6v Solar panel 4.8v Ni-Mh charger

Hello, This is my first post here and i'm wondering if anybody could help me as I'm a bit confused :/

I am planning on making a solar charger with 5x 200mA 6v panels, charging 4x 4.8v 2500mA Ni-Mh battery packs in parallel.

I want a 5v buck-boost regulator coming from the 6v panels (after the schottky diode) & batteries. The 5v buck-boost that I have found is here: (DC/DC Converter (Integrated Switch) - Buck-Boost Regulator - TPS63002 - TI.com)

I plan on having a 6v DC input jack to charge tha batteries also.

Right, on to my few questions...

1. As the buck boost maximum input voltage is 5.5v, would a normally open 6v 2A relay work for disconnecting the panels and buck-boost when 6v is applied to charge batteries?

2. Also, as the buck-boost only accepts up to 5.5v.. Would a 6v panel exceed this when a 4.8v battery pack is permanently connected?

3. As the battery pack is 4.8v (the max voltage is should ever be is 5.6v) the buck-boost regulator would be working in step-up mode when running from batteries. The specs say 800mA draw max, so should / is there a way of making sure that usb devices don't draw more than this without messing up the 5 volts?

Thanks in advance for anyone that can help =]

Killian

Great concept...

Did you know that 4 cells can rise to 6 V fully charged? Test whether the buck-boost refuses to accept that. If so then you'll have to put a diode in series, or attach a load, but only at the moment you connect the buck-boost. This is hard to do.

USB devices can operate in the 4.8V to 5.2 region. Daylight sun may supply enough power to maintain your bat pack at this level to drive the USB devices directly. (Although this may be risky.)

Baty pack V will drop gradually as they drain. Once the pack drops below 5V (1.25 per cell), the buck-boost will demand increasing current from them in order to produce 5V output. You may get only brief operation until low V cut-off. In other words a boost may only help you for short times.

About your questions.

1.

Certain power jacks have an internal switch that opens when you insert a plug. And reconnects when you pull out the plug. (Example, to choose incoming power rather than battery power.) If you hook it up right then you won't need a relay.

Or if you use a relay...

You cannot include a relay coil in the charging loop. It has too high ohms.

It needs a few volts across it to supply the mA to close the relay. It will draw that amount all the time it is closed. It will open at a lower mA.

Different relays have different open and close voltages. You'll have to devise a voltage detection circuit (or current detection) that opens and closes the relay when you want.

2.

Your solar panel will automatically drop to match the battery pack V. (Within limits.)

If your batteries develop scratchy contacts then solar output will instantly soar to 2 V per cell having no load.

3.

IC's that use 5V can operate from 4.75 to 5.25 (typically). Undervoltage may result in damage.

You ought to add a low-voltage disconnect of some kind.
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