Converting AC track power to 5 VDC

The ground is the negative side of the power, and if you use the TO-220 package, you won't need a heatsink for a small DC fan.

Here's the RS version.....http://www.radioshack.com/product/index.jsp?productId=2102856&locale=en_US

Do I need the mounting hardware kit or does the VR just screw to the 220?

If you need a heatsink, that one just screws to the TO-220 package.  OTOH, as I said, for a small DC fan, you shouldn't need any additional heatsink.  I run them bare up to 150-200MA, then I use the heatsink.

Yes, I was a bit confused about the 220 notation. Apparently it's the type of case around the regulator. RadioShack lists other things with 220 notation, but they apparently ATTACH to the 220 on the reg.  .......like heat sinks.   Going to get the parts now and build it. I was hoping the fan would arrive today, but....nope.

I know that John knows what I'm going to try and do with the fan, but for the others who have given advice on the regulator.......I picked up a Lionel Industrial Smokestack recently. Very cool accessory. While the smoke unit is not fan driven, it produces a decent smoke volume (and the lights on it are great). I found that when I blew air on it from below, it sucked more smoke out of the stack. So.....I'm going to put a computer fan under the layout and have it blow air from under and through the stack. I have high hopes for the result.      I was going to attach this fan to the last open spot on the terminal strip for my Miller signs, but.....I have my eye on another sign. So, this regulator should do the trick.

We'll see how it all shakes out. 

The fan will probably only draw a couple hundred MA at most, so I suspect you're fine with no heatsink.

How do you know when you do require a sink. How warm is too warm? I'm planning on putting this under the layout with the fan, so I won't really be able to go under there too often and actually touch it.  Although, maybe my placement will actually help. Attached near the fan....some of that air will undoubtedly flow over the regulator.

Well, the regulator has temperature protection, so it'll just shut down if it's too warm.  You could also put the regulator in the input stream to the fan to give it some forced air cooling.   The operating junction temperature is 150C, obviously if the case gets up to more than about 100C, it'll probably go into thermal shutdown as the internal temperature is higher.

All you ever wanted to know about the LM7805T/LM340T three terminal regulator: LM7805T Datasheet

I ran a test just now on my 6 amp diode. I was running about 12-13V AC with the diode attached to one of the leads. When I tested the output off the other end of the diode and the other lead from the KW, it read about 5.68V DC. Should there be that much of a dropoff? This is without the regulator in there yet. 

Also, should the stripe end of the diode be on the transformer side or VR side?

John,

I think I know what the problem is. I returned the VR to RadioShack and swapped it out, figuring maybe I got a bad one. As I was walking away from the drawer, I glanced at the label on the spot where the VR was. It said 7-25V input. Well, as I said, I was getting 5.68 at the other end of the diode. So, I'm thinking it's not enough to power this VR. Any way around this (other than using my Miller strip)?

I just took off the 6amp diode and replaced it with my full wave rectifier. With the diode (at 13 or so ACV in) I was getting 5.68. With the rectifier at 14V, I got almost 12V DC.Not sure I understand what's going on here, but maybe the VR will now work if it's got higher input voltage.

The capacitor across the output of the diode from the transformer will change the results a bunch, did you try that?  The regulator will also have more difficulty regulating with no filtering, it's designed to have the input capacitor.

The reason I specified the single diode was that if you feed the regulator excess voltage from the full wave bridge, at high input voltages you'll exceed the maximum rated input voltage.  Obviously, the input AC voltage determines what you need here.

John,

No, I hadn't hooked up the cap yet. The diode itself halved my voltage to the pt. where the regulator couldn't do anything. I'll try it with the cap, though. Thing is, I generally don't go above 16V with my accessory transformers.    I read over those stat sheets on the regulator last night and could see why I was having problems.   

John,

If the stripe end of the diode is the + end , which lead from the cap will connect to that? A little shaky with that part of it. I'm thinking + because the flow is + to -. Is that the correct orientation for it?

The cathode (stripe) of the diode faces the load, the anode faces the power supply, aka transformer.  The capacitor connects to the cathode side of the diode and common after the diode has rectified the AC voltage.

I'm sure the diode will be plenty with the cap and 12VAC or more on the input.  Glad you got it all working

What I discovered in my bit of research last night was the "smoothing" effect of the cap. Supposed to produce a "pulsating DC" without it . Exactly what I was getting. It was difficult to get a reading, but it was way low, mainly because the input was less than 7V. 

Leaving the capacitor out to just do some quick tests on the setup made more of a difference than I thought it would. 

Here's an update to my original posting for this topic.  I bought the circuit boards with the 5 volt USB output and it worked very well.  No overheating and the output is stable so no flickering on the LCD photo frame.  The whole thing is mounted on a flatcar and the photo frame is powered by track voltage.  However, any disruption in track power causes the photo frame to shut down and, even though it powers back up immediately, it has to be reset to display the photos again.

Do I need a 5 volt battery to keep the circuit active or some kind of capacitive circuitry to maintain the 5 volts until the track power is restored?

How long is the interruption? Are you talking about dirty track or removing power? For dirty track more capacitance across the power supply bridge should work. maybe a 4700uf condenser with a small resistor in series to reduce arcing. For long shutdowns you would need a battery i would think.

Dale H

Dale,

Thanks for your quick reply.

The power interruption is only momentary due to dirty track, a crossover, a remote operating track.

The 470 uf capacitor I'm using now I guess is not storing enough energy to cover a short interruption since the photo frame shuts down instantaneously.

What value small resistor do you recommend?

