Touch voltage

[studentspark]

I don't really understand touch voltage, I'm embarrassed to say...

When I ask people they usually say , stop over thinking it , the RCD will take care of it..

Im not to keen on leaving everything to an RCD, Ive found a few that don't work as expected..

I know the principle. And exposed conductive part can't reach a dangerous potential >50v

What I don't get is how the voltage can't rise above 50v.

Touch Voltage2

• studentspark
• 5 Feb 2020

230v is supplied to a class 1 switch.
There is a live earth fault. Chap No1 is touching it.
So the shock potential would be between his finger and his resistance to the floor.
The CPC will take the fault current around the EFLI path and trip the fuse within 0.1 - 5 seconds.
The Zs of 0.72Ω suggests the fault current will be 319A
Now if 319A appeared at the switch, even for 0.1 seconds, I would have though that would kill you.
That is a low impedance path though
Say his resistance to the floor is 1000Ω and the resistance back to the transformer is 1000Ω
so 2000Ω. 230/2000 = 0.11A.
But these resistances are never a constant, could be more or less, and 0.1A can still be dangerous. I also presume that when ADS was being designed, one of the calculations was not the possibility that you are wearing wellies.

Some other impedance must be involved.

So a 319A fault current is making its way around the loop, in the CPC, and for a brief moment, raising everything to a fault potential, until the OPD trips.

Chap 2 is touching the faulty switch and a switch with no fault.
He could be between 0v and 230v
But the CPC are commoned back at the met.
Now it this was in a special location we would want < 1.66Ω between them. If not, it needs supplementary bonding but 1.66Ω that quite a lot, but thats 50V


Chap 3 is touching another faulty switch, but also an extraneous CP with a resistance to the met of 1000Ω, but this is not bonded.
Chap 3 is in the trouble.?

So touch voltage must be limited. but how does it work.
You have 230 at the switch, but the metal work is limited to a rise above 50v

As in C- in the drawing (Should say Extraneous CP) a resistance of 23000Ω between you and 230v would give a fault current of 0.01A. That seems straight forward.

What resistance is in the way of the 230v to limit it to 50v

For a 6 amp circuit R <50/30=1.66Ω

So if two Exposed CP were measure to have 1.66Ω between them supplementary bonding is not needed. But that is with the voltage at 50v

So after all that waffle, simple question how is the 230 voltage restricted to 50v?

Oh and another bit....
That drawing on the right, if the fault current is only 0.23 amps, it won't trip the fuse, but 0.23 amps will be on all metal work, and you can feel that. No RCD

I hope that make some sense...

Thanks

 

[flameport]

Now if 319A appeared at the switch, even for 0.1 seconds, I would have though that would kill you.

That current flows through the cables, not the person.

Say his resistance to the floor is 1000Ω and the resistance back to the transformer is 1000Ω
so 2000Ω. 230/2000 = 0.11A.

Voltage at the fault location isn't 230V - the wiring has some resistance and if current is flowing through those wires, there will be significant voltage drop.
If the two wires were equal resistance, the voltage at the joint between them would be half of the supply voltage.

Earth resistance to a transformer will be far less than 1000 ohms.

Chap 2 is touching the faulty switch and a switch with no fault.
He could be between 0v and 230v

Not so, because:

But the CPC are commoned back at the met.

As they are connected, the voltage between them will be much smaller, depending on the resistance between those two parts. Smaller resistance = smaller voltage.

Chap 3 is touching another faulty switch, but also an extraneous CP with a resistance to the met of 1000Ω, but this is not bonded.
Chap 3 is in the trouble.?

Yes - because the extraneous part will remain at the same potential and the switch does not.
If equipotential bonding had been installed, that situation would have been avoided, both items remain at the same potential (voltage) because they are connected together.
Equi potential = equal voltage.

 

[JohnW2]

What I don't get is how the voltage can't rise above 50v.

It can rise above 50V ** - but what that diagram is saying is that if (with a 'just compliant' Zs - 1.66Ω in their example) it does rise to above 50V, then the OPD will operate 'immediately' and clear the fault.
[** for example, given that R2 is, with T+E, usually greater than R1, a L-CPC fault ('dead short') at the end of a length of T+E should result in a 'touch voltage' of more than half the supply voltage ]

The point is that if (in their example) Zs were greater than 1.66Ω, then the 'touch voltage' might have to rise above 50V before the OPD operated 'immediately'.

