try to draw all of your 150 service amps.

modeezy - That power supply is what I use for a plug in. No resistors needed. It works great. Just wire it up, plug it in, and vape away.

 

[Kewtsquirrel]

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[Laredo7mm]

Lets start out with some known values first, and work from there:
Wattage can be calculating using the formula W=VI, V is voltage, I is current (in amps)

First, your standard DSE901 battery. At peak (We always want to use peak numbers) you're looking at roughly 4.2v and 150mA, which means your output is roughly .67watts

Now, we know that 901's can be safely run at much higher power levels, Magnum mods that run on unprotected CR2 batteries sit at roughly 6.5v/.9A, which means the wattage output is roughly 5.85W at peak.

However, some users have reported a burnt taste at this level, and claim the sweetspot is around 4.5v/.9A, or 4.05W.

The USB wallwart that you would like to use is outputting 4.5v/1.6A, or 7.2W, thats quite a bit of power so lets see what we can do to get that down.

We want to reduce your total wattage by a value of 3, and since we know the amperage level, we can deduce from this the desired current drop.
If W=VI, and both W and I are known, we must simply solve for V. So V = W/I, or V = 3/1.6. This gives us a current drop of 1.875V, now we can plug this back into our original equation and check the wattage output.
W=(4.5-1.875)*1.6

This would give us a 4.2 watt output @ 2.6v/1.6a, perfect! Now we just need to figure out how the hell we can drop the voltage by 1.875v!

When it comes to resistors, and dropping voltage, I'm not your guy, but you should be able to use a potentiometer to do this, just keep adjusting it until your output voltage is 2.6v or so.

More bad info. If you feed an atomizer, that has 3.7 ohms resistance, 2.6 volts, how is that going to draw 1.6 amps?

 

[Kewtsquirrel]

I'm not following, maybe you can explain it better, this is my first stab at dropping the voltage on something like this to not burn an atty.

How is the atomizer "drawing" anything? The power supply is outputting 4.5v @ 1.6amps, you're wiring that directly to the atomizer, how would it *not* receive 4.5V @ 1.6amps?

 

[Laredo7mm]

modeezy - I think it is your best interest not to follow any of Nerf's advice.

 

[Kewtsquirrel]

modeezy - I think it is your best interest not to follow any of Nerf's advice.

Your advice has been pretty amazing, I bet modeezy knows exactly what to do now.

Nice explanation, btw.

 

[Laredo7mm]

I'm not following, maybe you can explain it better, this is my first stab at dropping the voltage on something like this to not burn an atty.

How is the atomizer "drawing" anything? The power supply is outputting 4.5v @ 1.6amps, you're wiring that directly to the atomizer, how would it *not* receive 4.5V @ 1.6amps?

Because that is not how electronics work. If this is your first try at dropping voltage then why are you giving advice?

The atty does draw current when voltage is passed through it. The atomizer is noting but a resistor. What happens to resistors when you pass voltage through them? They put a load on the system. Load is current.

Ohms law. V=IR. I=V/R. If V = 4.5 and R = 3.7, then I = 1.216 amps.

Just because the power supply has the capacity to put out 1.6 amps, does not mean that is what is going to be used by the atomizer.

 

[Laredo7mm]

Your advice has been pretty amazing, I bet modeezy knows exactly what to do now.

Nice explanation, btw.

Your advice is even more amazing since it is consitently wrong. How many posts have I proven your advice to be junk? This has to be at least the fourth one.

Why do you feel that you can help people out when you don't even know the basics?

 

[Kewtsquirrel]

Because that is not how electronics work. If this is your first try at dropping voltage then why are you giving advice?

The atty does draw current when voltage is passed through it. The atomizer is noting but a resistor. What happens to resistors when you pass voltage through them? They put a load on the system. Load is current.

Ohms law. V=IR. I=V/R. If V = 4.5 and R = 3.7, then I = 1.216 amps.

Just because the power supply has the capacity to put out 1.6 amps, does not mean that is what is going to be used by the atomizer.

So what happens when your amperage capacity is less than that? If you're running USB off of a PC, your maximum amperage is only 100ma, does that somehow get increased to 1.35a? (v = 5, R = 3.7, I = 1.35?)

 

[Laredo7mm]

So what happens when your amperage capacity is less than that? If you're running USB off of a PC, your maximum amperage is only 100ma, does that somehow get increased to 1.35a? (v = 5, R = 3.7, I = 1.35?)

If your power supply can only put out 0.1 amps at 5V, then use your P=VI equation to get 0.5 watts.

See my reply in post #4

The power supply will only put out what the attached load will draw. If the power supply can supply more amperage than what the circuit will draw, the power supply will only supply the power needed to run the circuit.

If the attached circuit can draw more power than the power supply can put out, then the power supply will burn up unless it is design to limit its output current and not self destruct.

 

[Kewtsquirrel]

If your power supply can only put out 0.1 amps at 5V, then use your P=VI equation to get 0.5 watts.

See my reply in post #4

So your current can be up to the maximum output of your power source, but not over, meaning that the V=IR equation cannot be used to solve for current in cases where your power supply is limited?

Edit reply: An atomizer at 3.7ohms resistance will almost always draw more current than the supplies that everyone uses can put out, none of the batteries that anyone uses have a rating of over 900mA, USB hubs are 100ma (or 500ma in the case of some powered hubs)
I can see the USB hubs having protection to limit their output, but what about batteries?

