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High Efficiency Relay Driver

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azulene

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I work with PVs.

I have an NI USB 6509, a USB powered digital driver box for use with LabView, it basically allows me to control 96 channels of digital on / off signals.

Each channel can hold up to 24mA at 5V.

The only relays I could find that trigger at such low voltages / currents are latching relays. This is because, I assume, they use a magnet on their actuator. I am using a TQ-2-L-5V.

I hook it up to the following circuit and for the life of me cannot get it working...

**broken link removed**

From the site, I think the above circuit is made for 12V but I have run many simulations and cannot get it working at any voltage properly.

Could anyone please help me with the above circuit or suggest another that I could use with the pile of latching relays I bought?
 
There's no DC path through the relay (220µF is in the way). Can you figure out how to fix this?
 
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I've never formally studied electronics and build things by making a series of sub circuits from bits I find here and there. Then I chuck it into Spice (sometimes) then put it into Altium or Eagle and either etch myself or send it off to China..

So from what you are telling me there is some sort of wire routing problem and the capacitor ought to be in a different spot.

I know this circuit is meant to reverse the voltage by charging the capacitor which discharges through the relay when the external voltage is cut off.

I know it may look lazy and obvious to you but I have been trying to crack this simple problem for about 20hrs to no avail.

My background is in electrochemistry, they always ask me why I do stuff like this.

If the solution is simple I would really appreciate it..

Kind Regards, Azulene :)
 
I may have spoken (typed, actually) too soon. The schematic is labeled as a "pulse latching relay", so in theory it should work as drawn (relay is momentarily energized while the capacitor charges). Not sure why it doesn't work for you.

Have you modeled this circuit in Spice? If so, can you post the file here? I could do this, except I have no relay component (I guess an inductor of the right size would do).
 
1) Your transistor as shown can only discharge the cap via the relay coil. There is no means of charging the cap.
2) The circuit requires connections to both a positive supply terminal and a negative terminal.
3) Another resistor is needed, from the transistor base to whatever drives the base.

Change the circuit so that (a) the emitter of the transistor goes only to the supply negative and (b) the collector of the transistor goes only to the bottom of the relay coil. The coil should then energise briefly when the transistor turns on.

If the latching relay is a single-coil type (as shown) you will have to reverse the coil connections to reverse the relay contact state.
If it's a dual-coil type you will need one transistor for the 'set' coil and another one for the 'reset' coil.
 
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Since you are using the DO from the board to drive a transistor which in turn powers the relay I don't understand the concern for relay current unless you specifically want a relay of the type you are using. You could come off the DO ports and using a small switching transistor like a 2N2222 drive a relay with a 500 mA coil easily. Is there any specific reason you have chosen the relay you are using?

Ron
 
I may well be out of my depth here but ...

I am guessing that the capacitor in the circuit is designed to prevent a continuous current through the relay coil. However if you are controlling the relay with a microcontroller presumably you can program it to supply a short pulse and dispense with the capacitor.

How much current does the relay coil require and what is its resistance?

Does each successive pulse latch the relay on and then off or are there separate coil connections for on and off?
 
Since you are using the DO from the board to drive a transistor which in turn powers the relay I don't understand the concern for relay current unless you specifically want a relay of the type you are using. You could come off the DO ports and using a small switching transistor like a 2N2222 drive a relay with a 500 mA coil easily. Is there any specific reason you have chosen the relay you are using?

Ron

My board has loads of relays and I wanted to get past using a separate power supply for tidiness and noise issues. Are you saying I could use a higher current relay (my DO ports are rated at only 24mA) and a transistor to squeeze out more current without needing another power supply? I would be interested in such a circuit if you could kindly direct me to what it is called or where I could find it. Thanks in advance :)
 
I got to thinking more about this. If you are going to be using many DO outputs and you want the relay(s) on while the DO is logic high I would just get a few chips like the ULN2000 series. Each chip can drive 7 discreet relays. They are simple to use and when doing DIO they are great. Maybe if you better explain your app in detail we can be of more help.

