Timeline for Using DC current to operate a fish feeder that takes 2 AA batteries
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 7, 2015 at 8:26 | comment | added | Criticizing Israel not allowed | Let us continue this discussion in chat. | |
| Mar 7, 2015 at 8:25 | comment | added | Yeti | @immibis looks like the best solution here is to go get a regulated adaptor (which I doubt I'll find in my country). I'll try using a 2.2kΩ resistor and see how it goes. | |
| Mar 7, 2015 at 8:22 | comment | added | Criticizing Israel not allowed | @Yeti but from the first measurement (5.4V with the motor running) it looks like the power brick simply won't output the right voltage unless it's supplying about 1000mA. (Which is normal for a simple, not-so-modern, unregulated power brick). Which means if you go with a resistor in parallel, you'd probably need it to waste a fair bit of current to keep the voltage down. | |
| Mar 7, 2015 at 8:20 | comment | added | Criticizing Israel not allowed | @Yeti I don't know what you mean by "when I complete the circuit using the multimeter". | |
| Mar 7, 2015 at 8:12 | comment | added | Yeti | @immibis when the motor is running the voltage is 5.4v. When I complete the circuit using the multimeter the voltage shown is 3.9v. | |
| Mar 7, 2015 at 6:08 | comment | added | Criticizing Israel not allowed | @Yeti What voltage do you measure when the feeder is connected? | |
| Mar 7, 2015 at 2:39 | comment | added | Yeti | I don't have the registor yet. I went ahead and connected the feeder to the adapter anyways. The reading was at 90mA constant and motor rotated in 5 seconds. This is similar to when I used 3 AA batteries. The problem is that now I hear a clicking sound when the motor rotates (whoops!). So now I'm sure that the feeder must only operate at 60mA. Anything more than that rotates the motor too fast and feeding is incorrect. I'll try to get hold of the registor. Is there anything else I need to do/buy? | |
| Mar 6, 2015 at 16:28 | comment | added | Greg d'Eon | Good answer; +1. FYI, 36 ohms is the point where V^2 / R will be greater than 1/4 W - but as Phil says, you probably won't be near that point just to make it work. | |
| Mar 6, 2015 at 15:38 | history | answered | Phil Frost | CC BY-SA 3.0 |