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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.
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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.

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.
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You should be looking at something along these lines ( see attachment )

However, if your 9V is completely independant from the Arduino and the battery is draining, then you have something wrong with that external circuit.

Does the 9V system operate correctly if you remove the Arduino and control it manually with a push button switch ?


* drawing1.jpg (17.72 KB, 387x292 - viewed 32 times.)
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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?

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.

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.
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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!

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?

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.

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.
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I think that you need to split the analysis of the problem into 2 sections :

section 1 ( s1 ) : Arduino side : 5 V

section 2 ( s2 ) : nichrome high current 9 V

My thoughts :

1. if the 9V battery connected to s2 is draining, and s2 is not connected to s1 in any way  ( through the relay 'paddles' does not count as connected ), then there is a fault in the s2 circuit.  My guess is that the nichrome wire you are using, does not have sufficient resistance and the 9VDC current is basically draining to ground as fast as possible as soon as the relay is activated.

I think of this problem as follows :  in an AC circuit, the current relies on a push - pull - push - pull ( alternating ) type of flow, so there is usually very little resistance to hamper ( or restrict ) the flow. Think of a large syringe connected to a plastic pipe, all filled with water, and a small water wheel positioned in the middle of the pipe.  The push - pull causes the water to constantly change direction ( in / out / in / out ) and turns the wheel.   In the DC circuit, the components rely on a certain specific amount of current to flow to make them work. The same syringe is only able to push - just once ( until recharged ) - so if the plastic pipe has no resistance ( a.k.a restriction / flow limiter / current limiting resistor ) it is going to allow all the water in the syringe to be expelled very quickly. Place a water flow valve ( or current limiting resistor ) in the plastic pipe, and the water from the syringe is going to last a lot longer because you are limiting the speed that the water can flow out.

2. the 5V circuit ( s1 ). You are using an Arduino. Do not think of the output pins as being there to 'drive' your relay or motor, etc. They are there only to act as a trigger.  There is a limit to the amount of current that can flow through the pins. Think again about the syringe. If the water wheel is spinning faster ( powered by an external source ) then it will suck the water out of the syringe faster than the syringe can move, and the syringe implodes / cracks.

The arduino has a current limit per pin ( I think it is about 30mA ) but your relay coil requires 40mA ( just a guesstimate ), so the coil is going to try and suck current faster than the pin can provide - implosion is imminent.

Use a transistor ( uses very low current from the pin ) to drive the relay, which then controls the 9V circuit, just like it would if the relay was a light switch.
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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.

1) 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?

2) 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)?

3) 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-

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!

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?

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.

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.
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OK. I must state that my comments are based on my own experiences, and I have no training of qualifications in this field, but here goes :

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?  

because you are draining the battery very quickly as there is no flow restriction on the water pipe from the syringe. DC power is one-way only, and once used is gone forever, until you recharge or replace the battery.

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.

Current will not flow until there is a closed circuit from positive to negative - the less resistance between the pos and neg, the faster the current is going to flow between them, and the faster your battery is 'flat'


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.

1) 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?

So you have 2 problems - a 9V battery that appears to be short circuiting and draining away all it's power very fast, and a relay that has a coil drawing 89mA from an output pin that can only give 30mA - something is about to break - implosion is imminent.


2) 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)?

I would expect it to happen faster than that.


3) 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-

Yes, a resistor in the circiut will reduce the current through the wire. But take a look at the specs for the specific wire you are using. You will likely find something like : 9V 500mA generates a temperature of 250 celcius.  Assuming you want 250C heat from the wire for the launch, you need to restrict the current to 500mA to achieve the required temperature from the wire.  In fact, if you pass too much current through the wire, and exceed the specs for the wire, you will likely reduce the life span of the wire and it will burn out very quickly.

battery+ > nichrome > resistor > battery- will reduce the current flow - maybe to a point that is optimal for the life span of the wire and slowing / limiting the drain from the battery to the current actually needed by the wire.
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I really appreciate your help, that clears up a lot of my questions.

