Any other solutions you guys can come up with it? I'm sorry for being impatient but I pretty much have only one more week to finish this before it gets troublesome.

Sorry if you feel that the response time is greater than your needs. We have a lot of people that would like answers to their dilemmas, and comparatively few that are able or willing to attempt to provide answers.

Your situation is difficult, as you don't have ready access to even simple components like low-value resistors, your project enclosure has no room, you are using linear regulators where switching DC-DC converters are necessary, and you seem to wish that we could wave a magic wand and make your design all better with no effort.

You've taken a basically workable design, and added a terrifically imbalanced load to the power supply, which is causing a great deal of stress in the way of power dissipation as heat. WHY that is, I don't know; there are no specifications about the amount of current required by the UNO board. Perhaps you are powering something else from the UNO board, or perhaps the regulator and L2722 are having to "fight" a ground somewhere; we have a rather limited amount of information at this end, as you have not given us a lot to work with.

 

I apologize if I've come off like that. I understand that my situation is limiting and I apologize for this. I've just put alot of effort into this and just before I finish I get this huge problem. Its just frustrating and I apologize if that comes off when asking for help.

Now I thought something was odd when you said to use the 10A scale if measuring the current draw by the uno. I just didnt think there was no way it could draw that much. I mean the board is designed to be powered by USB and I'm not using it to power heavy components.

Now, I'm not sure if this is correct. I connected +8V to the positive lead of my meter, the negative lead goes to the UNO's VIN. Virtual ground to the UNO's GND. I hooked up the SPI connections and the mux's controls as well as powered the IR receiver. This is really the min I need, there is an RGB led I wanted to power but I can easily live without.

As far components go. I dont mind ordering but I cant really order one simple resistor each time, shipping times and costs wont allow it. I apologize for that but it is my first project.

As far as the lack of information, I again apologize. Please let me know if I can do anything to help that out. Meter readings, photos, etc..


So at this state the op-amp doesnt get hot but it is getting warm. All LEDs on my breadboard and UNO light up correctly. The meter reads 45mA.

 

You said that the AC/AC wall wart is rated for 500mA, and 24 volts. You didn't say what voltage you are reading across the input of the voltage regulator with all loads connected and operating, but it will need to be at least a couple of volts higher. A 24VAC transformer rated at 500mA after rectification would have about 24/.707107 - 1.4v average across the filter capacitor C4, or around 32.5v - but at lower current draws, it will be significantly higher.

Power dissipation (as heat) in the regulator will be ~= (Vin-Vout) * current_out. You will need a way to get rid of that heat. You've said that you have a small heat sink on it, but that may not be enough.

Let's for instance say that your total load is 100mA. ~10.5mA more is used by the regulator's resistor divider R11/R12, so we'll just say 110mA for ease of calculation.

P(regulator) ~= (32.5v-16.625v) * 110mA = 15.875V * 0.110A ~= 1.75 Watts! That's why your regulator is getting warm. See, even that 45mA is quite signficant. The efficiency of regulation would be less than 50%, as your Vout should be 16.625 (according to previous calculations) and so 16.625 * 100mA = 1.6625 Watts; 1.6625/(1.75+1.6625) ~= 48.7% efficiency.

 

The voltage across the capacitor is at 35V. The voltage output from the regulator is 16.2

I dont understand how the conversation switched from op-amp heat to regulator heat. I understand your calculations but the regulator's heatsink doesnt get that hot. I mean I can keep my fingers on it with not that much problem. I expect the regulator to get warm within reason but not the op-amp.

Secondly if I were to use a switching regulator instead, wouldn't that create a noise problem? The whole reason of me using an AC-AC wallwart and regulating it myself instead of just getting an AC-DC switching one was because of noise. The research I've done says that its not a great idea.

Let me thank you again for all your help though, I'd just be staring at this thing confused.

 

Switching regulators are noisy. If the switching frequency is high enough, and filtering good enough, the noise should not be audible.

I'm just a bit puzzled why the L2722 is getting warm.

Not sure what the rest of the circuit load is on the L2722, but for now, let's assume that the load is balanced; so only the quiescent current of the L2722 is dissipating power.

That leaves the 45mA imbalance from the UNO's ground that the L2722 has to supply.

8.3125v offset * 45mA ~= 374.1mW power dissipation in the L2722 - plus whatever other imbalance is left from the rest of the circuit, which probably isn't much.

While ~375mW is not a lot of power, it still needs somewhere to dissipate that heat - otherwise, it will just continue to get hotter.

The output may be oscillating; I don't know. Without a 'scope, it will be difficult to tell. Oscillations will add noise - but the noise may be at a frequency where you can't hear it or see it. I have seen simple 7085 regulators oscillating in the MHz range because they did not have suitable bypass capacitors across their inputs and outputs to ground.

It could also be that the UNO board doesn't have very good filtering on its' input, and the L2722 is constantly trying to charge/discharge the UNO boards' filter cap(s).

Speaking of bypass caps - do you have caps from the L2722 supply pins to its' virtual ground pin?

[eta]
If the L2722 does not have a heat sink, with 375mW of power dissipation it will get 26°C hotter than the ambient temperature. So, if the ambient temperature is 25°C/77°F, the IC will heat to ~51°C/124°F.

