Ti 4050c alternatives, only that please
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Efficient way or not. Taking it on a road test... so to speak, is my way of telling if it's efficient or not.
If I can get 6 1/2 to 7 1/2 hours with moderate to heavy vaping on ONE 14500, 900mah battery than I would say it's efficient. At least in my eyes.
Mind you I have used the resistor hack along with the diode to bring the voltages to (my selected range at this point in time) 4.5v to 6.4v
If I can get 6 1/2 to 7 1/2 hours with moderate to heavy vaping on ONE 14500, 900mah battery than I would say it's efficient. At least in my eyes.
Mind you I have used the resistor hack along with the diode to bring the voltages to (my selected range at this point in time) 4.5v to 6.4v
Just routed the board out yesterday. Had to go into work today so I wasn't able to wind the inductor or assemble it yet. The simulation looks great from 2.75-4.2Vin. Output should be in the 4-12 volt range.
Oh yeah, .50 x .65 includes the adjustment pot mounted on the board.
Your probably right, my concern is getting that lower voltage through the trick/hack, where as the diode/rectifier approach allows the chip to work as is, and for a situation like yours is the only extra component needed.
Not knocking your work, just discussing an alternative. This is how we all learn.
Lotsa choices too spanning many various voltage drops.
Someone could get about ~4.5V with only the addition of this one component.
1N5821-TP Micro Commercial Components (MCC) Schottky (Diodes & Rectifiers)
The whole diode thing is kinda interesting. In a general sense (say off a battery) if we get a 0.6V drop that is due to a resistor, is it different than the same drop with with a diode? Seems like it. The diode is holding back the current (the check valve analogy), while the resister is using some, burning it off as heat to drop the voltage. (I think).
Just tossing out some random thoughts.
As usual, always a pleasure.
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Hey WillieB, I'm am not an E.E. but I have some training in electronics. What I think is happening with the resistor hack is that we are using it to adjust regulation of the regulator and not merely drop the voltage through the resistor. (Mind you, I said think) I could be completely wrong but... Sounds good? NO?
Even though the PTN4050c's datasheet does not provide an internal block diagram I tend to believe that it has some sort of a pull-up resistor on the Adj pin, which in combination with the external resistor gives a voltage divider. It's simpler for the average user just to put a resistor and get the desired voltage than create a refference voltage. By adding the other resistor between Adj and V+ we just modify the division factor, that's why we can get lower than 5V. BUT it's also possible that driving the module out of it's intended range we lose performace, and instead of 96% we get 85-90%.
About the diode in series with the atomizer: the "forward voltage drop" is about 0.7V in silicon diodes but can go as high as 1.2V, the diode opens and lets the current flow, but that voltage drop multiplied by the intensity of the current is a form of lost power, the energy can not vanish, it's just transformed, in heat.
About the diode in series with the atomizer: the "forward voltage drop" is about 0.7V in silicon diodes but can go as high as 1.2V, the diode opens and lets the current flow, but that voltage drop multiplied by the intensity of the current is a form of lost power, the energy can not vanish, it's just transformed, in heat.
Diode operation is often described as:
Diode behavior is analogous to the behavior of a hydraulic device called a check valve. A check valve allows fluid flow through it in only one direction
Like check valves, diodes are essentially pressure- operated (voltage-operated) devices. The essential difference between forward-bias and reverse-bias is the polarity of the voltage dropped across the diode.
Inquiring minds want to know.
Sorting through the diode equation is far beyond me. Using Ohm's Law with a resistor is rather simple.
So how does this compare to a resistor with the same drop? Are you say since heat is involved the loss is identical? That's the question.
I had always thought that the voltage drop was due to the current having to do work to get through the diode.
Thus the dropping would be in the form of heat.
The current "working its way" says it all, a diode when forwarding current has an internal resistance (small) that's why the voltage drop. As I said energy can not just disappear, it has to go somewhere, if you get less electric energy than it means some of it was transformed in other types of energy: light (LED) or heat.
No one's denying some heat, or loss. The point is, at the same drop, is there an advantage to using a diode over a resistor?
