Need advice on dual power design

[Arlet]

Agreed that the ground looks somewhat confusing. Instead of drawing ground wires, my preference is to insert ground symbols at every connection. That way you don't have to trace the wire to see where the net is going, and reduces overall clutter.

[cbmeeks]

Aslak3 wrote:

A cosmetic observation, but wouldn't the schematic be clearer if the GNDs were joined up?

That's a good observation. When you look at something for a long time you sometimes don't see the "forest for the trees". I will add a wire to the GND's.

Aslak3 wrote:

What's the current draw?

Not sure yet. I'm debating on a few decisions. Mostly, how many VIA's I want to use. And, I'm going to have a couple of micro-controllers so I'm not sure if I want to bother with a UART or try to do serial with the mcu.

Aslak3 wrote:

Don't forget the heat-sinks on those 78xx

No worries there. I know those things get hot.

[jac_goudsmit]

I don't intend to steal the thread, but besides the fact that the MCP1700 has a max input voltage of 6V and the 7805 has a minimum input voltage of 7V, is there a benefit of putting the regulators in series?

For my own L-Star project I use two LM1086 LDO's. I based the circuit on a schematic from Parallax where the 5V and 3.3V regulator are in series. But I was thinking: both these LDO's have a 35V max input voltage so there's no need to put my 3.3V regulator's input on the 5V supply. If the 3.3V regulator would get its input directly from the same source as the 5V regulator, it wouldn't damage anything, and the current that goes to the 3.3V circuit doesn't have to go through two regulators (and generate heat in both). It would also leave more headroom on the 5V rail, because the current through the 5V regulator wouldn't be shared with the 3.3V circuits.

Any thoughts, anyone?

===Jac

[cbmeeks]

Well, I liked the idea of feeding the MCP1700 with the regulated 5V out of the 7805. I also like the TO-92 package of it which takes little board space and doesn't require a heat-sink.

Having said all of that, the schematic I built came from others I found online...you know, that whole "shoulders of giants" thing.

But I'm confused on what you mean about the current generating heat on both regulators. Do you mean current flowing through?

I have to admit, I'm also confused on the datasheet for the MCP1700. It says 250 mA for current over 2.5V. I could potentially have TWO Propellers on my board (one on-board and one via an expansion bus).

So I might have to rethink my design! I definitely want a 3.3V regulator that can handle a decent load.

Do you still recommend LM1086?


**EDIT**

Looks like the LM1086 can handle 1.5A of current. Yeah, I'm thinking I should use something like that.

I purposely didn't want to copy the L-Star. Even though it's open-source hardware. :-/

[jac_goudsmit]

cbmeeks wrote:

But I'm confused on what you mean about the current generating heat on both regulators. Do you mean current flowing through?

Yes. I'm not an EE but I imagine if the current to the 3.3V circuit goes through both regulators, it generates heat on both, whereas if the current only goes through the 3.3V regulator and the 5V regulator is just sitting there not feeding anything, it's not going to generate any heat either.

Quote:

I have to admit, I'm also confused on the datasheet for the MCP1700. It says 250 mA for current over 2.5V. I could potentially have TWO Propellers on my board (one on-board and one via an expansion bus).

I don't know what else is in your project but if you have an expansion bus, 250mA for the entire 3.3V rail doesn't sound like a lot of headroom. I seem to remember the maximum supply current for a Propeller is specified at 100mA but I may be wrong; of course you're not going to switch all your pins to output and short them to ground, but you never know what you're going to put on your expansion bus later.

Quote:

So I might have to rethink my design! I definitely want a 3.3V regulator that can handle a decent load.

Do you still recommend LM1086?

Yeah, I built several boards with them now (Propeddle and L-Star). They get less warm than (say) a 7805 because they're more efficient but I don't know if there are better choices out there. Like you said: shoulders of giants and stuff. I also liked that they can supply up to 1.5A which will come in useful for some of the expansion boards that I have in mind. What I didn't like was that they're not pin-compatible with the 780X series, which bit me on the first Propeddle boards (I had to put botch wires to fix it).