Anything small. 18 ohms,33 ohms etc half watt. Just slowing the charge a bit to reduce wheel sparking. Polarity of the condenser is important. Capacitor size depends on current draw of the device.

Dale H

Get a BCR that are used to replace the 9V battery

in MTH engines much capacitance in a small package.

A 5V supercap costs peanuts and will power that through any track interruption.  I put one on my Oscar Mayer Wienermobile because the audio would cut out, now I can take it off the track and it keeps playing for over a minute with no power!

JW,Thats a good suggestion John gives. I put 5 farads, 5, 1 farad button type capacitors, on LED lighted cars on the regulated side of the 7805 regulator. Remember,the regulated side,not across the bridge, these super caps are 5 volt rated. Lights stay on in conventional over 2 minutes when power is removed and slowly fade. Even long after shutdown you can see the cars dimly glow in a dark room. My wife used to think I left the power on,but it was off.

You do not need a resistor on these button type capacitors in series like I suggested for the 4700uf one, they have some internal resistance of their own. First time on the track it will take some time to charge up,but after that,they stay charged pretty easily in an operating session.

Dale H

Hi y'all.  I've been on vacation for a couple of weeks and haven't provided an update on my quest for a reliable AC to 5 vdc power supply.

I bought some 5.5v 0.33 farad disk capacitors to hopefully provide constant power to the photo frame even with dirty track and "dead" spots like switches and uncoupling tracks.  I added one 0.33 farad capacitor (no resistor) across the power supply output, let it charge up for a minute or so and then removed the AC track power.  Unfortunately the display did not stay on and I lost the photo mode on the photo frame.

I bought 10 of the 0.33 farad capacitors.  Do you think I need to add 4 more for a new total of 1.65 farads?  I would like to have the display stay on for at least 5 seconds after power is lost.

This is really trial and error. I would use 1 farad ones. Keep adding and see if it works.

Dale H

Keeping the display on for 5 sec may not be possible. The display

draws much current and when the voltage drops below 5v it shuts off.

There is another parameter than voltage and capacitance for a

capacitor and that is ESR which is series resistance of the

capacitor. With high load current and high ESR, the voltage

to the display may drop too low for the display to operate.

There are capacitors with ultra low ESR, under 1 ohm, but

you must check for them.

What are you using for a DC source? Switching regulators

draw less current at high voltage inputs. With TMCC track

voltage input, rectified, to 25vdc, the input current, for 5vdc 1 amp

output, would be onle .2 amps.so a 1000 uF cap on the input would

keep the output at 5v for some time.

The LM2596 regulators only cost 10 for $12 on eBay and only

took a week to come from China.

Not sure where you are on this, but keep in mind that supercaps have a predictable but steep voltage drop as they deliver current.  This is different than, say, a recharge-able battery which holds its own for a while before dropping.  Specifically, the discharge rate of a cap in Volts / Second = Current in Amps / Capacitance in Farads.  Earlier you said the module draws 1 Amp.  So if you have a 1 Farad cap, the voltage drops 1 Volt/Sec.  I don't know when your display module conks out and loses its photo mode setting but let's say it's at 4V.  This means you'd need about 5 Farads of capacitance to achieve your 5 second goal.

Separately, as others mentioned, you have the ESR issue with low-cost supercaps.  pa mentions "ultra low" ESR of under 1 ohm.  You need maybe 10 times lower than 1 ohm.  Supercap ESR is like a resistor between the cap and your module.  So if you had a 1 ohm resistor, with your 1 Amp load, the voltage drop (=Current x Resistance) would be 1 Volt and your module would only see 4V and immediately conk out.  If the ESR was, say, 0.1 ohm your module would see 4.9V.  Supercaps with 0.1 ohm ESR are out there but a tad more expensive than those at 1 ohm or 10's of ohms.

Finally, as a general comment, one problem with using unregulated supercaps is the wasted stored energy if it becomes ineffective below, say, 4 Volts given your design.  Skipping some arcane equations, using a 4 Volt assumption for module conk-out voltage, your supercap only delivers about 1/3rd of its stored energy as it drops from 5V to 4V.  Yes, the architecture is simple and perhaps cheap, but arguably ineffective.  Reminds me of the old design adage, "Fast, cheap, good...pick 2."

Two directions I'd look at are:

1) boost converter regulator modules that generate 5V from a lower voltage.  For example, search eBay for "5V boost module" and I see one that claims 1 Amp of 5V from 3V for less than $2 shipped.

2) batteries.  I'm thinking of the MTH PS2 module which uses 2 NiCd cells to run the engine for, coincidentally, about 5-10 seconds when power is lost.  The 2 x 1.2V = 2.4V battery voltage is boosted to, coincidentally, 5V and is delivering probably a similar amount of current as your module. Like the MTH circuit, I'd add some kind of timing logic to cut power after the battery runs for 5 sec (or whatever you need) so that the battery saves its charge for real work.  Note that while battery charging circuits can be "fast, cheap, good...pick 2", if you don't want to mess with them you might even get by with some primary alkaline cells.  That is, on each major outage, you need to deliver 1 Amp at 5V (i.e., 5 Watts) of power for 5 seconds.  That's an energy of 25 Joules.  A pair of AA alkalines stores over 10,000 Joules of energy so you could get maybe a thousand "cycles" before having to change batteries.  In this case I'm reminded of one of the Lionel engines I have which uses a 9V alkaline to temporarily keep the sound/electronics running during power dropouts.  At first you'd think what a nuisance to change batteries but it is infrequent enough that it's not a bother.