Wiring Diagram for Potter&Brumfield PM-17AY-120???

Nov 26, 2003 39 Replies

What are you missing?

1) You are fighting against the advice you have been given. That advice is 100% correct. You are, in effect, saying you know better than the people who give you the advice AND the National Electrical Code. And you are wrong. Unfortunately, you could end up *dead* wrong. You are placing a killing level current on ground if you follow your scheme.

Current should never be on ground intentionally. Ground is for safety.

Hot--->relaycoil--->brokenground--->you--->realground.

The coil is going to try to draw 128 mA through you if you get between the ground and it.

It is illegal and could kill you or someone else.

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Don't know if you are reading it wrong. See my earlier reply, where I gave you the solution. Here's the existing and new circuit analysis:

Hot----coil---ground. You said it draws ~128 ma: "But, the way I see it, maximum current through the coil is only 128ma". Assuming that is true, the relay coil impedance is E = IZ; 120 = .128 Z; Z = 120/.128 or 937.5 ohms. I told you to use a 1000 ohm resistor in series with the coil.

The new circuit: hot---coil---resistor---hot. Figuring the coil as 1000 ohms, and the resistor as 1000 ohms, the voltage will divide equally.

Using the exact numbers:

240 = I * (937.5 + 1000); 240 = I * 1937.5; 240/1937.5 = I so I = .1238 At I = .1238, the voltage across the coil will be E = IZ; E = .1238*937.5; so E = 116.129

The resistor will have to dissipate some heat. The formula for that is P = I^2*R; P = .1238*.1238*1000, so P = 15.32644 watts. My recommendation was to use 4 250 ohm, 10 watt resistors in series to spread the heat out. You always want to give yourself a good margin on wattage. 4 250 ohm resistors will give you 40 watts capability - and spread the heat out "geographically". You could use a single 25 or 50 watt resistor, but you would not gain the spreading out effect.

Let us know how you make out.

The coffiee hasn't set in yet, so forgive me if I'm "out to lunch."

Isn't the Z of the ac coil R + jwL. (w is little omega) Then adding another resistor R1 gives Z = R + R1 + jwL.

What you are doing works fine for DC, but with AC, the inductance of the coil adds to the total impedance as well.

And with regards to your other note, the impedance of the coil of an open relay is less than when engergized, so more than 128mA will flow on inrush.

Hopefully I'll agree with myself after coffee :-)

True, you *can* get 120V on the frame to ground if the ground line from the frame opens.

This is where you are mistaken. The relay contacts opening will not remove the 'hot' from the relay coil. If they did, then you could never get the thing to start in the first place. The relay coil 'hot' is probably from the on-off switch *upstream* of the relay contacts. Normal operation: Close the switch, energize the coil, picks up relay, closes contacts, energize motor. With open ground, close switch, put 120V on coil, no current through open ground, but frame now @ 120V. You touch frame, bad day at the workshop.

If the coil can pass 128 mA, then you most certainly *can* be dead. That is enough current to contract the muscles in your had (can't let go) and if passing from arm to feet, constrict your chest (can't breath). And heart can go into ventricular fibrillation. Granted, several things would have to 'align' to give you all this, but that's why its a code violation.

A series resistor to drop the voltage from 240 to 120 could work. The data sheet that Alan posted a link to earlier seems to show for 220V operation, 'use the 120V relay with a series resistor'

Easy, no violation, no arguments.

daestrom

Two things:

1) I know that impedance and resistance are the same for a resistor, but am concerned it is not the same for the coil. It has been a very long time since my EE courses, so I am a little fuzzy on this. I suppose though that for 60hz, it doesn't matter what the voltage or current are, the impedance for the coil is constant, so a resistor that matches it in one circumstance will match it in all (for a constant hz)?

2) If the coil draws 128ma at 120, it has an impedance of 1000ohms. No? So why are people telling me to use a 3000ohm resistor? That would cut the current at 240v down to 60ma, and might not be enough for the coil to work properly.

John, DO NOT WIRE THE RELAY COIL TO GROUND!!!!!!

The easiest thing to do is to wire a lamp socket in series with the relay coil and screw in a small lamp bulb. Start with a 30 or 40 watt bulb and work your way down ;to a minimum of 15 watts. Besides reducing the voltage to the relay coil this gives you a visual indication that the saw has power.

I'm still hazey about the on/off switch is wired to the relay, usually this type

of circuit uses two push buttons, one normally open and one normally closed.

ARM

I guess I am being obstinate for a few reasons

1) As I read the spec sheet, you can put a resistor in series with a DC coil, not an AC coil. I don't know if they have a good reason for that, or just want to sell 240v coils. 2) If I were to use a resistor, it seems to me that it should be a 1k to double the impedance, not a 3k. The coil draws 128ma at 120v, so it's impedance is 1k, no? So, when people advise me to use a 3k resistor, I have to be skeptical. No one has explained why they say 3k, so I could be misunderstanding something here. 3) While I am not saying it is a desirable practice to send current down a ground, my oven puts 7a down the ground (as do millions of other electric ovens) and it meets code and seems reasonably safe; so it just doesn't seem that terrifying.

