Even a dead battery will still have some juice in it, so I think the method of you measuring the current is incorrect or you have a bad meter.
1, each pin has 30ma, enough for lighting a LED or triggering a switch. In your case, you need to use a transistor as a switch amplifier. 30ma X 100= 3000ma. most transistor does amplifying current up to 100 times.
2, you can't. This is a battery discharger I built. a 600mah battery will need more 6 minutes to drain out if discharged with 6A current. My simple Ni-mh battery discharger mah calculator Arduino project Mark II - General Electronics - Arduino Forum
3, I have no idea what a nichrome does. Most of the time, you need a resistor to protect the component, ie, if you connect a 9v to a LED, you will burn it and damage the LED. Sometimes, you don't, if that is a motor, 9v will run it faster than 6v. putting a resistor with a motor will just waste energy on the resistor.
jwllorens:
Then how come I am reading no current when I measure the contacts on the battery itself? And how come the relay works the first time it activates and not subseuent times until a new battery is used?I am very confused. It makes sense what you are saying, in fact the relay should never be working (yet it works the first time I plug in a fresh battery to the auxillery circuit, not subseuent times) and my meter is reading no current when I measure the contacts on the battery itself.
The nominal coil current is 89.3ma, so I will try rewiring with the transistor. Maybe I can pull a transistor out of this old broken coffee maker.
A few questions though.
If the coil needs 80ma to create a strong enough magnetic field to move the paddle securely against the contact, and the output pins on the arduino are only providing 30ma, is the coil going to "suck" more current through it than the digital pin wants to provide and thus break the arduino? Or is the arduino just going to provide an "impotent" amount of current that wont reliably move the paddle because the current is weak, thus the magnetic field is weak?
Can a battery really drain itself in a couple of seconds if it the circuit is basically completed with a wire with low resistance? Such as taking a 20gauge wire, attaching one end to the positive terminal on the battery and the other end to the negative terminal, would that really drain the battery almost immediately (in around 5 seconds)?
If I put a resistor on the 9v (nichrome wire) circuit, wont it reduce the current flowing through that circuit so the nichrome wont heat quickly and thus the rocket wont launch immediately, defeating the purpose of the relay to interface with a higher voltage circuit in the first place? Also, doesn't nichrome provide an inherent amount of resistance anyways? So, say I have: battery+ > nichrome > battery-. Would the amount of current going through the nichrome be the same or less if I did battery+ > nichrome > resistor > battery-
arduinomagbit:
because digital pin output current in ma is about 30ma. you need higher current than that to have your relay on all the time you intended. The battery is not dead, the relay without enough current is ! google: relay transistor circuit!jwllorens:
Why not? It is a very small relay, the coil itself is rated for 5v (the "paddles" are rated for much higher). Why do I have to use a transistor to drive the coil when the coil operates at a minimum of something like 2.5 volts? Is the voltage or current coming from the digital pin unstable or something? Some more info about the relay though, the relay itself is rated at a pretty high voltage, but only rated for AC. It doesn't say anything about what it is rated at for DC. I have seen most relays are rated at about half the AC voltage for DC voltage, so I assumed since this relay is rated for about 100v AC then it would handle 9v AC just fine (though I know a transistor probably would have worked fine, I just didn't have one on hand.) But that gets me thinking, is it a backflow problem?Also, that doesn't explain why the battery that is running current through the "paddle" part of the relay is almost instantly going dead, dropping from 9v to 7v with almost no detectable current (even when I pull the battery off the connector and measure the contacts on the battery itself, to make sure it isn't a weak solder joint or something, the battery is definitely going dead.) This circuit has killed two batteries almost immediately, to the point where they will give a voltage reading of 7.5v on their contacts but no current whatsoever. Interestingly enough, when I snapped the two "dead" batteries into each other, so one dead 9v was plugged directly into the other dead one, and left them for about a minute, one (but only one) of the dead 9v batteries "restored" itself to 9v across the contacts and suddenly produced a current. Is this something to do with backflow?
arduinomagbit:
you can't just wire up the digital pin on the arduino into one end of the coil in the relay!! you need to use a transistor to drive the relay, and use that digital pin to turn on the transistor.jwllorens:
Hello. I am fairly new to this but I have run into a bit of a problem.I have my Arduino Uno set up to do some stuff with some sensors and make some decisions, and when certain conditions are met, a digital pin is turned on.
Now, this pin outputs a current at 5v, with slightly less amps than the 5v connector on the board. I need to drive another circuit that requires about twice the amps and voltage, so I set it up with a rely (I really don't know what I am doing here).
So, I wired up the digital pin on the arduino into one end of the coil in the relay, and wired up the digital ground to the other end of the coil. Then, I wired up the positive end of a 9v battery directly to one end of the paddles in the relay, and the negative end goes to the higher current circuit I am driving. The other connection in the relay then goes to the higher current circuit.
The higher current circuit is very simple. It is a piece of nichrome wire, I believe, all it is doing is setting off a model rocket. The power from the digital pin in the arduino board is not enough to set off the rocket, but powering it directly from Vin doesn't seem like what I want to do because I don't want to suddenly drop the voltage going to the arduino. So I am using this relay to have the arduino be on its own battery but interface with another more simple circuit with another battery.
So everything seems to work, except this one thing. I plug in fresh 9v batteries, attach my volt meter to the leads coming from the relay, and then activate the relay. I hear the physical click, the voltage jumps to 9v on the leads coming from the relay, seems to work. I try to launch my rocket after the countdown and junk that my arduino is doing, and nothing. I go and test the voltage again on the leads from the relay, and it has dropped to 7v when the relay is activated, and the amps have fallen to nothing.
So simply by testing the voltage that first time, I am somehow draining the battery completely, and it is only happening to the battery going through the high voltage part of the relay which is a very very simple circuit assuming i hooked up the relay properly. Any thoughts? I am doing something wrong, I just don't know what it is.