Go Down

Topic: atmega328 glitching on and off (Read 1 time) previous topic - next topic

Docedison

All of what I wrote in my last post is I believe correct... for a device that needs 200 - 300 mA Max current from a 24VA transformer and yes you do need a 24VA transformer to supply 24 VAC @ 1A load current. A "Half Wave Rectifier will give you 1/2 (approx) of the 24V ac or about 12 Vdc. I suggested a large enough diode that the surge resistor might not be needed and that there would be 9 to 10 Vdc available for the regulator.
The best method (IMO) is a switcher and I had very good results with the part i mention next... I made about 100 of them (Not me I just did the design and PCB layout) but they were really quiet and trouble free.
I built a lot of 24Vac based switchers in 2003 - 2006 I used an LT1170 because I needed about 3A for 10 - 15 seconds to power a 10W transmitter. It was an easy device to use... all the information you will need is in the data sheet. There are three varieties of this device, LT1170 - 1 - 2 for 5A @60 V max input voltage (LT1170 2.5 A (LT1171 and 1.25 A for the LT1172 I used the 5 lead TO-220 and the 5 Lead SMT package and all worked very well. This is the link to the data sheet.
http://www.stanford.edu/class/ee122/Parts_Info/datasheets/LT1170.pdf
an ac transformer and a bridge rectifier work very well with this device for a stable and adjustable buck mode power supply. If you need a lot of current. If you are just doing a small control device then my last post should have more than enough information. I would add that a Tranzorb should be included for transient protection on the secondary of the transformer to protect the diodes. I also included poly fuses to prevent fires in case of component malfunction. I also used tranzorbs on the rectifier output as these were used in Kansas on a Rural Power Grid where KV level spikes were common and some of long duration... 1/2 second spikes were measured... Usually though when the Power company did a "Load Drop", A frequent nightly occurrence as most of the big Agricultural companies used power at night when rates per KW were cheaper.

Doc 
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

winner10920


Joes

ok and if the current in secondary windings is 1A for example how can you calculate the current in the primary winding at 230V?

winner10920

Power is always conserved so power out equals power in, minus any losses
For simplicitys sake say 23v secondary drawing 1 amp, so 23w power used
230v/23w=.1 amp
In reality it will be a little bit more do to various losses

Go Up