Perhaps a resistor is a better way to go, and if it is, I will do that. Before I had a wiring diagram for the starter (which is what I tried to get here) an industrial electrician told me that in general, going to ground was much better solution than a resistor. I will have to clarify with him why he said that.

I appreciate your help though.

Well... not quite. The START switch is momentary contact. After the momentary close, current goes to the coil through one of the relay contacts. So, if the ground opens, the coil loses power, the relay opens, and there is no voltage to the coil or frame. I thought all START switchs worked like that; maybe not, but I know mine does.

I said "almost impossible". Ever get a shock? I imagine everyone who works with electricity has had plenty of them. Every go into VF? I haven't, nor have I spoken to anyone who has. (Now of course I am talking about 120v; all the reported electrocutions I have ever hear about have been from 4000v or more.) The figures I have seen is that it takes 3 seconds at 100ma for VF, and that normal shock from a 120v hot is about 10ma because a person's resistance is pretty high. 3 seconds is a pretty long time to be in contact with a table saw while getting a shock, even if it were 128ma. Only this is not a normal shock from a 120v; this is a shock after the coil has caused a voltage drop, so it should be even less. Even if I were in salt water, I couldn't get a shock of more than maybe 60ma because of the coil. And this completely ignores the alternate ground directly through the frame to ground, which is certainly better than through me; for if I am in salt water, the frame has to be also. So, I probably wouldn't get any shock at all.

Bear in mind that the same things have to 'align' for this to be dangerous as have to 'align' for my oven or dryer (which quite legally send current to ground) to be dangerous, and it just doesn't happen. (The oven and dryer are actually much more dangerous, as you are likely to have damp hands when using them, but there still are no deaths from millions of appliances. I have never used my saw with damp hands, nor have I ever seen anyone else do so.)

Would you like me to ground the saw frame to the water pipe it is next to? A 128ma ground loop is not a big deal. (now attack me for that statement).

Actually I was just looking at lights today! I figure I need a 15w, and they have candellabra bases, which I couldn't find cheap sockets for. Any suggestions?

That is correct. The ground connection breaking would have the same effect as pushing the STOP switch; cutting power to the coil.

The specs for this relay says that the 120 vac coil has a dc resistance R of 120 ohms. Since it passes 128 mA at 120 vac, the impedance Z is approx 1Kohm (937). The inductance portion (reactance) can be calculated. Think of a right triange with the hypotanuse as Z, the horisontal leg as R and the rectanace on the vertical leg. I get 930 ohms for the reactance portion. This is the same as long as the frequency is constant. So you need to find out how much resistance to add to R to double Z to use it on

240 vac. I come up with about 1508 ohms. Probably 1600 would work just fine. However, it needs to dissipate the I squared R losses.

The chart for this relay says to use a 2.7K to 3.3K resistor on a

110 VDC coil to use it on 220 VDC. This does not apply in your case.

hth

I want to thank you both for the most thoughtful answers in this whole darn thread. I guess it is the in rush that is concerning me; I am afraid they had a good reason for suggesting resistors with DC but not with AC.

It doesn't say you cannot use a resistor on AC coils. They don't have a relay speced at 220 V DC, so they tell you how you can do it at that voltage with their 120 V DC relay.

Agreed.

Agreed, with some comments. At one time, code allowed the connection of neutral and ground within the range. That is no longer allowed. In any event, if your oven is putting

7A to ground, it was not wired to code at the time it was installed, or it is defective. It shouldn't even be putting 7 A on neutral. The neutral to an oven serves controls and lights, which don't draw that kind of current. Code never allowed even the low current that lights or controls draw to be placed on ground. What it did allow was a circuit wired without an equipment grounding conductor for some specific appliances - ranges, ovens and dryers - and appliance manufacturers connected the neutral to the frame of the appliance inside that appliance. I agree with it not seeming that terrifying - but that's one reason we have the code. What you or I don't think of as terrifying can nevertheless kill us. The experts establish the code as a practical safeguard. The code doesn't allow intentionally putting current on the ground conductor. We may deal with installing 200 ampere services, wiring things hot of necessity while doing that or some similar situation. So we could sneer at 128 mA - but medical experts tell me it can kill me, and GFCI's are set to trip at ~5mA. So I figure I better follow the code. Besides, the last thing I want is a shock when I'm feeding some wood through the saw. My name's not Lefty, and I want to keep it that way.

statement).

John

You are completely misinformed about electrical safety & electrical hazards. Many electrocutions occur at 120 volts (& 208, 240, 277, 480). There are no rules such as you described. The amount of current and the time required depend on many variables.

Do not connect the relay to the machine frame. The hazard willl exist if the ground connection becomes a high resistance. It does not need to open completely. Others have explained this to you in detail.

The other ideas, such as series resistors will work BUT you need to mount them, space them from wiring, and insulate or guard them appropriately. Unless you are educated in designing machine controls, I do not recommend this.

