Grounding Issue in DIY Circuit

Hello everyone, I'm having a bit of a problem getting my LM317 soft-start circuit to function properly.

If you'll take a look at my schematic, I've built everything on the schematic up until the relays and diodes get involved (specifically where the output forks into two). In order to check that I soldered the board correctly, I hooked an LED up to where the fork would be to make sure the soft-start worked.

Here's where things get funny: If the LED is connected to the rectifier's negative, the LED behaves as I expect it to, with a slow turn-on. However, if I keep the LED connected to the same positive point, but hook the led directly to ground, nothing happens. In fact, using a multimeter showed that no voltage would flow to ground from the positive, but that 14 volts were flowing between the same positive point and negative.

I need voltage to flow to ground, this is a ground safety interlock for a welder I'm working on... does anyone have any suggestions?

Thanks for reading :)

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reakter (author) 5 years ago
The purpose of this circuit is to verify that ground actually works, it won't activate the relays unless some current flows to ground.

The relays control a fairly high-amperage circuit, so to avoid tripping breakers I made this as a slow-start that eventually cuts a current-limiting resistor out of the circuit after about 2 seconds.

I chose to do this because I didn't know about schmitt triggers, nor did I think those thermally regulated resistors sound particularly reliable.. I think they're called thermistors
But NOTHING CAN FLOW TO GROUND ! The 317's circuit is COMPLETELY isolated from it !
Ok so you have the negative lead from B1 going to your first relay and the positive lead from B1 goes through your regulator and to the relays. You need your ground to be tied in with the negative lead from B1 if your going to get any current flow through your other 2 relays. Dumping the end of those 2 relays to the ground connector of your mains outlet isn't completing the circuit in any way. Now if that ground point is connected to a ground plain and you have the negative side of B1 connected to the ground plain as well you wouldn't be having these problems.
reakter (author)  mpilchfamily5 years ago
I sense there was a misunderstanding on my part of electricity based on your reply, I was under the assumption that electrons flowed from the + side, and would try to either go to - *or* ground.

I still think electrons will follow that rule right? They'll go to either ground or -, whichever offers the least resistance?

I see now that electrons flow from -, but damn that's confusing, I always wired my circuits thinking (for instance) that the resistor goes before an LED to "slow" the current before it hit the LED... I guess that's still true, just the resistance happens after the LED. Same difference?

Anyway, you're suggesting that I tie the negative lead to ground to power the circuit? If potential rushes from high to low via -, why would it want to go to the positive lead of the rectifier (B1)?
In most DC circuits ground is tied to the negative lead of the power source. In a DC circuit electrons flow from negative to positive. You have to have a complete loop/closed circuit for the electrons to flow. Your trying to use the grounding wire in the 120 mins and that is leaving the circuit open so no electrons will flow.

In a simple resistor, LED, and battery circuit it doesn't matter which end the resistor is connected at. The flow of electrons will be slowed either way. Think the circuit like a stream of water. The battery is the water pump pushing the water through the loop. A sponge would be like a resistor. Now the pump can only push the water as fast as it reaches the intake valve. The sponge slows the flow down making it reach the intake of the pump slower so the pump can't push the water along any faster then the sponge will allow. No matter which end of the pump you place the sponge it will slow the flow of water. No if you open the loop up to allow the water to flow out or in other words to ground the pump has no water to push along any more. Thus you have an open circuit and no available electrons.

Your ground in the circuit needs to be connected to the negative lead on the rectifier otherwise the circuit is open and no electrons can flow through the other 2 relays. Just connect it up and whatch it work.
BTW. AC and DC circuits behave differently. Yes the current will take the shortest and easiest path possible. With a DC circuit it tries to take the shortest path between the 2 poles of the power source. In an AC circuit it takes the shortest path to the neutral line or ground. don't confuse the way AC and DC work. You may be trying to control an AC device but your using a DC circuit to do it.
reakter (author)  mpilchfamily5 years ago
Oh I didn't confuse AC and DC, thanks for the advice though.

So... here's my concern: I know there will be a flow from high potential to low, but the thing I need to know for sure is that ground actually works.

I was trying to get some kind of flow to ground so that the relays would operate.. no ground=no bueno.

Will tying ve- to ground ensure that this works?
If the ground is grounded and tied to the negative lead of the rectifier then it will be fine.
Volts don't "flow" anywhere.

Do you REALLY have 120V going into b1 ?

Relays don't LIKE slow starting like that. You'd be better using a schmitt trigger arrangement - and an RC network

Why have you got two relays in series in the lower pair ?

Which part are you calling the rectifier ? B1 ?

I don't see how your relays CAN turn on - you don't show a connection from the 0V on the supply to your gnd. You should tie b1 -ve to gnd.
reakter (author)  steveastrouk5 years ago
Hello there, please take a look at the comment I left for mpilchfamily, I pretty much forgot to change the value on AC1 & 2 to be ~16v AC, and I think I had a misconception about electricity (electrons flow from - to +, I thought it was the opposite).
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