 

I do not. What I posted on the schematic is what I drew up looking at my breadboard.

What type/value would you recommend? I have .1uF, 22pF 100pF, and some larger electrolytic ones.

 

Use the 0.1uF caps.

I added to the end of my last post. If you have a thermometer, you might see how close I came to the temp rise calculation.

 

I looked and looked and cannot find a thermometer lol. Only one I found was dead.

So with bypass cap in, it still behaves as before. I let it run for a while and after about a min or two both the L2722 and the regulator get hot. If I do get a switching regulator instead I'll still have the L2722 problem right?

Or since wasted power is the problem there, couldnt I just get a transformer closer to the output I want? Then I wouldn't have to regulate down that much.

 

That should have been "bypass caps" as in plural.

If you don't have the space for heat sinks, you are going to need to go to a more efficient solution.

Reducing the input voltage will help the power dissipation in the regulator greatly. However, it won't help the power dissipation in the L2722.

 

Wait caps? You said to put a cap between its supply pin and its ground pin. Looking at its data sheet
http://www.datasheetcatalog.org/datasheet/stmicroelectronics/1321.pdf

That would be pins 2 and 4 right? Or am I misunderstanding?


I can try and make some space for a bigger heatsink for the regulator but do they have heatsinks for DIP packages?

 

Caps from pin 2 to the virtual ground, and from pin 4 to the virtual ground.

No, they don't really make heat sinks for DIPs that small. They depend on the designer making wide enough traces on the PCB to dissipate the heat.

 

Haha I see, totally misunderstood that. That makes alot more sense though, shoulda read it with my brain and not just my eyes.

So ignoring the regulator problem because that seems to be better understood and with a couple of workable solutions.

What do I do with the op-amp problem, though. What about looking for other solutions to the original problem? Or do you think that oscillation is a probable? Enough to bet a few bucks and some wasted time on?

 

OK, a "quick and dirty" fix.
You said you have some 330 Ohm resistors, right? Are they at least 1/2 Watt?

If so, you can connect one of them from the output of the L2722 to the negative rail.
That will cause the L2722 to have to source 24.6mA current through that resistor to keep the virtual ground exactly between the two rails when the UNO is not connected.

However, when the UNO IS connected, the L2722 will only have to sink 20mA via the UNO to keep the virtual ground between the two rails, as the 330 Ohm resistor will be sinking the other 24.6mA. Either way, the 330 Ohm resistor will be dissipating ~200mW power; so you will need at least a 1/2 Watt resistor (or larger) to keep it from getting too warm.

The overall power dissipation will be the same; but it will be "spread" between the resistor and the L2722. The more you can spread out the heat, the less of a problem it becomes.

By the same token, you could add some resistance between your 500mA supply and the rectifier bridge. Since I don't know what your maximum current requirement is, the optimal resistance you will need remains a mystery.

 

As far as the regulator heat...
Find out how much current the whole enchilada is drawing when everything is fired up/turned on.

Then calculate how much resistance you'll need to drop about 12v-14v at that current.
For example, if your current flow into the regulator's IN terminal is 100mA, then:
R=13v/0.1A = 130 Ohms.

Then calculate how much of a power requirement the resistor will need:
P = 13V * 100mA = 1.3 Watts - then double it to 2.6 Watts. Then select a resistor that is rated for that or more power. The closest standard wattage rating >= 2.6W is 3 Watts.

 

Sounds like a good idea but sadly the resistors I have are 1/6W. Most of the ones I have are 1/4W with the only 1/2W are 1.5k. Luck is never on my side for this project.

It looks like I'm going to have to make another order. Is 330 ohms the optimal value?

 

What values and Watts do you have? Number on hand for each?

For example:
5 - 330 Ohms, 1/8W
3 - 1.5k, 1W
It is possible to connect resistors in series and parallel to make other resistances, or to simply increase the wattage rating. For example, four 1/4 Watt 330 Ohm resistors wired in series AND parallel (two series strings, two resistors per string, wired in parallel) would be rated 1 Watt.

If you had four of those 1.5k resistors, you could wire them in parallel and have 375 Ohms. That would work out to a constant 21.7mA load on the opamp (it's actually about the optimal resistance to use). 361 Ohms would be perfect, but you don't need perfect.

 

I have 330s and 220s at 1/6W
Alot at 1/4W, the closest to the above values is 120s
1.5k at 1/2W

And its a good idea to test , long term I'll have to order anyways. I usually order either 5 or 10 of each resistor though the 330 are good for LEDs so I have close to 20. Anything you see used in the schematic I posted is most likely 1/4 W.

 

Connect three of the 120 Ohm 1/4 Watt resistors in series. That will give you a 360 Ohm, 3/4 Watt resistor, and that's perfect.

 

Okay so maybe a stupid question but...

Do I connect op-amp output to Virtual ground AND connect it to the resistor network that goes to V-. If so then it seems like the op-amp gets just as hot, its kinda hard to tell without a proper thermometer though. I put my finger on it and after a while it begins to feel hot, maybe I'll call up my pops and ask to use his IR thermo.

 

The L2722 output IS the virtual ground.
The three-120-Ohm-resistors-wired-in-series-to-make-a-360-Ohm-resistor should be connected between the L2722 output (virtual ground) and V-.