fluid-pumppowersensor
I have a PIR sensor that when triggered, gives out a 3.3V signal. I'm trying to use that to switch on a 5V water pump, and I'm assuming I will use a transistor. The sensor only puts out 10mA, while the pump draws around 1A. How would I switch on the pump with the sensor?
Thanks!
I wouldn't recommend connecting the panels in series in case the 20V panel pumps too much current through the ones that don't work very well. To decide how to connect them (if it's possible) details of those panels would be required.
However, for the 20V panel I would say you need a switch mode buck regulator that takes the 20V and reduces it efficiently to 12V to power the two 12V motors. At 0.8A total draw from the two 12V motors the power requirement is 9.6W. The panel and regulator should be able to supply 90% of 20V x 0.8A = 14.4W.
This leaves about 4W (conservatively) for the 6V motor which you imply requires 0.6W. You'll need a second step down buck regulator for this device and it sounds feasible.
Switch mode buck regulators - there are tons of them but I can only recommend a design type such as the LT3976: -
For a slightly more efficient regulator there is the LTC3780 which will work providing 12V down below 12V from the panel: -
Both regulators can supply 12V at 5A.
I wouldn't bother thinking about a linear regulator because it will waste about 8W in heat.
It's possible that you can search along these lines and get some kind of module/PCB that you don't have to build. Maybe someone can recommend something?
A simple sensor used in such an application is a magnetic reed switch.
(source)
The simplest normally-open reed switch has two contacts, which are separated when there is no strong magnetic field incident on the switch. When the switch is exposed to a strong enough magnetic field, the contacts close and a circuit can be established through it. In your case, the float would have a small magnet on it, which passes close to the reed switch when the float moves.
The advantage of this approach is that no current flows, and hence no power is consumed, while the switch is open. Power requirement of the sensor is marginal even when the switch is off - just sufficient to overcome a pull-up resistor at the microcontroller sense pin, for instance.
simulate this circuit – Schematic created using CircuitLab
Best Answer
Use a logic level (Vgs-threshold <= 2V would be good) MOSFET (N Channel for a low side power switch, it's the easiest).
The following diagram is something I put together to teach my robotics club how to do basic power circuitry, in this case to drive a 12V DC PC case fan. In your case, the fan motor is just a 5V motor, but still is essentially a fan/motor. The protection diode shown is still important.
In this case the "switch" is your 3.3V sensor output. The PIR sensor output may not be high for very long, you should perhaps use a 555 timer and some buffers/inverters to keep the signal to the MOSFET high for longer, with the RC constant as something reasonable for your application.
Depending how your 5V rail is supplied to the pump, you may want to put some bulk capacitance such as a few hundred microfarads, or even a few thousand, to prevent it from drawing the line low during current transients as the pump turns on. This same advice holds true for using servos, and many other motors. Always buffer the power supply with local bulk capacitance!