However, I am having a little trouble understanding what is going on in the diagram you linked.  


The part that is confusing to me is the transistor and the diode.  I don't understand why the diode is needed there, or why you would ever want current to flow back towards the 5v output on the arduino?

Also, I do not have a 4001 diode, but I have a 4007.  Will the 4007 work in that diagram as well?

I also do not have that transistor, but I have this one:  http://www.scribd.com/doc/59717674/SM945
« Last Edit: August 13, 2013, 02:29:18 pm by jwllorens » Logged

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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. http://forum.arduino.cc/index.php?topic=182505.0

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.


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.

1) 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?

2) 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)?

3) 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-

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!

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?

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.

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.
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I really appreciate your help, that clears up a lot of my questions.

However, I am having a little trouble understanding what is going on in the diagram you linked.  


The part that is confusing to me is the transistor and the diode.  I don't understand why the diode is needed there, or why you would ever want current to flow back towards the 5v output on the arduino?

Also, I do not have a 4001 diode, but I have a 4007.  Will the 4007 work in that diagram as well?

I also do not have that transistor, but I have this one:  http://www.scribd.com/doc/59717674/SM945

for the diode, have a read here : http://en.wikipedia.org/wiki/Flyback_diode

Can't say that I understand it fully myself, but have taken it as a serious requirement from other posts that the guru's have commented on.

As far as I understand, the main difference between the 4001 and the 4007 is that the 4001 can only manage a 35V reverse voltage, and the 4007 a  700V reverse voltage.  I have used both in the past on relays.

As far as the transistor is concerned, sorry, but we'll have to wait for one of the guru's to look at that one.
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Nichrome wire is resistance wire (try Google) that get hot enough to glow, that's how it starts the rocket.

 A standard rectangular 9V battery just doesn't have the energy in it to do this for long. And you won't get 6 minutes at 6A from a 600mAHr battery. The mAHr rating is done at 10hr discharge rate, the energy you can get out goes down drastically the faster you discharge it. So 6A for a couple minutes, if you are lucky.
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Steve Greenfield AE7HD
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THe Nichrome wire is called a squib it usually is lost when the rocket motor is ignited. In my experiences a 6V lantern battery is used to light up the squib and the device has an interlock to prevent accidental firing of the squib when it is first connected.
The 1N4001 is a 50 PIV (Peak Inverse Voltage) diode commonly used as a low voltage rectifier for transformer operated mains power supplies. PIV refers to the maximum allowable reverse voltage that can be applied to the diode. A 1N4007 is a 1000V PIV diode and the slightly better choice of the two. The transistor mentioned ended in 945 and is likely a 2SC945 transistor which is a medium Beta (DC Gain) transistor which should work just fine. Connect the Emitter to digital ground, the Base through a 1K resistor to the Arduino output control control pin and the Collector to the relay coil and the Anode of the diode. The other end of the relay coil and the cathode of the diode to the positive supply. The relay operates sending a high current pulse from the "firing Battery" relay N/O contact to the squib which gets Very hot (Some have a small amount of incendiary material at the tip for an open flame to ignite the rocket motor.
My Son and I used to do model rocketry together so I have a passing familiarity with the hobby. There is an electrical lockout that senses when the squib is properly connected and will not allow the firing circuit to be armed If the squib is at fault (used or improperly connected).
Be most careful when testing a squib, a firing circuit or anything involving either the firing circuit or the squib the incendiary material is a direct fire hazard as well as a general safety hazard and should be handled with Great care.

Doc
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The diode never has to withstand more than the voltage used to turn on the relay, so a 1N4001 will be fine, if that is what you have.

As Docedison says, a 6V lantern battery. Or four D cell alkaline, or in a pinch for AA alkaline will work.
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Steve Greenfield AE7HD
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Not for long AA batteries even considering that the squib only needs to be hot for 100 mS or so certainly less than a whole second.

Doc
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