is battery drain linear with resect to Mah?

[Helmholtz]

Drop-out voltage is the lowest allowable difference between input and output voltages.
3.5 V drop-out voltage means that your battery voltage must be higher by 3.5V than the output voltage.

[daz]

Drop-out voltage is the lowest possible difference between input and output voltages.
3.5 V drop-out voltage means that your battery voltage must be higher by 3.5V than the output voltage.

I get that, but what i was asking was since the voltage will never go below maybe 7v, why would i want of need one with super low DOV like mick mentioned?

[Helmholtz]

I get that

Not sure.

3.5V drop-out means that the regulator will only work properly if the battery voltage is at least 12V for 8.5V output.

[daz]

Not sure.

3.5V drop-out means that the regulator will only work properly if the battery voltage is at least 12V for 8.5V output.

So if i got one like Mick suggested with a 100mv DOV, does that mean that a 8v regulator would shut off when the battery hits at 8.1v ? Sorry if that is way off but when it comes to math skill i am literally retarded.

[daz]

I discharged them fully overnite then recharged them and they no longer charge to 10.5v. Now they are at 11.6 ! And under load they only drop to about 11.4. But on the bright side they seem to sound more like the alkaline. Maybe it's me but i think i may be ok with them as is. Hard to be sure tho because since this all started the alkaline has dropped really low so i'm not even sure that sound like it did. 11.6v....crazy huh ? At least they will last a long time because even at 10.5 they lasted probably 3 or 4 hours or more. So 2 of them would get me thru the longest of band events.

[Enzo]

The regulator doesn't "shut off", it just stops regulating. If you only put 6 volts into it, it cannot make that into 8v somehow, just the 6v comes out. As someone said, usually it takes 12v in to make 9v out. If you use a low drop out regulator then it takes only 8.1v or more to make 8v out. The regulator prevents voltage from going higher than its rating, so an 8v regulator never puts out more than 8v. But it won't hold up voltage, so if your battery is at 7v, then 7v is all you can get from the regulator.

[daz]

The regulator doesn't "shut off", it just stops regulating. If you only put 6 volts into it, it cannot make that into 8v somehow, just the 6v comes out. As someone said, usually it takes 12v in to make 9v out. If you use a low drop out regulator then it takes only 8.1v or more to make 8v out. The regulator prevents voltage from going higher than its rating, so an 8v regulator never puts out more than 8v. But it won't hold up voltage, so if your battery is at 7v, then 7v is all you can get from the regulator.

Ahhh, ok, thanks Enzo. I'm not sure i need to bother anymore but i may anyways. If I can find a 9v with a 100mv DOV that would be good enough. I think since they charge up to 11.6v it would probably be a good idea even if theres not a big difference in tone. Just staying at 9v till it gets there and then recharging it by the time it hits 8 should give me good consistency in tone. Or even use a diode like u suggested on the - to get closer to 8v would be even better to get the most out of each charge cycle with no change in tone. I looked tho and it seems most have a much higher DOV so it may be hard to find. By the way, mick suggested diodes in series with the + to lower voltage. You said -. Why the difference?

[Enzo]

Um... WHen I mentioned a diode, it wasn't a series diode. I was describing a way to increase the output of a regulator by part of a volt. That is case 8.5v was more important to you that straight 8v. A three terminal regulator has a ground leg. If we add a diode to that ground leg, it lifts the whole regulator a half a volt or so from the voltage drop across the diode. And if it helps, regulators come on both 8v and 9v types.

[daz]

Um... WHen I mentioned a diode, it wasn't a series diode. I was describing a way to increase the output of a regulator by part of a volt. That is case 8.5v was more important to you that straight 8v. A three terminal regulator has a ground leg. If we add a diode to that ground leg, it lifts the whole regulator a half a volt or so from the voltage drop across the diode. And if it helps, regulators come on both 8v and 9v types.

But in series to ground? which side of the diode?

[Enzo]

PV 400BH.pdf

Here, look in this schematic, lower right. There are two 15v regulators with the added diode to make the outputs 16v.

[g1]

Do you have a link to the battery datasheet or info on the exact rechargeable that you are using? 11.6V for a "9V" does not seem right, may even damage certain equipment.

[Mick Bailey]

Why not use an adjustable regulator with a trimmer? I do this quite a bit and you can then get the exact output you need. Maybe something like this (note the dropout voltage is lower under low-load conditions);
https://uk.farnell.com/microchip/mic...sop/dp/2510297

[nickb]

Not so sure about that, not if same chemistry.
There is a built-in "voltage" value for every chemical compound and that does not change with "discharge" condition.

WE measure lower and lower voltage because battery gets worn/tired, internal resistance increases so we get less voltage at terminals, but "chemical" voltage does not change, thatīs why a worn battery wonīt pull current from a fresh one.
Again, IF same chemistry.

If you parallel two mythical identical batteries, then it would be OK. In practice due to manufacturing tolerances it's probably not a great idea. The model of a fixed voltage source and a variable internal resistance is wrong. The voltage changes too; just grab any used battery and measure the unloaded voltage.

I took three different used 9V batteries and connected then to a 9V power supply.

#1 Drew negligible current. This was a very tired battery.
#2 Drew 10mA, steady
#3 Started at 70mA, fell to 20mA after 30mins then steady

So you can see that the drain on the feeder battery is significant. Thus, in general, the only safe and reliable method is not to directly parallel

[daz]

Why not use an adjustable regulator with a trimmer? I do this quite a bit and you can then get the exact output you need. Maybe something like this (note the dropout voltage is lower under low-load conditions);
https://uk.farnell.com/microchip/mic...sop/dp/2510297

Sounds like a plan but the pdf shows typical application and it doesn't show a trimmer in the schematic. I googled it but can't find anything on hoe to do it, where the trimmer goes, what value, etc.

[Mick Bailey]

Looking at p12 of the datasheet, R1 could be 330R and R2 a 1K trimmer in series with a 1.5K Ohm resistor. This would give you a range of 6.8v to 10.5v output. It's roughly the same method used with the common LM317 3-terminal regulator, though Vref is slightly different. If you pick values using an online LM317 calculator you get pretty close. With SMD devices I usually solder them to a small substrate board if I'm doing anything like this - easier to mount and wire up. I bought a load of them for pennies each - through plated and gold flashed.