Just be aware that the relay will be on if the inputs are below +12V. Not just when the input is ground.
In other words, if the input goes from +5 to ground, then the relay will be on all the time.
Just be aware that the relay will be on if the inputs are below +12V. Not just when the input is ground.
In other words, if the input goes from +5 to ground, then the relay will be on all the time.
P.S. Welcome to the forums
Bean
Thanks Bean, sounds like that's probably not a good solution then...
So, if you were going to do it, how would you do it?
Welcome to the forums. It would help to know what your control signal is that you need to drive the relays. You said it needed to be low instead of high, but what is the voltage range of this control signal? Is it open-collector?
Welcome to the forums. It would help to know what your control signal is that you need to drive the relays. You said it needed to be low instead of high, but what is the voltage range of this control signal? Is it open-collector?
D'Oh! Yeah, it's an open collector, sorry, should have put that initially!
If it is open collector, then it should work as you have it...
Bean
Not as the board comes, because the signal line has to go high to turn the relay on. Since he has an open-collector output I was thinking that if it can handle the current, it could switch the relay in place of the on-board transistor, but I don't know the nature of this open-collector signal.
To the OP, what is the driver of this signal? Is it a FET and if so, how much current can it sink? What is the device that you're using to drive the relay board?
I suppose you could put an inverter, such as a 74HC04 chip on the input to reverse a relay's behavior. But you can do this inversion in code in the Propeller or BasicStamp by just swapping High and Low output in code. Using such a chip requires a power source to it.. that can be a problem.
There are safety issues with reversing the signal from High equal to coil on, to Low equals coil on. What happens when you first power up the whole system including the relay board? This needs to be looked at as signals may default in a way that something is running that is unexpected.
It would be odd to have a microcontroller provide an open-collector output. That role is is regularly a buffer chip or a driver chip of some sort.
There are open-collector buffer chips, such as the 74HC07 that do this AND the open-collector side will accept a much higher voltage (like a +12 relay) and connect it to ground. So you can jump from a 3.3 volt output to activating a +12 relay in one-step.
Open-collectors are a good thing for robotics. It pays to learn to use them, not just with relays.. but with other devices that desire a voltage jump.
I have a slight issue in that the open collector is within a mixing desk, (not a microprocessor - so I can't alter the code!) and I don't have the exact spec for it! I'm pretty sure that trying to use it to drive the relay directly would be a very bad thing to do, hence trying to modify this circuit to use a PNP transistor switched to ground, rather than an NPN which needs the signal to go high...
Bean? Did you mean that it should work with the modifications I posted a diagram of?
Hi!
Bean? Did you mean that it should work with the modifications I posted a diagram of?
Yes, I would think it would work WITH your modifications. As long as the open-collector input does not sink any current when it is off.
But it would have been much easier to just use the original circuit and add a pull-up resistor to the input.
This would turn the relay on when there is no input, and the relay would be off when the input is active. But since the relay has both NC and NO connection, just simply reverse them.
Bean's suggestion of a pull-up resistor is the best of all.
But for what it is worth....
The primary issues with an unknown specification open-collector are the usual -- volt limit and current tolerance.
Generally, the open-collector is just a transistor (or it would be called an open-drain if it was a MOSfet). It is nearly impossible to build a transistor that won't tolerate 25 volts or less in open-collector switching.
The issue of maximum acceptable current is much more critical.
Does one dare to guess?
I suspect 20ma is no problem and limits might be approached at 40ma if it is an IC. (Take a look at the specs for the 74LS07 for comparison http://www.ti.com/lit/ds/symlink/sn74ls07.pdf). Since driving a 12amp relay needs a bit of speed to engage, I suspect that 60ma is barelly enough to properly do the job. Many 12amp relays need more than that at 12 voil coils, a lot more.
What to do? Consider a Darlington transistor configuration (or its compliment) to do the inversion and to handle the extra current. I suspect the open collector in place can provide the first stage and if you need an NPN second stage a 2N2222 will handle 500ma easily. (For NPN, use its compliment the wN2907). Amplification would be a lot higher. The trade off is Darlingtons operate with a bit of heat.
Well, I did it, (as per the diagram I posted) and can confirm that it does work...
Thanks to all who have answered, it's been much appreciated... (and I'll definately be buying more of these kits, as the build quality is top notch...)
I suspect it is best to use the board unmodified and apply Bean's solution of a pull-up resistor if at all possible.
Modified boards may migrate into the hands of users that think they are unmodified and that the documentation is enough to work with them. Confusion and likely destruction of something might follow.
Yes, I would think it would work WITH your modifications. As long as the open-collector input does not sink any current when it is off. But it would have been much easier to just use the original circuit and add a pull-up resistor to the input.
This would turn the relay on when there is no input, and the relay would be off when the input is active. But since the relay has both NC and NO connection, just simply reverse them.
