Is this a good power regulation circuit?

Thread Starter

AG6HQ

Joined Aug 14, 2016
4
Greetings all!
I am going to fly a weather balloon with a payload that contains:
  • A Raspberry Pi Computer, which runs on 5v via a micro USB port
  • A DTMF Receiver (http://byonics.com/mdtmf), which runs on 12v via screw terminal ports
  • An APRS Transmitter (http://byonics.com/mt-1000) which runs on 6v via 4 AA batteries - I plan to add an external connector for external power on this device
For power, I plan to use qty 2, LiFe 6.6v Batteries (http://www.lifesourcebatteries.com/hcam6446.html).
I want to get this payload back so the APRS Transmitter will tell me where it is every 60 seconds, but only as long as it has power, so I am using these large batteries.
The DTMF Receiver will help me communicate with the Raspberry Pi, which will be collecting weather data and taking pictures along the way.
By law, I have a 4lb limit for the payload, but I am well under that so far.

So to try to provide these three different power requirements, I have attempted to cobble together a circuit that will provide the three values.
I've based it mostly on what I have found using google, so I thought I would run this past you guys for your advice.
I am not sure what all the electrolytic capacitors are for, or what the non-electrolytic is for (.1uF? what's that going to do?)

The Bridge Rectifiers are a trick I learned in model railroading, as a means to protect polarity. The LiFe batteries have a non-polarized plug, which could be reversed, but now that I take a closer look, the red is in the middle of the 3 wire plug, so if I connect to ground/B+/ground that should be ok too.

Thanks in advance.
 

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Veracohr

Joined Jan 3, 2011
772
The electrolytic capacitors on the input side are typically there to filter rectified AC in an AC-DC supply. You probably don't need such large capacitors on the input when you're starting with a DC source.

The 0.1uF is usually ceramic & doesn't have the ESR that electrolytics have, so can filter transients better.
 

paulktreg

Joined Jun 2, 2008
833
Major problems I can see:

7805 requires about 7VDC input to maintain regulation.
7806 requires about 8VDC input to maintain regulation.
7812 requires about 14VDC input to maintain regulation. Where is this coming from?
 

Thread Starter

AG6HQ

Joined Aug 14, 2016
4
Major problems I can see:

7805 requires about 7VDC input to maintain regulation.
7806 requires about 8VDC input to maintain regulation.
7812 requires about 14VDC input to maintain regulation. Where is this coming from?
paulktreg;
Thanks for the comment, but this is a disturbing revelation, which means I may not be able to use these regulators.

I'll be using the + from B1 and the - from B2 to provide 13.2v nominal. The batteries are stated 6.6v, but start out at about 6.9v and drop from there. I'm not supposed to use them below 6.3 volts, or there abouts. I have to re-check this value because when these batteries drop below a certain value, they get damaged.
 

Thread Starter

AG6HQ

Joined Aug 14, 2016
4
The electrolytic capacitors on the input side are typically there to filter rectified AC in an AC-DC supply. You probably don't need such large capacitors on the input when you're starting with a DC source.

The 0.1uF is usually ceramic & doesn't have the ESR that electrolytics have, so can filter transients better.
Veracohr,
Thanks for the reply. I'll remove the C1, C2 and there equivalents on the other two regulators.
Do you think I should keep C3 and C4?
 

crutschow

Joined Mar 14, 2008
34,280
Here's what I would do.
Put the two batteries in series to give 13.2V.
Use a low dropout regulator (such as a LT3080) to give 12V.
Use the LM7805 and 7806 to give the desired 5V and 6V.


Ditch the bridge as it will cause about 1.5V drop in the voltage.
If you need reverse polarity protection then use a P-MOSFET (drain to battery, gate to ground, source to load).
Here's a description of how that works, if you are not familiar with it.
 

ci139

Joined Jul 11, 2016
1,898
it looks your supplies +5V & +6V nodes are highly dependent on loading on the +12V , also it requires double the batteries for any near normal loading - if it's ok the +6V is floating on the air - who cares -- the supply -- it will do some 2-ce or more that good on variable /!\ not co-peaking /!\ loads -- but this (attachments) is your worst case scenario or what you apx. can rely on ;)
 

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Marley

Joined Apr 4, 2016
502
I would connect your batteries in series to give ~13V. Use an efficient switch mode buck regulator to produce 5V for the Pi and the APRS Transmitter (says 5V in the manual).
The micro DTMF has a series regulator on board so can simply run directly from the 13V battery.

You can buy small cheap DC-DC switch-mode buck converters on ebay. Saves making your own, which is not easy. Example

Using switch-mode a converter instead of a linear dropper will be much more efficient and will give longer battery life.

Edit: No bridge rectifiers, either. Don't know why they were there.
 

Marley

Joined Apr 4, 2016
502
I considered switchers also, but I'm concerned about the switching noise getting into the measurement data.
A good point. I would try it and see.
Possibly put the switcher in a small steel box (a peppermint tin is very good) with extra filter components if noisy.
These tiny DC-DC regulators run at quite high frequencies (note 4u7H inductor on the example) so are not so troublesome. No good for AM radio, of course!
 

ci139

Joined Jul 11, 2016
1,898
found some interesting links cabbage , LTC3788 , Li-S

usually to regulate power you need twice the target voltage coz E=u+U=ir+iR e.g. at max efficiency the half of energy dissipates on . . . voltage source - another half on your useful payload - also keeping in mind that near "empty" or/and in low temperatues the power output may reduce to 10%(empty) × 10%(at low T) = 1% of what you normally got
 

crutschow

Joined Mar 14, 2008
34,280
...............
usually to regulate power you need twice the target voltage coz E=u+U=ir+iR e.g. at max efficiency the half of energy dissipates on . . . voltage source - another half on your useful payload - also keeping in mind that near "empty" or/and in low temperatues the power output may reduce to 10%(empty) × 10%(at low T) = 1% of what you normally got
You don't need a voltage at twice the output voltage. You want the voltage to be no higher than what headroom the regulator needs to operate, for maximum efficiency.

You seem to be talking about the maximum power transfer theorem where half the power is dissipated in the source, which is the maximum power point, not the maximum efficiency point.
That's normally only used for things like RF circuits where maximum power is more important than maximum efficiency.
That's not the case here.
 

Marley

Joined Apr 4, 2016
502
One more thing: When you link your DTMF Receiver and APRS Transmitter to the Pi, remember that the Pi works internally at 3.3V and will be damaged if 5V signals are fed into its inputs. Level changing will be required.

I wonder if the Pi is a bit of an overkill for this application. It's quite power hungry too for a battery powered system. Have you considered something like an Arduino?
 

ci139

Joined Jul 11, 2016
1,898
(inertia checks)
little about Li-Fe-PO4 performance
price options - with mount tabs on demand (bit lesser capacity but bigger drain density)
(it would look a bit like an idiot contest :oops: i hope it does not) those guys can run on lesser no.of electrons -- can't advice you -- haven't used -- just read some specs :mad:
 
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