West-se

I've been trying to get my electric fan to work by using JS0 and a relay but Im having no luck. I got it connected like so,

MS1 v3.0 msns-e 029i

JS0 --> 85
12V -->30,86
Fan+ --> 87
Fan- --> chassis ground.

I've also added a diode as per the diagram at msns-e site



When I powerr up my ms, my LED's are dim and nothing works, I cannot connect to the MS via MT. But if I disconnect JS0 and power cycle its fine. If I remove JS0 from 85 on the relay and ground 85 the fan turns on.

pmrobert

JS0 is a direct connection to a processor pin, it can only source/sink about 10 ma. This is enough to drive a transistor which will then be able to drive the relay. The 2n2222a transistorcircuit that is shown in your message is what you want - JS0 would connect to where it says "X2,X4,etc". The 1K resistor is very important. Where it says "X11,X12.." is what you would ground your relay with. Please don't try to drive your relay (or anything other than a logic level high impedance device!) directly with JS0, it's easy to ruin the output ports in the processor by attempting such things. The reversed diode across the relay is very important too - when the relay is turned off (ungrounded) the energy in it's collapsing magnetic field has to go somewhere; that diode will conduct when that happens and keep the transistor from getting hit with that potentially destructive inductive kick.

-Mike

West-se

That is the relay driver schematic for the v2.2 is there any difference for the v3.0 ? There is no X2,X4 on the V3 as far as I can see.

pmrobert

My apologies - I wasn't clear on things. The v3 board's jumpers, pads, etc. are differently named vs 2.2. The goal that needs to be accomplished here is to make a processor pin (JS0 connects to a processor output pin) control a transistor which controls the relay. The processor pins can only provide or ground a very small amount of current - about 10 milliamps. This is plenty to "turn on" a transistor because the transistor has very high impedance on it's control line (the base) so very little current actually flows from the processor pin. The transistor is then used as the equivalent of a solid state switch to actually flow enought current between it's emitter and collector to allow enoungh current to flow through the actual fan relay's coil to close the points/mechanical switch inside the relay that CAN handle lots of current. So, processor pin controls transistor, transistor controls relay. You can see by the attached schematic that JS0 is actually connected to the PTA5 pin of the processor. This stuff can be a little difficult to initially get your head wrapped around but it'll click. If you need any more help understanding this, please ask!

-Mike

West-se

Thanks, after I read your original post it made more sense, I thought the pad (JS0) could provide enough ground but I understand now, your post was informative as I know understand better on what is actually happening with the diode and transistor.

Sean