I suggest that you buy a 240 volt relay or pull a neutral wire, and be done with it.

Ben Miller

Right. The DC resistance of the coil is 120 ohms - but I did not use that figure. Instead, I used E = IZ to figure out what Z is. (It actually does not matter what Z is, you can figure out the size resistor needed without knowing Z. See below)

Ummm - I got a little lost here. I didn't mention inrush or impedance energized vs de-energized in my other post, so I don't understand what you have in mind.

Try it this way, it's simpler and you don't need to know the value of Z (coil impedance) in the original: The unknown Z draws 128 mA on a 120 volt circuit. We first double, then halve the voltage with this circuit: hot---Z----Resistor----hot

Therefore, the resistor has to drop 120 volts, while the relay draws 128 mA. If we stuff a 1K resistor in there, it will drop 128 volts when .128 amps is drawn by the unknown Z. That's close enough.

In my first post, I recommended putting a voltmeter across the coil on the new circuit to see if the voltage was around 120, because 128 mA is a nominal figure. We don't know what the current draw really is, so some changes might be needed.

Everything you say is reasonable, except for one fact. My GE stove does feed 7a through the neutral/ground. When I first discovered it I thought the stove was broken, but GE sent me a wiring diagram and that is how it is designed. I don't like it, because you are quite likely to have wet hands when using the stove, but I am sure GE made lots of them. Just for fun I tested my 1 year old dryer. There was 4.1a in the ground. I haven't looked at the wiring diagram, but I expect the motor is 120v.

The saw is next to a water pipe that shows low reistance to true ground. If I temporarily went the ground route, would it be reasonable to ground the frame to the pipe? I know ground loops are discouraged, but it is only 64ma at most.

You can ask the question 9 thousand different ways, the answer is still the same. Here's what you've been told, copied from replies to you in the thread.

"I would NOT wire the relay coil circuit from one hot leg to ground"

"Don't wire it to ground."

"DO NOT use the ground wire under A NY circumstances"

"If the coil can pass 128 mA, then you most certainly *can* be dead."

"Do not connect the relay to the machine frame."

"Current should never be on ground intentionally."

"DO NOT WIRE THE RELAY COIL TO GROUND!!!!!!"

I'm sorry you don't get it. I don't know how to make it any clearer.

I thought about the 'momentary' type of setup after I wrote before. But there is *still* a problem with an open ground when you use the ground for the coil. Say the unit is running when the ground opens. As you say, the coil drops out and the unit stops. So far, so good. What's probably the

*first* thing you will try?? Push the start button again. Puts 120V 'hot' on the coil, through coil, to frame. If you're holding frame.....

There are 'maintaining' style that use light duty on/off switch to control relay that in turn controls power circuits. Interlock/safety devices just wired in series with on/off switch so any one can open circuit to control relay and shutdown unit. Seen 'em both. As this was not a large, complex machine, I guessed it could go either way.

Oh woe unto thee....

There are probably just as many, if not more electrocutions caused by 120V than any other. The wide spread use by lay persons is a major factor. People that work on 4160 and up almost *always* receive proper training in such things (of course, accidents can happen anywhere).

What you say about the coil being in series with your body is true. But remember, blood in the human body has about the same salinity as seawater. The only part of your body that has a high resistance is your skin. Cover that with some honest sweat and you'd be surprized how low the total resistance can get. One old study showed a hot sweaty sailer working in an engine room can have as little as 300 ohms from one hand to the other (right across the chest). The Navy had some old training film, 'The Deadly Shipmate' that demonstrated how *many* on board electrocutions where caused by 'good ole 120VAC'

Do you *really* want to bet your life on this? Rubber feet on the legs? Plastic castors to make it roll around easy? You 'just happen' to be holding another power tool that has a metal case and a good ground?? So many possibilities.....

I may be over stating things, (I do get teased about being 'Mr. Safety' sometimes), but I just wouldn't want anyone to get hurt. You do what you feel is right. I just wanted to give you some 'food for thought'.

As far as 128ma ground loop.... No, in this sort of application it isn't a big deal by itself. If it were in process controls or networking, that would be different. I was only concerned that you may not understand some of the safety implications of this code issue.

daestrom

I find that 'incredible'. The motor probably *is* 120V. In that case, the motor current should be from one 'hot' to the *neutral*, not the equipment grounding conductor. If you see 4.1A through the green, equipment grounding conductor, then you wired *that* up wrong as well.

A typical electric dryer will have *four* wires, two 'hot' from a two-pole supply breaker, a neutral for carrying the motor, timer, interior lamp currents, and an equipment-grounding-conductor (often green) for bonding the frame back to the service panel. Sounds like you only have three wires and are using the 'middle' as both 'neutral' and equipment-grounding-conductor. I don't think code has allowed this for many years now.

When you got the new dryer, you probably used the existing outlet. Did it have a three-prong pigtail or four? For proper grounding, it should have been a four-prong one.

Either that, or you're not measuring the current in the equipment-grounding-conductor correctly.

daestrom

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