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19V PSU charging 12V lead acid battery
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nuclearcat:

--- Quote from: KL27x on September 16, 2019, 08:50:26 pm ---
--- Quote ---Certainly not a buck. As far as a 19V PSU is concerned, the battery across its output is a dead short. It will dump its maximum output current into it. The output capacitors will rapidly discharge when the switched is turned on, possibly shortening their life after many such cycles.
--- End quote ---
Ok, geniuses.

You've gone this far. Add the frigging inductor. In this case you don't even need an inductor. Just a power resistor in series. Done.

--- End quote ---
Probably this circuit will generate less heat:
mikerj:

--- Quote from: KL27x on September 16, 2019, 08:50:26 pm ---
--- Quote ---Certainly not a buck. As far as a 19V PSU is concerned, the battery across its output is a dead short. It will dump its maximum output current into it. The output capacitors will rapidly discharge when the switched is turned on, possibly shortening their life after many such cycles.
--- End quote ---
Ok, geniuses.

You've gone this far. Add the frigging inductor. In this case you don't even need an inductor. Just a power resistor in series. Done.

Battery is your Q. Add resistor. You have a low pass filter and the power resistor drops most all the excess voltage. As I said, as long as the PSU is happy to provide the juice, you're good.

--- End quote ---

The point being made was that a buck converter requires an inductor.  Adding a resistor to limit current defeats the point of using PWM in the first place.

What do you mean by "Battery is your Q"?  Lead acid batteries are not particularly inductive, what am I overlooking?
magic:

--- Quote from: nuclearcat on September 16, 2019, 08:52:32 pm ---Probably this circuit will generate less heat:

--- End quote ---
That's a constant current source? Nope, it's a linear regulator which will generate exactly as much heat as any other linear regulator: 7V times charging current.
KL27x:

--- Quote ---The point being made was that a buck converter requires an inductor.  Adding a resistor to limit current defeats the point of using PWM in the first place.
--- End quote ---

No it doesn't defeat the purpose; the purpose being to safely charge a lead acid battery. It means you can adjust/limit the current, and without needing a full voltage regulator. I somehow doubt he is concerned with efficiency. The resistor is not there to limit the current to the battery. The resistor is there to drop the excess voltage that someone correctly complained is going to wreck the PSU. It's there to limit the current to the max of the PSU, say (19V-10V)/(MAX PSU Amps) =R. The PWM is to adjust the current to the battery.

You got your power source sorted out with a single power resistor, and now you can get on with charging your battery using your ADC/comparator and micro/logic circuitry and output FET. You can tweak the max current settings for any size lead acid battery you want. You can set it to automatically throttle back to keep the voltage of the battery at 13.8V once it reaches that mark.


--- Quote ---What do you mean by "Battery is your Q"?  Lead acid batteries are not particularly inductive, what am I overlooking?
--- End quote ---
Yeah, you're right. I mean to say "C" not "Q." Brain fart.

If adding a resistor defeats the purpose of PWM, then every LED that is PWM'd is "defeating the purpose," since it also needs current limiting or some sort to avoid shorting the power supply. The PWM gives the ability to turn down the current.
magic:

--- Quote from: KL27x on September 17, 2019, 06:28:09 am ---
--- Quote ---The point being made was that a buck converter requires an inductor.  Adding a resistor to limit current defeats the point of using PWM in the first place.
--- End quote ---
No it doesn't.

--- End quote ---
It does :)
It's the first time I see anybody call resistive dropper a "buck converter".
Buck converter is that thingy with a coil, a power switch and a rectifier.
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