EDIT: You want the 2003A which you can drive directly with the 5 volt TTL you have from each DO. Have you chosen the actual relays yet? What do they need to handle?

Ron
 
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1) Your transistor as shown can only discharge the cap via the relay coil. There is no means of charging the cap.
2) The circuit requires connections to both a positive supply terminal and a negative terminal.
3) Another resistor is needed, from the transistor base to whatever drives the base.

Change the circuit so that (a) the emitter of the transistor goes only to the supply negative and (b) the collector of the transistor goes only to the bottom of the relay coil. The coil should then energise briefly when the transistor turns on.

If the latching relay is a single-coil type (as shown) you will have to reverse the coil connections to reverse the relay contact state.
If it's a dual-coil type you will need one transistor for the 'set' coil and another one for the 'reset' coil.

It's a single coil latch relay. My opinion of the circuit was that on the leading edge of the pulse it charged the relay and flipped it, the current then fell as the capacitor became charged. On the tailing edge of the pulse (if it were long enough, say a few seconds) the capacitor discharged reversing the polarity and flipping the circuit the other way. Like I've said I have no formal qualifications in electronics and this is more of a waffley opinion of how I thought it worked.

I can switch the relay by charging the capacitor in series through it. When I pivot the capacitor to the other side of the relay (keeping the negative leg attached to the relay) I can get it to reverse state. It's sort of what I thought the transistor did..
 
I can switch the relay by charging the capacitor in series through it. When I pivot the capacitor to the other side of the relay (keeping the negative leg attached to the relay) I can get it to reverse state.
Yes, that clearly works.
It's sort of what I thought the transistor did..
Unfortunately not.
 
OK, got it working!

Works like I thought it ought to but it needs a PNP transistor instead of an NPN with the resistor on the other leg.

Thanks for the musing here, it actually helped :)
 
The correct circuit diagram is attached...

It runs at 5V, sub 25mA using a TQ2-L-5V relay, and that is just a short pulse for a switch, the current fades as the capacitor charges. The return switch is done on the capacitor discharge when external signal is cut.

I am very impressed with this simple circuit.
 

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Hmm, something looks fishy about that circuit.

How does the transistor not burn out from the full current of the power supply when it's turned on? (Unless it's never turned on ...)
 
Here's a concept I use:
>+ power through resistor to relay+ And capacitor +
>capacitor - to gnd
>relay - to low side driver

>cap charges during "off" time
>resistor is "enough" to hold relay
>on low side driver "on":
cap has added energy to pull relay
resistor allows current to hold it
driver only has to withstand:
pull in current pulse Then hold current

I use this for serial output drivers, since it limits current through them.

Resistor & capacitor values depend on cycle rate & relay coil VS supply power.

'mem: pull in power is >> hold power... I find it well worth tinkering out...
Good Hunting... <<<)))
 
Hmm, something looks fishy about that circuit.

How does the transistor not burn out from the full current of the power supply when it's turned on? (Unless it's never turned on ...)

Thanks for pointing that out...

I have double checked the wiring on my board and this is the final circuit I got to work.

I think the resistor limits current through the transistor.

Actually, your question has saved me a lot of grief, the resistor value ought to be 50k to get the 24 mA draw I am limited to..

I turned it on and went to lunch to check its stability, when I came back and switched it off it still worked and no components were hot, so it's stable..
 
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Is the + on your diagram actually the output from a pin on your microcontroller rather than the system power supply?

If so, what logic do you use to make the relay switch on and off?
 
Is the + on your diagram actually the output from a pin on your microcontroller rather than the system power supply?

If so, what logic do you use to make the relay switch on and off?

Yes, the + is the output from my DIO controller that can sustain 24mA per channel.

NI USB-6509

This is driven from LabView software that can just sustain a high state for the length of time I want the relay on.

If you haven't played with LabView or used NI stuff, I highly recommend it, zero learning curve, lots of fun, very flexible and fast.
 
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