I have a canakit adjustable power supply kit (kit# R182), that I'd like to slightly modify to be able to provide an additional +5v supply independent of the main output (for led's and a cooling fan).
If I am not correct, would there be an easy way to get a solid 5v out of this thing no matter what the main output voltage is? I understand it is tough without a schematic, I will get one up as soon as I can.
There is no way! To my understanding they should have used V0 rather than Vz in the first place... But this is a quite versatile regulator with a lot of options. However it willl provide one voltage only! (and the reference voltage Vref of course).
It sounds like my best bet would be to include a second regulator in the box, perhaps just a small wall-wart tied to the 120v input hooked with a 7805 or similar to provide the 5 volts. There is also a 24vAC transformer in the box as well...
Take care that you have to establish a common ground when you tap the same transformer with a different rectifier... It would be better to tap directly at C1 if possible.
Not much at all... Enough for a small cooling fan and a single red (25mA) power status led.
Tapping directly from the board at C1 would give me 24vAC, what would I use to get me 5vDC? another smaller transformer? or is there a (physically) small device that will do the trick?
So you need not really stabilized voltage at all!
At C1 you will have an untidy DC (I should guess around 30V, dropping when you have some amps load at the end..)
There are tiny DC/DC converters for around 1Watt (and $3 or so), transforming 24V to 5V. As they do not stabilize, you will have 5 to 6V eventually, this can be mended by an additional 1N4001.
C1 is the smoothing capacitor on the output of the bridge rectifier ; it is not AC as you claim. This is the correct location for your 7805 providing that the voltage across C1 is within the rating of the 7805 input. I haven't looked up the spec sheets but I think you might be exceeding its capability with an expected DC level of around 40 volts.
You might be trying to over-design things. If all the LED is doing is showing a supply then why not just use 2k2 resistor straight off the C1 voltage.
As to the fan, it's unlikely to take more than 50ma but at say 35 volts drop in the regulator (40 supply -5 desired = 35) you will be dissipating somewhere in the order of 2 watts. You should use a series resistor in the input of the regulator, say 390 ohms rated at 2 watts to save the regulator having to do all the work and it'll also act to reduce the input voltage of the regulator.
As I've said, my electronics knowledge is pretty much all self taught.
According to the schematic (I am in the process on transferring it to eagle), C1 is connected to ground and the AC input through the 4 diodes (is this the bridge rectifier you mention?). If AC is going in, and C1 is connected before the regulation, how is it DC at that point?
I think it may be time for some night-school
I just looked at the fan and it's rated at 5v/0.25A. If I can just get away with a few series resistors to handle things, that would be fabulous.
I looked at Farnell for DC/DC converters..as you will need 2W they are quite expensive >$10. So this is no solution...
1.5 W @5V would mean additional 6W @ 20V. This is the dissipation for resistors or simple regulators; 5 Volt regulators will in fact have other issues with input voltage > 20V!!!
So the cheapest solution migth really be a Chinese $1.50 5V cell phone loader from ebay or flea market......
The voltager across C1 is unregulated rectified DC. The four diodes form a rectifier bridge in which the upper pair of cathodes foprm the +ve terminal and the lower pair of anodes form the -ve terminal.
If your fan is pulling 250ma then I suggest all you need is a 150 ohm resistor stright off C1 +ve terminal. However the dissipation wattage is
35x35/150 or over 8 watts. This is quite a load. If you want to go this route then use a resistor rated at least 10 watts and have the fan blowing air over it.
You'd be much better off looking for a miniature 12 volt fan which will draw a lot less current. The series resistor required will then be of much lower wattage. Even better would be a 24 volt fan but these will be harder to find.
Sounds like my best best would be a small wall wort adapter as what deSilva mentioned... It would be fairly small, simple to wire up and stable (read: cool).
Thanks for the insight
It made me realize that as much as I have learned about electronics, I have MUCH MUCH more ahead of me