 

[Laredo7mm]

So your current can be up to the maximum output of your power source, but not over, meaning that the V=IR equation cannot be used to solve for current in cases where your power supply is limited?

Well it can be over if your power supply is designed to handle it. Power supplies can be current limiting, meaning that if a load is attached to it that draws more power than what it is designed to handle, it will only supply its max amps, but not be damaged my doing so.

Otherwise, something is going to give if you try to draw too much power from the supply. Some component inside of it will fry.

It is always important to size your power supply to your application. I think LI-Ion batteries can at least supply 2C. But it might be more and could be less. If you have a 1000mah batttery it should be able to supply a contsant current of twice its capacity (2C), or 2 amps. That is why it is important to use protected cells. They are protected for over charge, over discharge, and usually over current draw. Over current draw on an LI-Ion or LI-Poly is a very bad thing.

Edited to add: I have not done much research on the constant current output of LI-Ion batteries. I seem to remember seeing values higher than 2C. It would be interesting to do some measurements and see what the stock 150 or 180 mah e-cig batteries can actually deliver. I am pretty sure the 2C figure is safe, but DYOD.

 

[Kewtsquirrel]

Well it can be over if your power supply is designed to handle it. Power supplies can be current limiting, meaning that if a load is attached to it that draws more power than what it is designed to handle, it will only supply its max amps, but not be damaged my doing so.

Otherwise, something is going to give if you try to draw too much power from the supply. Some component inside of it will fry.

It is always important to size your power supply to your application. I think LI-Ion batteries can at least supply 2C. But it might be more and could be less. If you have a 1000mah batttery it should be able to supply a contsant current of twice its capacity (2C), or 2 amps. That is why it is important to use protected cells. They are protected for over charge, over discharge, and usually over current draw. Over current draw on an LI-Ion or LI-Poly is a very bad thing.

Ok, that makes quite a bit more sense, can most decent multimeters test for the max current supply? I'm still waiting on the slow boat from china for my new one, and my current one can only test to 150mA.

Furthermore, any idea what the actual amperage output is of the standard batteries that ship with our kits?
Another edit: The DealExtreme 14500 3.7v/900ma protected batteries that so many of us use show a max discharge of 1.5c or 1.35a, although it doesn't say if this is limited by the integrated protection PCB or a safe value, so using 2 in series could possibly be dangerous without further testing?

 

[Laredo7mm]

The other thing to keep in mind is when using boost converters to up your voltage from the 3.7 nominal to 5V, is that it puts more load on the battery. If your battery is putting out 3.7 volts, and you are pulling 1.35 amps at 5V, your power is 6.75 watts. That means your battery has to supply 6.75 watts at 3.7 volts, which would be 1.83 amps. Then factor in the inefficieny of your boost converter, say it is 92% efficient, then your circuit is really asking for 1.98 amps. It gets more taxing on your battery as the voltage drops.

I don't believe that this comes into play with voltage regulators used to drop the voltage, since the curcuit is calling for less power than what the battery is putting out.

 

[Kewtsquirrel]

Lets assume for a moment (since neither of us know if the 3.7v protected batteries have overcurrent protection) that these batteries do indeed cap output at 1.35a, how would that effect a voltage boost converter? As I understand it, boost converters essentially trade amperage for voltage, but if the amperage output is capped at 1.35a, would your available amperage actually decrease to the atomizer while increasing voltage, but for a net wattage gain of 0?

 

[Laredo7mm]

Ok, that makes quite a bit more sense, can most decent multimeters test for the max current supply? I'm still waiting on the slow boat from china for my new one, and my current one can only test to 150mA.

Furthermore, any idea what the actual amperage output is of the standard batteries that ship with our kits?
Another edit: The DealExtreme 14500 3.7v/900ma protected batteries that so many of us use show a max discharge of 1.5c or 1.35a, although it doesn't say if this is limited by the integrated protection PCB or a safe value, so using 2 in series could possibly be dangerous without further testing?

It could be if you are running them in series putting out 7.4 volts. At 7.4 volts, your 3.7 ohm atty would be asking for 2 amps, and the batteries are only going to want to give 1.35 amps. But like I mentioned before, I haven't done much research on LI-Ion batteries since I mostly researched and use LI-Poly batteries. The LI-Ions I do use are in series, but regulated down to 5V.

If you run the 14500 in series and then regulate the voltage down to 5V, then you should be fine, since you are then asking for 1.35 amps.

 

[Kewtsquirrel]

It could be if you are running them in series putting out 7.4 volts. At 7.4 volts, your 3.7 ohm atty would be asking for 2 amps, and the batteries are only going to want to give 1.35 amps. But like I mentioned before, I haven't done much research on LI-Ion batteries since I mostly researched and use LI-Poly batteries. The LI-Ions I do use are in series, but regulated down to 5V.

If you run the 14500 in series and then regulate the voltage down to 5V, then you should be fine, since you are then asking for 1.35 amps.

Gotcha, quick question though, you said you were using li-polys and I know that you've been experimenting with some ldo vregs, most of the li-polys I've seen that are reasonably priced are 3.7v cellphone batteries, are you running 2 of these in series?