Hello all,
I am currently busy with a project that involves powering a micro controller from a 7V solar panel. I would like for the Micro controller to be able to switch to a back up battery pack whenever the Solar Panel doesn’t provide enough output voltage to power the micro controller i.e at night preferably.
Can anyone please suggest how i can go about doing this.
Thanks.
Die-Ode
Thanks for the reply,
Is the circuit in the image link below along the lines of what you had in mind?
SRY, your link requires that I login.
What voltage (range) is your battery pack ? Is it getting recharged from the panel ?
Mee_n_Mac:
SRY, your link requires that I login.What voltage (range) is your battery pack ? Is it getting recharged from the panel ?
-Thanks for letting me know i have uploaded the image!
Yup, that’s a typical way to do it. The voltage at the junction of the 2 diodes will be whichever is the highest voltage, battery or panel (I assume the C1 is across it), minus the diode drop (~0.65V). D2 prevents the panel from acting as a load when unlit. D1 blocks the panel output from backfeeding the regulator and battery.
More elaborate schemes to switch (vs block via diodes) can be implemented, if the diode loss is a problem or if the charging scheme requires it.
Thanks for the reply Mee_n_Mac:
Just out of curiousity, what would happen when both the Cap and the Battery are both at 5V… will i get zero out??
3 words…
Par a llel
Ozwurld:
Just out of curiousity, what would happen when both the Cap and the Battery are both at 5V… will i get zero out??
Most likely they’ll never be exactly the same. In any case, you’ll get the higher voltage minus the diode drop (~.5 to .7 V depending on type). If both the battery and solar panel are at exactly 5.0 V, you’ll get 4.3 to 4.4 V but you really can’t know which device is supplying it unless you measure the current into each diode. The 1N4000 series are silicon diodes, and will have .7 V drop. The best you’ll get from the battery (regulated to 5.0 V) is 4.3 V. The PIC will be OK with that, but will everything else?
No, you'll get 5V - a slightly smaller diode drop. Slightly smaller since the current supplied to the PIC will come from both sources and so there will be less current through each diode and hence less drop across it. In a theoretically perfect world each diode would be the same and so the current through each would be 50% of the total, but in reality the V-I curve of even matched diodes aren't exactly the same. So I hesitate to guess at an expected current split, not that it matters in use.Ozwurld:
will i get zero out??