Bean
I just wanted to point out, in case the OP wasn't aware, that in doing so the relay will be energized when inactive and only de-energized when you activate it. So when the circuit being switched is off you'll be drawing ~80 mA of current. This should only matter if the circuit is battery powered or depending on duty cycle of the switching.
While relays are absolutely brilliant devices for isolation and jumps to higher voltages and to and from DC to/from AC....
Latcing relays can resolve that remaining bugaboo of needing a coil to be consistently powered in one position.
In nearly all cases, the latching relay is the low power cousin to the conventional relay.... very very environmentally friendly.
But there are some things to notice.
A. they require pulses.
B. the more powerful ones require two input lines -- one for each state. (This is actually a good thing as you don't need to have feedback confirm which state you are in.)
Comments
In other words, if the input goes from +5 to ground, then the relay will be on all the time.
P.S. Welcome to the forums
Bean
Thanks Bean, sounds like that's probably not a good solution then...
So, if you were going to do it, how would you do it?
Cheers!
Ian...
D'Oh! Yeah, it's an open collector, sorry, should have put that initially!
Thanks for the welcomes... :cool:
Bean
Not as the board comes, because the signal line has to go high to turn the relay on. Since he has an open-collector output I was thinking that if it can handle the current, it could switch the relay in place of the on-board transistor, but I don't know the nature of this open-collector signal.
To the OP, what is the driver of this signal? Is it a FET and if so, how much current can it sink? What is the device that you're using to drive the relay board?
There are safety issues with reversing the signal from High equal to coil on, to Low equals coil on. What happens when you first power up the whole system including the relay board? This needs to be looked at as signals may default in a way that something is running that is unexpected.
It would be odd to have a microcontroller provide an open-collector output. That role is is regularly a buffer chip or a driver chip of some sort.
There are open-collector buffer chips, such as the 74HC07 that do this AND the open-collector side will accept a much higher voltage (like a +12 relay) and connect it to ground. So you can jump from a 3.3 volt output to activating a +12 relay in one-step.
Open-collectors are a good thing for robotics. It pays to learn to use them, not just with relays.. but with other devices that desire a voltage jump.
http://en.wikipedia.org/wiki/Open_collector
I have a slight issue in that the open collector is within a mixing desk, (not a microprocessor - so I can't alter the code!) and I don't have the exact spec for it! I'm pretty sure that trying to use it to drive the relay directly would be a very bad thing to do, hence trying to modify this circuit to use a PNP transistor switched to ground, rather than an NPN which needs the signal to go high...
Bean? Did you mean that it should work with the modifications I posted a diagram of?
Thanks all for your help, it's appreciated...
Yes, I would think it would work WITH your modifications. As long as the open-collector input does not sink any current when it is off.
But it would have been much easier to just use the original circuit and add a pull-up resistor to the input.
This would turn the relay on when there is no input, and the relay would be off when the input is active. But since the relay has both NC and NO connection, just simply reverse them.
Bean
But for what it is worth....
The primary issues with an unknown specification open-collector are the usual -- volt limit and current tolerance.
Generally, the open-collector is just a transistor (or it would be called an open-drain if it was a MOSfet). It is nearly impossible to build a transistor that won't tolerate 25 volts or less in open-collector switching.
The issue of maximum acceptable current is much more critical.
Does one dare to guess?
I suspect 20ma is no problem and limits might be approached at 40ma if it is an IC. (Take a look at the specs for the 74LS07 for comparison http://www.ti.com/lit/ds/symlink/sn74ls07.pdf). Since driving a 12amp relay needs a bit of speed to engage, I suspect that 60ma is barelly enough to properly do the job. Many 12amp relays need more than that at 12 voil coils, a lot more.
What to do? Consider a Darlington transistor configuration (or its compliment) to do the inversion and to handle the extra current. I suspect the open collector in place can provide the first stage and if you need an NPN second stage a 2N2222 will handle 500ma easily. (For NPN, use its compliment the wN2907). Amplification would be a lot higher. The trade off is Darlingtons operate with a bit of heat.
Thanks to all who have answered, it's been much appreciated...
Modified boards may migrate into the hands of users that think they are unmodified and that the documentation is enough to work with them. Confusion and likely destruction of something might follow.
I just wanted to point out, in case the OP wasn't aware, that in doing so the relay will be energized when inactive and only de-energized when you activate it. So when the circuit being switched is off you'll be drawing ~80 mA of current. This should only matter if the circuit is battery powered or depending on duty cycle of the switching.
Latcing relays can resolve that remaining bugaboo of needing a coil to be consistently powered in one position.
In nearly all cases, the latching relay is the low power cousin to the conventional relay.... very very environmentally friendly.
But there are some things to notice.
A. they require pulses.
B. the more powerful ones require two input lines -- one for each state. (This is actually a good thing as you don't need to have feedback confirm which state you are in.)