4
\$\begingroup\$

I've got an automatic litterbox that 'needs' to be hacked for better control. As it ships, it comes with a 120VAC to 24VAC (300 mA) wall wart that drives the two motors in it. I'm trying not to change out the original power supply, as I don't want to change the motors.

What I'm looking to do is get 5VDC for a microcontroller. My first thought was to simply toss a rectifier circuit and a 7805 and I'd be done. However, it seems that the 7805 wants a max Vin of 20VDC. Next thought was to find a small transformer to step the 24VAC down to around 12VAC and then pass that to the rectifier and 7805.

To that end, I started looking around for a transformer that I could use. Sensibly enough, everything I can find seems designed for a 120VAC primary. I came across this small tranformer, which I'm guessing is an audio transformer.

Am I right in presuming that the resistance of each coil is directly proportional to the number of windings? Thus, if I were feed 24VAC into that transformer with a primary of 115 Ohms and a secondary of 69 Ohms, I think I would get 14.4VAC out. Is that correct? Also, can I get away with using what is probably an audio transformer for this purpose?

\$\endgroup\$
6
\$\begingroup\$

You can't rely on using resistance measurements to get turn ratios, because they almost always use different gauge wire for the two windings. The primary winding is usually several hundred turns of some fine gauge stuff, while the secondary is a few dozen turns of something heavier, to carry the necessary current.

Probably the easiest solution is to get a 2nd wall-wart that supplies a few hundred mA of regulated 5V, and you're done.

However, if you really want to DIY it, you may be able to find transformers with 24V secondaries that are center-tapped, in which case you may be able to use one leg to the c.t. to get 12V AC for your diode-and-7805 approach.

| improve this answer | |
\$\endgroup\$
4
\$\begingroup\$

If you want to get from 24VAC to 12VAC you need a transformer with a turn ratio of 2:1. Since the input of a power transformer is normally line voltage (120V in your case) such a transformer would be rated for an output of half the line voltage (60V).

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

If you turn 24V AC into a half or full wave, you will end up with 27V to 33V. 7805 will only go up to 20Vdc AND event it you make it work at 20V it will get very hot.

You can turn to an adjustable one like a LM350 for example, you will need a few resistors more, but those have a higher input. I still suggest you put a heat sink .

Using an audio transformer might make the voltage drop. but they are not made for high current tranfer. Ratio is directly proportional from the input impedence to the output.

Depending how it was made. if you load the secondary too much, it will probably plunge down on the voltage side.

usual way is to use a switching regulator. Look up Digikey. Today there are a multitude of very easy to use switching that uses off the shelf inductors.

They come in all type of inputs and outputs and produces zero heat.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

you may well end up with close to 20V or less DC after recifiying the 24V AC anyway depending on what rectifier design you use.

Even failing that there are tons of regulators that will tolerate input voltages higher than 20VDC and i would recommend you look at switching regulators.

Using a linear regulator to produce a 15V DC drop is going to generate a ton of heat, you'd need a decent heatsink on the regulator.

| improve this answer | |
\$\endgroup\$
  • 3
    \$\begingroup\$ first, normally if someone says 24VAC, they often mean RMS, which means if you measure it, it will be much higher. Second, 15V drop does not mean alot of power dissipation, I doubt the microcontroller pulls much current at all. \$\endgroup\$ – Kortuk Jun 7 '10 at 15:01
  • \$\begingroup\$ The power dissipation of a linear regulator most certainly is directly proportional to the difference between Vin and Vout. For the 7805 Pd = (Vin - Vout)IL + VinIG. Even for low IL with Vin - Vout = 15 even if hes only draws 100mA for the uC and some LED's or whatever thats 1.5W + VinIG of power he needs to dissipate as heat. As for the rectifier, 24V RMS is ~34V peak to peak, using a half wave rectifier with decent smoothing, not voltage boost and silicon diodes we would except (17 - 0.7)/PI = 5.18V DC. Using a full bridge rectifier we would expect (2 * 17 - 1.4)/PI = 10.37V DC. \$\endgroup\$ – Mark Jun 7 '10 at 16:21
  • 2
    \$\begingroup\$ I will do this as 2 points. first, power dissipation, I am not saying you are incorrect that it is horribly inefficient, but most through hole linear regulators can easily dissipate 1.5 W without an overheat issue. On that note, a PIC microcontroller is pretty rare to push above 50mA unless you are driving I/O or using a high end 16 bit or 32 bit processor at a very high clock rate. \$\endgroup\$ – Kortuk Jun 8 '10 at 1:08
  • 4
    \$\begingroup\$ Second, you have done the RMS conversion incorrectly. This is an error I will admit to having made a few times. 24V RMS is going to become 24 times square-root of 2, increasing in magnitude. it will be magnitude 33.9411255, peak to peak of double that, almost 68V. down the page at en.wikipedia.org/wiki/Root_mean_square they have an example where they point out peak to peak voltage in the US is around 340V, which is double 170, which is 120rms * root(2). \$\endgroup\$ – Kortuk Jun 8 '10 at 1:14
  • 1
    \$\begingroup\$ a bridge rectifier does not output the Peak to peak voltage of its input, it rather outputs the zero to peak voltage. \$\endgroup\$ – SingleNegationElimination Aug 18 '10 at 13:36
1
\$\begingroup\$

I ran across this switcher in my RSS feeds this morning: LMZ14201H: 1A SIMPLE SWITCHER® Power Module with 42V Maximum Input

It looks pretty simple, just a handful of discretes. It should dissipate quite a bit less power than a 7805, since it's a switching regulator.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

I'm going to recommend the easy solution, as JustJeff pointed out -- grab a second wall-wart that gives you an appropriate voltage to feed the 7805 regulator.

If you insist on getting your low-current 5 VDC power from 24 VAC, probably the best way to do it is to add a bridge rectifier followed by a switching regulator.

Would the 3-transistor Black regulator; or some 2-transistor Black regulator work for you?

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.