So I have something that I want to power that requires 2.4-4.5 VAC 60Hz 200mA…originally I though this was DC and it was no big deal but it’s AC. Is there a quick way to drop voltage from the wall to something this low? Like something I can get from RadioShack or Home Depot?
The first thing I’d do is ask a double-E.
Chain variacs. (I’ve never seen one that would go that low—they have issues with current as you go lower)
Chain fixed transformers (2 110—24 is about 5V)
Chain a variac and a 110—12 and set the variac to 110/3 or 110/4 (verify load capability of variac at that setting).
Find a 2.4 to 4.5V DC wall wart (hopefully held together with screws) and remove the rectifier and capacitors. Mark it with a label so you don’t accidently use it for something else.
You want heavy transformer-type wall warts, not lightweight switching-style power supplies.
This is dubious: Get a 5V DC supply. Knock it down 0.7V with a diode. Use a 555 and a FET to create a 60Hz square wave. Filter it to something quasi-sinusoidal with a coil or wound toroid. Dubious because of noise issues. Could do something more elegant with a microcontroller.
Rewind a transformer (lots of work).
I’ve seen old HP AC laboratory supplies, but usually for very high AC outputs (like 1000VAC).
You might want to dig through the adapter box at the hive. you might get lucky and find something that low. The lowest I KNOW I’ve seen is 6V AC.
Easiest way I can see to do it:
Find a potentiometer, make sure the resistance isn’t low enough to over-current your source. Put the source across each end of the potentiometer. measure your source from one of the outer legs to the center until the voltage is approximately 4.5vac. Put your device where you were measuring your voltage. At this point you may need to adjust the potentiometer to bump up the voltage a bit depending on load.
I’m not sure how low that big transformer the doorbell is plugged into will go, but you could check it out. I’d imagine there is plenty of magic smoke available so be careful…
This is a little bit more voltage than you asked for (5V) but this might work?
What is it that you are trying to use it with?
FYI, I am a double E. Actually twice.
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Tye’s idea is basically a voltage divider.
Again, I’m no double-E.
But you can design a voltage divider for a standard 12V out transformer that will sit at 4.5V unloaded and 2.4 loaded.
With 13.8V AC input (typical 120V----12V transformer output):
Z1 = 30 Ohm resistor (min 4.3W)
Z2 = 15 Ohm resistor (min 1.35W)
Gives open voltage of 4.6V (no load).
With Z-Load of 12 Ohm (R=V/I = 2.4V/0.2A = 12 Ohm), this gives
Load voltage of 2.5V.
However, this sucks a minimum of about 6W to deliver 0.5W to the load.
If you can cheat a bit and assume it only pulls 0.1 A normally (which may not be a good assumption),
your Z-Load is 24 Ohms, so you can double the Z1 and Z2 and halve the wattage requirements of the resistors:
Z1= 60 Ohm at 2.2W (11v x 0.19A loaded and less open)
Z2= 30 Ohm at 0.7W (4.6V x 0.15A open and less loaded)
I can get 1W resistors easily, so I’d probably build the first one like this:
Z1=30 Ohms = Six 4.7 Ohm resistors in series (6W)
Z2=15 Ohms = Two 30 Ohm resistors in parallel (2 W) or Three 4.7 Ohm in series (3W)
Z1=60 Ohms = Four 15 Ohm in series (4W)
Z2=30 Ohm = One 30 Ohm (1W) or Two 15W in series (2W).
I’d use Metal Oxide, not wire wound. The j value components at 60Hz should be pretty low (ask EE!), but why risk it?
http://www.parts-express.com/cat/metal-oxide-resistors/302 (This place is up I75 in Springboro. Not sure of Cincinnati stores with 1W resistors.)
And seriously, I am a mechanical engineer, not an electrical, so “caveat maker”.
One last idea. You could possibly find a transformer (old-school, not a wall wart) with a center-tap or multiple taps. The 6-0-6 type of 120–12 is pretty common, but 6-3-0-6 configurations are out there.
Like this, except with a 110V primary: http://www.electro-tech-online.com/general-electronics-chat/120631-question-about-multi-tap-transformers.html
These have been great suggestions! Thanks … for the love of Tesla, I’m going to need to do some more research.
One other suggestion: Use your favorite digital components and power supply to get a PWM signal of the correct frequency (60 Hz I assume?), and then feed it through a transformer. Transformers act as very selective filters that are resonant at a specific frequency - so the DC component and all the harmonics should drop out, provided you find a transformer tuned to 60 Hz, and a nice 60 Hz sinusoid should emerge.
If you set it up so that the peak voltage of your PWM roughly equals the ratio of the transformer multiplied by the peak-to-peak voltage you need (which for your case is the RMS voltage * 2 * sqrt(2), or about 6.8 to 12.73 Vpp), your output should be what you need. However, it could be tough to find a transformer that has the ratio you need.
Sorry to spoil all the fun,
This one requires a little soldering
So they do make an AC-AC walwart! Event better!
An ac to ac wall wart is a transformer with a case around it.
An ac to dc wall wart is a ac to ac wall wart with a full bridge rectifier and capacitor after it.
oops, I just saw John Clark’s post of the same 5vac Jameco walwart, sorry!
And just to state the obvious—It’s 5V at 400mA. It doesn’t meet the max 4.5V requirement. However, a voltage divider for 5V ac should be a lot less lossy than one for 12V. If it is unregulated, which is likely, it’ll probably be a bit over 5V at 200mA.