Oh and as you may have seen, I just used three 47uF electrolytic caps (one on the power input, one on the 5V and one on the 3.3V) and haven't had any problems with oscillation. I think the LM1086 datasheet recommends putting two caps between the two regulators when they're in series, but that just seemed silly to me so I used one bigger one instead.

===Jac

[GARTHWILSON]

jac_goudsmit wrote:

I don't intend to steal the thread, but besides the fact that the MCP1700 has a max input voltage of 6V and the 7805 has a minimum input voltage of 7V, is there a benefit of putting the regulators in series?

I think you just answered your own question. You cannot connect the MCP1700's input directly to the 9V without damaging it. Connect it instead to the output of the 7805.

Quote:

For my own L-Star project I use two LM1086 LDO's. I based the circuit on a schematic from Parallax where the 5V and 3.3V regulator are in series. But I was thinking: both these LDO's have a 35V max input voltage so there's no need to put my 3.3V regulator's input on the 5V supply. If the 3.3V regulator would get its input directly from the same source as the 5V regulator, it wouldn't damage anything, and the current that goes to the 3.3V circuit doesn't have to go through two regulators (and generate heat in both).

All these regulators we're talking about are linear regulators, so the amount of power each one will dissipate, that is, turn into heat, is the voltage drop across it times the current through it. Putting regulator A in front of regulator B will make regulator A take some of the heat load off of regulator B, because now regulator B doesn't have to handle the entire voltage drop from, in this case, 9V down to 3.3V. Let's say the load presented by the 3.3V circuit is 50mA. If regulator B were connected to 9V (if it could handle it without damage), the heat it would produce is (9V-3.3V)*50mA = 5.7V*50mA = 285mW. But if its input is connected to 5V, the heat it would produce is (5V-3.3V)*50mA = 85mW, so it runs much cooler. Regulator A will dissipate the other 200mW, which is ok because the 7805 in a TO-220 case probably has the reserve dissipation capacity. Let's say the load presented by the 5V part of the circuit is also 50mA, so the 7805 is carrying 100mA total. Its power dissipation then is (9V-5V)*100mA = 400mW, and it is able to dissipate far more than that even without a heat sink, if we're talking about room-temperature air. (There is a small amount of ground pin current which adds to the heating, but it's not really enough to worry about.)

The LM1086 has a TO-220 version available, so it can get rid of heat much better than the MCP1700 can.

[cbmeeks]

I'm sold on the LM1086. I'm going to redesign that power circuit using them.

Thanks again for all of the tips and information.


**EDIT**

Updated. I left the caps at 100uF. Mainly because it seemed like it should be OK and the fact I have many of them in stock.

[jac_goudsmit]

cbmeeks wrote:

Updated. I left the caps at 100uF. Mainly because it seemed like it should be OK and the fact I have many of them in stock.

The 100uF caps won't hurt, even if they're a bit overkill.

There's a serious mistake in your schematic: you drew a switch and a (possibly external) power supply on the left side, but the schematic program is going to interpret this as part of the project. And you connected what you intended as the positive side of the power supply, to VCC. So instead of the switch interrupting the current between the Power In socket and the 5V regulator (which is what you want to do), you now have it shorting the "power in" socket to the 5V rail!

If you design a circuit board from that innocent-looking schematic, be prepared to see some magic smoke when you close the switch

===Jac

[cbmeeks]

Hmm.

Do you have an example of a correct schematic? I'm not sure how it's supposed to look.

What my intention was is to have one of those large power switches like you would see on the side of an IBM 5150. So flipping that switch would connect the power from a 9V DC PSU (via a barrel jack) and supply power to the circuit.

Thanks.

[jac_goudsmit]

cbmeeks wrote:

Do you have an example of a correct schematic? I'm not sure how it's supposed to look.

This is the power supply in the head revision of my L-Star project. If you want the Kicad version, go to https://github.com/JacGoudsmit/L-Star.

It's essentially the same as your schematic...

===Jac

[LIV2]

Your Power input and 5V rail need to use separate power components on the schematic or your PCB design software will consider them to be on the same net.
See below for more info