A while back I built a 5V box that was just too harsh and hard on my Joye510's. I was asking about resistors when Caesar recommended a IN40XX 1A diode instead, he assured me it would work fine, it did. That's a tiny little thing and it did hold up. I don't think there is resistor of that physical size that could work. A .5W, which I don't think would hold up, is about twice as long and thicker.
Oh well we spent enough time on what's a not very important distinction. Be movin' on now.
That is a good question. I misunderstood and thought the issue was if diodes would NOT give off heat to do the job of dropping the voltage. I was only saying that they would give up some heat. As for one being better than another, I could only guess. I prefer to use a resistor across the adj/vout pins only because it is one less part to take up space.
The resistor at Adj pins is huge, kilo ohms, at a few volts, the current through it is a few milliamps, so no heat, 0.25w is more than enough.
The resistor in series with the atomizer, at 2-3 amp, and in series with another resistor (the coil) which is 1.5-3.5ohms and must be with a very small value otherwise the voltage drop would be across the resistor and not the coil. So the diode IS way better.
The formula for series resistors is: r1/r2=v1/v2 and v1+v2=v, that is: at v volts across both, the voltage on the first (v1) and the voltage on the second (v2) have the same proportion as the resistances (r1 and r2). Hope it helps.
The resistor in series with the atomizer, at 2-3 amp, and in series with another resistor (the coil) which is 1.5-3.5ohms and must be with a very small value otherwise the voltage drop would be across the resistor and not the coil. So the diode IS way better.
The formula for series resistors is: r1/r2=v1/v2 and v1+v2=v, that is: at v volts across both, the voltage on the first (v1) and the voltage on the second (v2) have the same proportion as the resistances (r1 and r2). Hope it helps.
Resistors and diode all have wattage ratings , diodes drop voltage across them so they do produce heat as a by-product , you can calculate the effective resistance of a diode by knowing the current and voltage across it , then use ohms law and this will give you the resistance or effective resistance of the diode .
I believe there is a spec on diodes that state this , not sure which one it is , I would have to look one up.
A diode by it's nature will have a high res. in one direction and a low res. in the other , so when it is biased ON it is a low res. Measuring diode dynamic resistance
I believe there is a spec on diodes that state this , not sure which one it is , I would have to look one up.
A diode by it's nature will have a high res. in one direction and a low res. in the other , so when it is biased ON it is a low res. Measuring diode dynamic resistance
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Forgive me if this is a dumb question, but i'm about to order batteries for one of these mods and have not used IMRs before. Since this chip requires a min V in, will it automatically shut off when the battery gets low enough so I dont have to worry about the IMRs dropping to low?
Was looking at the AW 18650 2900 mah protected versus the AW IMR 1600 mah. Is it worth paying double for the protected? It seems by your post that the IMR might work better. I just see all the posts saying only use protected batteries so not sure which would be better.
Don't know if anybodies tried these
DC-DC Power Boost Step up Adjustable Module LM2577 C22 | eBay
I've got some on order. Gonna give them a shot...
They might be too big for your application though. 3.8cm L x 5.2 cm W x 2.0 cm H
Still pretty small seems to be plug and play
input 3.5v-30v
output 4.0-30v
3amp max
built in pot
It all seems to be there
DC-DC Power Boost Step up Adjustable Module LM2577 C22 | eBay
I've got some on order. Gonna give them a shot...
They might be too big for your application though. 3.8cm L x 5.2 cm W x 2.0 cm H
Still pretty small seems to be plug and play
input 3.5v-30v
output 4.0-30v
3amp max
built in pot
It all seems to be there
Well FWIW with AW Li-Ions stick with AW's own 2600Mah, over their overpriced, re-badged 2900 Panasonic. Yes it will last longer in a flashlight, but the 2600 maintains a slightly higher, vaping range voltage and the 2900 having more life under 3.4V is useless anyway.
IMR's have a different chemistry which doesn't need protection, which means one less layer of electronics the current needs to fight, hence helping with some initial loaded voltages too.
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