It's like the commonly-held misconception that an RCD 'limits' current through a person to 30mA. That is nonsense - the current through a person is 'what it is' (dependent upon Ohm's Law) and, equally, the touch voltage is 'what it is' (again, Ohm's Law). However, what the RCD or OPD do do is to limit the duration of those currents >30mA or 'touch voltages' greater than 50V

Kind Regards, John

 

[studentspark]

but what that diagram is saying is that if (with a 'just compliant' Zs - 1.66Ω in their example) it does rise to above 50V, then the OPD will operate 'immediately' and clear the fault.

Thanks so much for taking the time to explain.

What I don't understand, 'Is does rise to above 50v'. But its at 230v - is not like you have a dial and slowly turning up the voltage.
So its not about limiting the voltage its the time it takes. And to get a very quick trip , you need a very large current.

* for example, given that R2 is, with T+E, usually greater than R1, a L-CPC fault ('dead short') at the end of a length of T+E should result in a 'touch voltage' of more than half the supply voltage ]

Don't understand, I know the ratio CPC - R1 is 1.67 , so each has a different resistance. That relationship to the supply voltage , I don't get. Its Ohms law, but Im not sure what numbers are going in. (R<50/Ia) Plus you will have many parallel paths.

I obviously need to do some reading up on the science of current flow, and voltage

That current flows through the cables, not the person.

. I understand the very low impedance of the copper, v the higher impedance of the person touching a live metal part. would mean the current would prefer to go down the CPC
Some must go through the person it they were touching a live part. Otherwise no one would be getting shocks,

 

[JohnW2]

Thanks so much for taking the time to explain.

You're welcome.

What I don't understand, 'Is does rise to above 50v'. But its at 230v - is not like you have a dial and slowly turning up the voltage. So its not about limiting the voltage its the time it takes. And to get a very quick trip , you need a very large current.

Exactly. As I wrote ...

.... It's like the commonly-held misconception that an RCD 'limits' current through a person to 30mA. That is nonsense - the current through a person is 'what it is' (dependent upon Ohm's Law) and, equally, the touch voltage is 'what it is' (again, Ohm's Law). However, what the RCD or OPD do do is to limit the duration of those currents >30mA or 'touch voltages' greater than 50V

With a Type B MCB, it is required to trip magnetically (i.e.'immediately') with a current no greater than 5 x In (e.g. 160A for a B32) (although, in practice, it might trip magnetically at lower currents). With that example, one therefore has to ensure that a 'dead short' at the end of the circuit will result in a current of at least 160A, and that is achieved by ensuring that the Zs of the circuit is no greater than 230/160 (volts divided by amps = Ohm's Law).

Don't understand, I know the ratio CPC - R1 is 1.67 , so each has a different resistance. That relationship to the supply voltage , I don't get. Its Ohms law, but Im not sure what numbers are going in. (R<50/Ia).

If, for convenience, we take R1 as 1Ω, then, with that ratio, R2 is 1.67Ω. If there is a short at the end of the cable, then those two resistances are in series (hence a total resistance of 2.67Ω) across the 230V supply, resulting in a current of 230/2.67 flowing through them (Ohm's Law), i.e. about 86A. That means that the voltage across R1 will be about 86V (86A x 1Ω, Ohm's Law) and the voltage across R2 will be about 145V (86A x 1.67Ω). The potential at the point of the short (the 'touch voltage') will therefore be 145V above the neutral potential at CU. [Note that these {'easy for illustration'} R1+R2 would not be low enough for ADS with, say, a B32, since Zs would be 1.67Ω+Re ].

I hope that's fairly clear but, if you are still having difficulties getting your head around it, I'll try to produce a diagram or two for you.

Plus you will have many parallel paths.

Yes, and that obviously complicates the situation and calculations. If there are parallel paths to earth from the point of the short, then the overall resistance to earth will be less than R2, so that the touch voltage will be less than one gets (as above) by using R2. However, the principle remains, and one needs to design a circuit which remains safe even if there are no parallel paths (i.e. 'the worst case).

Kind Regards, John

 

[JM2]

You need to read (and understand) Kirchhoff's Laws - use DC and a 100v battery source by way of examples.
Then think of touch voltages as a node voltages (take the person out).

 

[Gasguru]

And why not Thevenins and Nortons theorem while you're at it...

 

[JohnW2]

And why not Thevenins and Nortons theorem while you're at it...

Why not?

In fact, the answer the OPs questions, one doesn't really need K's Law, let alone the others - Ohm's Law is essentially enough!

Kind Regards, John

 

[EFLImpudence]

It's not a very good diagram, is it?

Surely no.1 is not related to supplementary bonding and touch voltage. Is the floor wood?

