Hi everyone
I have a transformer outputting a rectified and filtered 17VDC, and want 12VDC to power a mains powered fast battery charger.. is the best way to experiment with a series power resistor to drop teh voltage down? although i need all the current i can get..
otherwise would using something like the opa549 be viable?
Thanks
I have a transformer outputting a rectified and filtered 17VDC, and want 12VDC to power a mains powered fast battery charger.. is the best way to experiment with a series power resistor to drop teh voltage down? although i need all the current i can get..
otherwise would using something like the opa549 be viable?
Thanks
SimpleSwitcher
Or- you can use a switching regulator. These are alot easier to design and build that one might think. Look at National Semiconductor's website at the SimpleSwitcher series of integrated switching regulators. The simplest I can think of would take only one diode, one coil and two caps.
If you're stuck on doing a linear regulator, please keep in mind the power that will be dissipated: (17-12)V x 5A = 25W dissipated as heat.
Hope this helps!
Steve
Or- you can use a switching regulator. These are alot easier to design and build that one might think. Look at National Semiconductor's website at the SimpleSwitcher series of integrated switching regulators. The simplest I can think of would take only one diode, one coil and two caps.
If you're stuck on doing a linear regulator, please keep in mind the power that will be dissipated: (17-12)V x 5A = 25W dissipated as heat.
Hope this helps!
Steve
A buck regulator would be better but it will also cost more.
The diodes, resistor, and transistor (darlington in a TO220 case would be good) will be simpler, easier, and cost less. You could use 3 diodes before the regulator (you coud also use a higher value zener and put one of the diodes after the transistor for reverse polarity protection on the output) to drop 2.1V and give you 14.9V at the collector of the transistor. Now the transistor only has to disspiate (14.9-12)*5=14.5W. You'll only need a total of 6 parts (3 diodes, 1 zener, 1 resistor, 1 transistor) plus a heatsink and can build the circuit easily (point to point assembly) by hand. Plus you can get each part for about a dollar
For the buck regulator you can either buy an off the shelf product or design your own pcb, order and buy all the parts, and build it. Don't forget the time it will take you to do all this.
In the end it's still your choice! Good luck
The diodes, resistor, and transistor (darlington in a TO220 case would be good) will be simpler, easier, and cost less. You could use 3 diodes before the regulator (you coud also use a higher value zener and put one of the diodes after the transistor for reverse polarity protection on the output) to drop 2.1V and give you 14.9V at the collector of the transistor. Now the transistor only has to disspiate (14.9-12)*5=14.5W. You'll only need a total of 6 parts (3 diodes, 1 zener, 1 resistor, 1 transistor) plus a heatsink and can build the circuit easily (point to point assembly) by hand. Plus you can get each part for about a dollar
For the buck regulator you can either buy an off the shelf product or design your own pcb, order and buy all the parts, and build it. Don't forget the time it will take you to do all this.
In the end it's still your choice! Good luck
I think the point he was trying to make was that he would rather "hold onto those 25Watts" and not dissipate them as waste. If one simply looks at any of the SimpleSwitcher Datasheets, they will not only see what the chips can do, but also schematics, design procedures, and suggested parts vendors. The design procedures are very cut-n-dried, nuts-n-bolts, rahter than alot of useless theory. They get right to the point.
I have constructed several buck, boost and inverting SMPSs using these chips and follwoing their design procedures. I have found them to be very helpful in understanding how they operate and how changing components affects the circuit operation as a whole.
I have constructed several buck, boost and inverting SMPSs using these chips and follwoing their design procedures. I have found them to be very helpful in understanding how they operate and how changing components affects the circuit operation as a whole.
Furthermore, another disadvantage of the linear regulator is that those 25W are wasted power from the transformer, thus the transformer VA rating available for powering the load is actually reduced. The buck regulator does not suffer from that drawback as the input current is *lower* than the output current, thus the VA rating of the transformer is kept.
N-Channel said:
I realize that a buck converter would be the best solution but I'm saying that a simple linear supply (even though it wastes a fair amount of energy) would be easier to build and less expensive. It would also seem to be a more suitable project for Optical who doesn't sound like he has a lot of experience with power supply design
What better place to start than with one of the simplest linear supply designs? Unless they give you a proven pcb layout chances are good that most people without any experience won't come up with a very good layout. The whole make your own SMPS route is still more complicated than you're making it sound Don't forget that you've probably done this quite a few times already!Alternatively, jack's solution seems like the way to go.
I've designed a battery charger using the UC3906 to give a bulk charge of 5A to an Optima red top gel battery. My rail voltage is about 18V, and I'm using a TO3 darlington as a pass transistor. It works quite well.
Keep in mind, if you're trying to charge 12V batteries, you'll need higher than 12V rails. To put any decent amount of charge into them, you'll need a fairly large transformer and a big filter cap as well, unless you use the approach most commercial chargers do, and use rectified, unfiltered DC.
Keep in mind, if you're trying to charge 12V batteries, you'll need higher than 12V rails. To put any decent amount of charge into them, you'll need a fairly large transformer and a big filter cap as well, unless you use the approach most commercial chargers do, and use rectified, unfiltered DC.
Mdern switching regulator ICs now available are almost as easy to use as a classic 7812. Everything is built into the IC, including switching element, diodes, control circuit, feedback, etc...
As it has been pointed out, all you have to do to get a working regulator with one of these ICs is to choose a suitable coil, input and output capacitors.
In comparison, building a regulator around a LM723 is more complex (I killed several LM723 ten years ago when I built my first high current linear supply, some junction inside the IC was geting reverse biased in a destructive way during turn-off).
As it has been pointed out, all you have to do to get a working regulator with one of these ICs is to choose a suitable coil, input and output capacitors.
In comparison, building a regulator around a LM723 is more complex (I killed several LM723 ten years ago when I built my first high current linear supply, some junction inside the IC was geting reverse biased in a destructive way during turn-off).
Lead acid cells are 2.21 Volts per cell at room temperature. To charge a 12V (6 cells) lead acid battery to full capacity requires about 13.5 VDC not 12. Most automotive alternator regulators actually run closer to 14V.
Several IC houses make specialty regulators for charging lead acid because the correct charge voltage is temperature dependent.
At "full charge" the charger should switch to a constant current mode to avoid over charging.
Several IC houses make specialty regulators for charging lead acid because the correct charge voltage is temperature dependent.
At "full charge" the charger should switch to a constant current mode to avoid over charging.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.