I’m new to soldering and electronics. I brought a couple of the DFR0571and DFR1015 buck modules to power a RPi from a 9v battery or solar setup (12-14.4v)
Question I have is, do you solder cables on the component side or the bottom side as the pads don’t seem very big on the bottom.
Another question, can I have an anderson cable with a diode connected to positive and a 9v battery snap with another diode on positive, feed the positive of buck and use a common ground? Only one or other will be connected.
In the above, is there a way to switch between the higher voltage when available and fall back to 9v if solar is unavailable?
If the wire is small enough insert through the hole before soldering. On the DFR0571 the pads are both the same size anyway. It does not make any difference which side you use but in the interests of preventing board damage and your self confessed iffy soldering skills I would suggest the side with the larger pads. Possibly a bit of soldering research and practice may not go astray first.
A little sketch of what you have in mind would help. Also what sort of current would these diodes be handling.
Did you mean a typical 9V box battery? The Duracell datasheet suggests you’d only get an hour of use at 0.25* (9 / 5) * 0.85 = 0.38A, the Pi can draw up to 1.8A at 5V at heavy load. and the voltage would drop quite a bit. I’d suggest a larger battery of some kind.
Do you have access to a larger SLA battery?
Also, what was the diode you were using? Just worried about voltage drop.
James has already said the 9V battery will be nowhere enough if it is one of the small box type as in DMMs etc. I would suggest a 12V lead acid Gel or AGM type, Has to be above 7.5V (or about 8V to allow for Diode voltage drop) for the DfRobot devices. If you anticipate using the battery for extended periods between charging fit a low voltage cut off device to prevent over discharging or the battery won’t last long.
The 2 diodes are in the same TO220 3 pin package with a convenient common cathode. Whichever supply (solar or battery) has the higher voltage will be the active one and will have all the load until they become equal then they will share until one drops below the other.
The battery should not be allowed to go below 10.8V absolute minimum. Cut off device should be set for 11V.
The lower voltage source turns on the higher voltage source eliminating the voltage drop across the p mosfet’s body diode. If the voltages are similar then it acts the same way as the diode solution. Additionally, the voltage source can be selected by a microcontroller but there are a few things to be careful about. Turning on the lower voltage source will create a short. The enable lines shouldn’t be connected directly to the microcontroller. If you’re interested in this solution let me know and I’ll provide more details.
With this in mind I tried to keep a solution as simple as possible.
The diodes I recommended are schottky so a bit of voltage drop I think is a small price to pay for simplicity. Great believer in the KISS principle.
Yeah agreed, KISS approach is almost always better, no matter your experience level. I learnt through a try and fry approach, works well if you’re working with a few cheap components and don’t mind the frustration. I still finger test chips and regret it every time.
And you can get our latest projects and tips straight away by following us on: