Propeller power supply.

TonyA

In the schematic for the Propeller Education Kit Lab Series (PE Platform & Setup) on page 4, which shows the voltage regulators power connections.

I am wondering why the 3.3V regulator is being fed from the output of the 5V regulator?

Would it be better to just connect each regulator to the (+)9V?

Also, why do we need such a large electrolytic cap (1000 uf) on each regulators output?

Any info would be appreciated, thanks.

Tony

(I had to edit the above, they use (2) 1000 microfarads electrolytic caps, not millifarads)

Post Edited (TonyA) : 7/24/2007 7:07:57 PM GMT

deSilva

The global power dissipation will be always the same, e.g. when you need 0.5 A for 5V and 0.5A for 3.3 V you will need 1A at the - say - 9V input = 9 W.

2.5 of that will be used in the 5V devices and 1.65 in the 3.3 V devices, leaving 4.85 W to warm up the regulators.
Using regulators is series will distrubute this heat between them (at the possible cost of a stronger 5V-regulator). Parallel regulators might require an additional heat sink at the 3.3 V regulator. The choice depends on how much power you need from one voltage and the other. E.g. it COULD have been slightly less expensive to choose a parrallel configuration with a heat sink at the 3.3. V regulator and a low power type for the 5 V regulator...

10,000 mF sound a little bit overdone indeed, but I think it's 10 mF only. You add these capacities when in doubt whether your regulators are fast enough to react to fast changing current request from the connected devices. 10mF would allow you to deliver additional 1A for 10 msec which is rather uncommon. In general you choose a capacity of at most 1 mF.

Post Edited (deSilva) : 7/24/2007 6:39:33 PM GMT

Mike Green

Have a look at the Protoboard's schematic. Parallax uses a 1000uF electrolytic capacitor across the servo power bus. They also have a small inductor in series with the servo power supply. This helps isolate the logic from noise and power spikes from the servos.

Bergamot

1000 microfarads != 1000 millifarads

EDIT: Mike Green got there first.

TonyA

Thanks for the replies.

In my original post I meant to say "1000 uf" not "1000 mf"

So in series the 3.3V might get too hot?

I thought it might be more practical to use the "parallel" design, each regulator connected separately to the 9V.
(But I'm not an engineer, and just learning about this stuff.)

So, the 1000uf caps (connected to the output of each regulator) would be good if using a servo, but not necessary for normal input/output stuff?

Thanks,
Tony

Mike Green

Tony,
The minimum amount of capacitance for a regulator input and output depends on the regulator. Some regulators also have maximum capacitance limits (mostly very old types). Check the datasheet for recommendations. The amount you "need" depends on what you're going to do with the power, what the instantaneous power requirements of the parts of the circuit are. Motors are often the most power hungry, but, if you're driving lots of LEDs, that might be the controlling factor. The capacitor helps the regulator supply large amounts of current for a couple of milliseconds until the regulator (and its power source) can catch up. You also need small capacitors (like 0.1uF ceramic) across the Vdd/Vss pins of nearly every IC in your circuit to supply current at the device for a fraction of a microsecond until the current can get from the regulator across a few inches of PC board and/or wiring to the IC. Typically, if you have several small PC boards, there's a 1-10uF tantalum capacitor across the power leads where they come into the board.

TonyA

Ok, I see what you're saying.

I'm using a 3.3V, 500 mA reg. (which I know is good), and then a 5.0V, 1A reg.

Is this 1A ok? Or should I use a 500mA 5.0V reg.?

Thanks,
Tony

Mike Green

The current rating of a regulator (in your case 1A) gives the maximum amount of current the regulator is designed to supply. If you try to draw significantly more current than that, the regulator will shut itself down to prevent damage to itself. Typically, long before that happens, the regulator will get hot (from the [noparse][[/noparse]Vin - Vout] x Iout power loss) and turn itself off to prevent damage from the heat.

If you want to draw anywhere near the maximum current rating of the regulator, you will need to limit the input voltage and provide a good heatsink. A low dropout regulator will typically let you use an unregulated supply only 0.5-1.0V above the output voltage. For a 500ma regulator, that's 1/4 to 1/2W of power that has to be dissipated. If you were using a 7805 regulator with a 2.5V minimum input voltage drop at 1A, that would be 2.5W of power to be dissipated. That's a lot of heat for a small circuit board.

TonyA

Yeah, I am using the 7805, with 1A. I should get a 500mA reg.

Thanks again, I appreciate it.

Tony

Mike Green

You don't have to get a low dropout or a 500ma regulator. You just have to understand how your regulator operates, what its requirements are ... the care and feeding of regulators. The 7805 is a perfectly fine regulator, cheap, readily available most anywhere. You just have to use the right output capacitor and heatsink it as needed for the power it has to dissipate. Get the datasheet if you don't have it. Be sure to read it and look at the examples. Search the web for more examples of its use.

deSilva

The question is how much current you plan to draw, not what type of regulator you use. The rule with a 7805 - as Mike already explained - is to draw much less than 1 A except you add a heat sink!

It helps - a little bit - to screw the regulator to a metalic area on the PC board, but not much - the main effect comes by the screws I think

You can use a 3A type in a TO3 case giving you all safety you might need..

TonyA

I see, it's really about how much power my circuit needs (how much current it draws). If my circuit is small, and requires much less than 1A, the reg. will then need to dissipate heat, and need a heat sink.

I'll check the datasheet for the 7805.

Thanks again.
Tony

Skogsgurra

Andr

Mike Green

Not quite ... The amount of power the regulator needs to dissipate depends on two things: 1) What's the voltage drop across the regulator (Vin - Vout); 2) What's the current drawn by the circuit (Iout) at the regulated voltage (Vout). Power to be Dissipated = [noparse][[/noparse]Vin - Vout] x Iout

The reason most new circuits are being designed with low dropout regulators is that the minimum Vin - Vout is much lower than in the older regulators like the 7805 where it's a minimum of 2.5V (which all gets turned into heat).

TonyA

Ok, I'm understanding a lot more with that equation.

Very interesting. Thanks again. The equation cleared it up for me.

(About the Hydra Book, yes page 79. That's what got me thinking. The Hydra uses a completely different type ("parallel") power design. The reasons explained in the book seemed to make sense to me (only on a very practical level, since I am just learning about this power supply stuff).

He also made a point I could immediately understand, saying if one regulator (in the·"parallel" design)·failed the other reg.·would keep working in this case. In the "series" design they would both stop working if one failed.

I thought maybe there are reasons for such different power supply designs, since one is a complete system with peripherals and an expansion slot, etc. and the other just a chip and some test circuits (i.e. the Propeller Edu Lab stuff).

Tony

Post Edited (TonyA) : 7/24/2007 9:48:18 PM GMT

deSilva

I thought I made is all very clear in my first posting; seems I was too optimistic


There is always the same amount of heat, proportional to your request of current (@TonyA !)! You can - a little bit - control where the dissipation takes place.
(1) Most of it in the 3.3 regulator (parallel solution )
(2) Less of it in the 3.3 regulator and the rest in the 5 V regulator (serial solution)

Of course this makes no sense at all when you just use the 5 V regulator to save a heat sink!
When you are in need for a little bit of 5V and you are within the range of inexpensive regulators (sub 1 A) this might be an option
When you mainly draw 5V - maybe MANY Amperes- an additional (LP?) 3.3 regulator after the 5 V regulator will certainly do no harm.

But I know of no really convincing reason for the serial approch...

TonyA

I see. (I know it took 20 posts, but I am actually learning a lot from it. I feel dense... [noparse]:([/noparse]

Thanks again. I appreciate it.

Tony