The floor appears the same in all cases yet no specifications are given for it.


What am I missing in 'C'? What does it represent?

Not withstanding the error - 0.01A, and it's hardly Ib, is it?
Why is the exposed-c-p 23000Ω?

 

[JohnW2]

It's not a very good diagram, is it?

Taken alone (which is perhaps not fair) it certainly seems to be somewhat of a confused/confusing abomination (quite apart from the errors) - but there may well have been some accompanying text which made things much clearer.

Fortunately, however, in contrast with the diagram, the OP's questions and 'difficulties in understanding' are very clearly expressed - so fairly straightforward to address.

Kind Regards, John

 

[studentspark]

It's not a very good diagram, is it?

Yes sorry, This is a diagram I did my self, and there are a few mistakes in it. It helps me understand better, when I draw things out, but not anyone trying to decipher it...

the OP's questions and 'difficulties in understanding' are very clearly expressed

Im glad I've made one thing clear.


Another drawing. Apologies for the mistakes in advance

TV3

• studentspark
• 6 Feb 2020

two drawings of the same thing, one just simpler .

Its still the touch voltage thing...

This time the Zs is to high so the MCB will take 30 seconds to operate .

So in those 30 seconds what happens at the switch where the fault appears.
What voltage could it be at?

Voltage is the constant isn't it. we can affect the resistance and the current flow?
Or have I got the very basics wrong.

It can rise above 50V **

Limit the touch voltage to 50 v - Can we do that then.


Also What affect on the tripping times would the MPB and other conductors add

Thanks

 

[EFLImpudence]

Yes sorry, This is a diagram I did my self, and there are a few mistakes in it. It helps me understand better, when I draw things out, but not anyone trying to decipher it...

Oh ok. In that case it is good then.

I thought it came from a publication.

 

[JohnW2]

Limit the touch voltage to 50 v - Can we do that then.

The short answer, in practice, is 'No' (in the sort of situations you are considering), just as nothing can 'limit' the current through a person receiving a shock to 30mA - all one can do is limit the duration (of 'touch voltage' or current flow).

If there is a short circuit between L and CPC (or connected exposed-c-p) at the end of a length of cable, then the only way that the voltage at that point could be <50V relative to the MET (hence other CPCs) would be if the resistance of the path through CPC to MET (modified by any parallel paths) was less than about 28% (assuming 230V supply - I'll explain the calc if you wish) of the resistance of the L path (i.e. R1). As you've said, on the basis of just the cable (i.e. ignoring parallel paths) (other than for 1mm² T+E), R2 would be considerably higher than R1 - so no hope of it being only 28% of R1!

Also What affect on the tripping times would the MPB and other conductors add

If an extraneous-c-p provide an extremely low resistance to 'true earth', then main bonding will effectively reduce Ze, hence Zs, and therefore could potentially increase PFC and hence reduce tripping times - but I doubt that that would often be a significant effect in practice with TN installations.

With TT (when one would not normally expect an MCB to trip at all as a result of an L-E fault) main bonding could make a big difference - and (as I've often described) my house is a prize example of that (thanks to main bonding, I have Zs figures low enough for ADS, even though its a TT installation)!

Kind Regards, John

 

[studentspark]

You need to read (and understand) Kirchhoff's Laws -

It all adds up to O,
Im struggling with Ohms law , Kirchhoff's Laws that one takes some grasping.

And why not Thevenins and Nortons theorem while you're at it.

Some night time reading thanks.

 

[EFLImpudence]

This time the Zs is to high so the MCB will take 30 seconds to operate .

Ok. Not likely but go with that.

However, touch voltage refers to the voltage between two e-c-ps.
That's why I said Chap 1 is not relevant.
I'm a bit confused what is actually being asked and discussed.

So in those 30 seconds what happens at the switch where the fault appears.
What voltage could it be at?

The switch will be at 230V wrt earth.
If you touch that and earth, you will get 230V.

If you touch that e-c-p and another e-c-p it depends on the resistance between them
Negligible resistance because of bonding and they will both be at 230V wrt earth but 0V between them.
It will still depend on whether you are standing on an earth or not.

Limit the touch voltage to 50 v - Can we do that then.

Yes, bond them together V=I(p)xR.
20Ax0.01Ω=0.02V 20Ax0.1Ω=2V 20Ax1Ω=20V 20Ax2.5Ω=50V
390Ax0.01Ω=3.9V

Also What affect on the tripping times would the MPB and other conductors add

Well, it is supposed to be ≤0.4 seconds so your example does not really apply.