Way to switch between two power supplies

[acerdisma]

Dear all,

what is the most power efficient way to choose between two power sources?
To be more precise, I have a solar powered circuit with a dc regulated output and a rechargable battery, and want to be able to automatically select the best of those two upon voltage level conditions (i.e.always keep a 3V output).

Any ideas/suggestions welcome!
Thanks a lot!

 

[laundry]

Hi acerdisma
Are you going to build the circuir your self?.if so you could use a voltage comparator and a couple of transistors,out put to switch a relay.

 

[acerdisma]

Yes laundry, I am to build the circuit...

could you please elaborate on your suggestion?

Thanks!

 

[xnuke]

You could use auctioneering diodes. Whichever DC supply has the higher output will be selected.

xnuke
"Do you think you used enough dynamite there, Butch?"
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[geekEE]

The auctioneering diode idea will work, but only if you can adjust your supplies up a bit to maybe 3.7V to account for the approximately .6-.7V diode drop. Otherwise you'll drop the output to about 2.3V. Probably too low for a 3V-powered circuit. Also, if the circuit draws a significant amount of current, beware of the power dissipated across the diodes.

 

[VisiGoth]

laundry has the best idea for best performance, but it is a and reliability penalty as compared to the simplicity of xnukes twin diodes.
However, voltage levels may not be the best indicator of which source to use. You may be trying to makesure the two solar sources do not feed each other. It should be OK and even desireable for both to supply the load. Xnukes diodes would optimally share as a function of the source impedance of the two supplies under the rare condition of nearly matched voltage source. But, the highe voltage supply might sag down until there is benefi from the lower volatage supply. If you can not live with the diode drop you would have to get more elaborate.

 

[acerdisma]

to make things more clear, this is a crude visualization of the situation

...................solarCell
..................|..........|
....dcConverter.......rechargableBattery
..................|.........|
...................\......./
..............powerSourceSelectCircuit
......................|
...................LOAD

A collegue suggested to you use p-channel mosfets driven from a low-power comparator through npn transistors, but i do not really know how to design this circuit. Can you please help?

 

[VEBill]

If the 'dcConverter' is required, then how do you live without it on the right hand side (battery side) of the circuit ?

 

[itsmoked]

VE1BLL; THAT is a VERY good question...

 

[acerdisma]

Sorry for the misconception guys...in my design context the rechargableBattery block is effectively a block comprised of a charging IC (which can regulate the output)+rechargeable cells.

 

[itsmoked]

Do you charge the battery and run the load at the same time?

Then u stop charging the battery and still run the load?

Normally you break this up differently....

 

[VEBill]

The normal approach is a simple serial line of hardware.

Solar panels > 'Battery Block' > VR (if req'd) > Load.

Why bother routing around the 'battery block' if it will automatically regulate its own charge level?

The only concern might be to make sure that your VR (dcConverter) is highly efficient or switched-off when not in use. You don't want it to drain the battery flat on a cloudy day.

 

[acerdisma]

VE1BLL,

If I use the serial line architecture you propose, do you think adding a dc converter after the solar cell will enhance the effieciency any any way? I am asking this first of all because the solar cell is rated at 6Voc and 400ma Isc, hence need to step this voltage down to 5V (for IC to work) plus I read that this will in a way increase the ability of the circuit to extract the max power out of the solar cell (track its maximum point of operation).

Thanks in advance for any comments

 

[VEBill]

If the solar cell is 6 volts max open circuit, then you can probably simply hook it to your (presumably '6-volt') battery. Check the solar cell documentation to make sure...

What exactly is the dc converter for ?

Do you need 5 volts or 3 volts ?

 

[acerdisma]

The dc converter is to scale the voltage down to IC levels (the charging IC needs 5V to work).

The 3V I need for my load.

I think that plugging the cell to the battery is not the the most efficient way to charge the battery.

 

[VEBill]

Perhaps it is time to find out what sort of battery you're using... ? Often, with lead-acid batteries for example, you do not require a 'charging circuit' and thus, in your case, you could also dispense with the 'dc converter'.

<clunk> <clunk> [the sound of two unneeded circuits hitting the junk box...]

Then you'd be back to simply paralleling the 6v solar panel across the 6v battery (like everyone else in the world), and then use a suitable low drop-out, efficient three terminal regulator to provide the 3v for the load.

You'll need to confirm that this will work with your gear.

Watch out for the quiescient current draw of the 3v regulator and load. You should be able to calculate the expected battery life.

 

[acerdisma]

the battery is 2xNiMH cells in series (@1.2V, 2000mAh).

Don't these batteries require a constant current charging algorithms to charge properly?

 

[VEBill]

So, you've got the following:

6 volt solar panel,
5 volt chip in the charging circuit,
2.4 volt battery pack, and finally
3 volt load.

That's a lot of different voltages for such a simple application. You're going to burn up all your solar energy as heat in all the various circuits (6.0 to 5.0, 5.0 to 2.4, 2.4 back up to 3.0)...

I'd use a 6 volt lead acid battery and a voltage regulator for the 3 volts. That'll get you maximum efficiency up to the 3 volt regulator.

 

[acerdisma]

your suggestion sounds good, but an SLA battery is quite bulky and heavy for my application (portable with a form factor that of AA size cells)...

any other suggestions

 

[VEBill]

6 volt solar panel, simple shunt regulator* (perhaps a 3.6 volt zener diode or something similar), a 3.6 volt battery pack**, a simple series diode (to drop about 0.6 volts), your 3 volt load.

Parts list: Panel, zener diode, 3 cells, diode, wire.

* the efficiency of the shunt regulator doesn't matter because it tosses away energy that you don't need (once the battery pack is fully charged).

**higher voltage than 2.4 volts to avoid having to step the voltage back up to 3.0 volts again. The batteries should be selected so that they are not fussy about how they are charged (it's a solar panel after all - not really reliable). Fussy batteries are the wrong batteries.

All this assumes that your 3 volt load can handle some minor variation in the supply voltage (3.6 to 2.7). If not, then you'll need to bump up the voltages (another cell to 4.8 volts perhaps) and use a carefully selected 3 volt regulator.