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> Question About Relay Drivers
xgabrielx
Posted: December 21, 2011 08:20 pm
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I have a 2.4v supply off a circuit board which I need to use to switch a 5v relay. I'm aware that I can use a transistor to use some of the power from the circuit being switched by the transistor to operate the relay, but the circuit has delicate components, and the relay will be switching a pretty large current. Is this ok, since, in effect, the delicate circuit will be sharing a ground with a pretty large current in comparision? I hope that made sense. It made sense as I was typing it.
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johansen
Posted: December 21, 2011 08:38 pm
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Think of your ground wire as an inductor, and the energy stored in it has to go somewhere when the relay opens and your "pretty big current" falls to zero.





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xgabrielx
Posted: December 21, 2011 08:57 pm
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I'm not sure what you mean. I'm a total noob as far as electronics goes. The probleme is, I need to use the 2.4v supply off the circuit board (And I'm sure it can provide only a very small current) To switch a seperate 6v circuit with a current of 5 amps (though only for a second or two) The circuit board cannot operate the relays coil, so I was going to use it, to switch on a transistor that would allow the 6v circuit to operate the relay coil. I assume that is ok but I notice that one leg of the 6V circuit is connected to the same point as the output from the circuit board in this circuit. I'm not quite sure if that's a problem or not. Again, I ain't making much sense. Maybe I should do a diagram...
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MacFromOK
Posted: December 21, 2011 09:43 pm
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A diagram would probably help, but you can mitigate the inductive kick of a relay (or other inductive loads) by adding a diode anti-parallel (reversed) across the coil.


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johansen
Posted: December 21, 2011 10:07 pm
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makes sense.
btw, 5 amps isn't really big.


A diode across the relay coil, and don't forget the capacitor from 5vdc to ground near the capacitor and transistor is all that is needed.
On the other hand if your 5 volt supply is connected to the circuit though very long wires, when your 5 amp load turns off, that 5 amp current that is still flowing causes the capacitor located near the relay, diode and transistor switch to be charged up to something more than 5 volts. one microfarad is all you need in this case.


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xgabrielx
Posted: December 21, 2011 10:32 pm
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user posted image
What I mean is, as you see in the picture, the circuit board shares a leg with the 6v supply. The circuit board runs off 2aa batteries, so probably can't take 6v, but the only share one leg..I dunno. I got this circuit somewhere..I dunno where. Bottom line is, I get 2.4v off the circuit board. I need to use that to switch a 6v supply @5 amps. The only relays I can find that can handle that current all have 5v coils. This solution was presented on some forum. (Like I said I don't remember which. It could even have been this one, but I doubt it)

(Obiously, the output from the relays switch shares the same ground as everything else, here. I dunno if "ground" is even a good term, since I have no clue WHAT is on the circuit board.)

This post has been edited by xgabrielx on December 21, 2011 10:33 pm
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johansen
Posted: December 21, 2011 11:05 pm
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looks good smile.gif


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xgabrielx
Posted: December 21, 2011 11:31 pm
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Oh, good. I was just concerned seeing one terminal of the 6v supply in effect connected to a circuit board that would die if I tried to power it off 6v. I know sharing a leg like that doesn't necessarily mean I'm directly hooking up 6v to the circuit board, but I don't really understand why not. To me, it just looks wrong. I'll try, anyway. It's all cheap stuff I'm using here, so I'm not concerned about seeing smoke. Just about wasting time.
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Jimthecopierwrench
Posted: December 22, 2011 12:01 am
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QUOTE
doesn't necessarily mean I'm directly hooking up 6v to the circuit board, but I don't really understand why not.


Well, lets look at what's happening. The output side of your NPN transistor (when not conducting, ie. 'off') looks like the cathode of a diode - so even if you connected a positive voltage directly to it (up to the point the transistor itself breaks down anyway) the low voltage (base) side will not see it.

If you measure the voltage at the collector (as you've drawn the circuit) with the transistor in cutoff, you'll actually see 6V through the relay coil.

When you drive the transistor into saturation (turn it on) the collector will appear (oversimplified) as if the diode were magically turned around with the colector looking like the anode of a diode connected to ground. The collector will "supply ground" to the load - referred to as sinking current.

So if you measure the collector voltage of the transistor when it's on you no longer get 6V - you get a value close to ground. The voltage is 'dropped' across the energized load.


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xgabrielx
Posted: December 22, 2011 12:36 am
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You lost me at "Quote". Lol.
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