Line voltage monitoring - preventing induced voltage for fault detection

Thread Starter

cdebackere

Joined Nov 2, 2022
10
I have an eletrical cabinet with five 230 VAC signals to monitor (after 5 contactors/differentials). They are all the same phase of a 3 phase system.

I take the N of that phase at the input of the cabinet, and the five L throughout the cabine.

I connected the signals to a 230VAC analog GIRA input module on KNX.

Now when I switch of a contactor, the loss of voltage is not detected. I checked and the reason is : there is a 'residual' voltage of 100~130 VAC on that L. Probably induced by the many wires and common paths in the cabinet/installation. But I'm not sure about that.

The input module spec indeed states: 0-7V is low 7-250V is high. So it functions correctly.
Product page with datasheet link: https://katalog.gira.de/en_GB/datenblatt.html?id=658747&car=1

Question:
- how to get these 5 line voltages in, without induced voltage?
- or what is the best way to connect?
Should I put a coil around the wire before the module to prevent voltage to raise over 7VAC?
Or should I put an optocoupler or relais before as interface?

Note: 1) I would like to have minimal power consumption to read the signal (99,9% 'high')
2) the module also takes DC signals, so relais/opto output can be 12VDC

Thanks,
Christof
 

Thread Starter

cdebackere

Joined Nov 2, 2022
10
The problem is that when the 230VAC CB/contactor is switched off, then at the load side of the CB/contactor, there is still something like 100 to 130 VAC when I measure. The equipment itself doesn't do anything, since there is no power. But the residual (induced?) voltage of 100 to 130VAC results in the monitoring input to consider the input as still be 'high' or present.
So the required automation actions don't get triggered.
 
Last edited:

Ian0

Joined Aug 7, 2020
6,318
It just needs a load to dissipate the induced current - a resistor would do. Obviously the input resistance of the monitor isn't low enough.
 

Thread Starter

cdebackere

Joined Nov 2, 2022
10
Is there a way to calculate what resistance is needed? Or does that depend on the cabinet EMC/coupling? So by trial and error?

What should be the starting point? a current of 1mA? Hence R = 115V / 1mA ~ 100kOhm?
This would result in a constant power leak of 0,2mA nominal ~0.5W per signal. Or 2,5W for the 5 , and thus 7,5 EUR electricity cost per year.

There is no solution without power 'leak'? Or can I reduce to 0,01mA?

Cheers,
Christof
 

Ya’akov

Joined Jan 27, 2019
6,610
A low impedance DMM input is from about 2KΩ to 5KΩ. Remember the ghost voltage source, induction between conductors, can provide almost no current, so while a 5KΩ resistor actually connected across a 110V line would dissipate about 5W, the actual current available is tiny.

Empirical testing will let you determine the actual power use.
 

AlbertHall

Joined Jun 4, 2014
12,158
Maybe you could use a class X, 250VAC, capacitor across the line and neutral. This will draw current to reduce the stray pickup but will not dissipate any power, whether the contactoe is on or off.
 

Ya’akov

Joined Jan 27, 2019
6,610
But it will dissipate that 5W when the contactor is on.
Note: TS supply voltage is 230V.
That would be 10W, then. But there is no reason to leave the resistor connected when the contactor is on. Whatever is turning on the contractor can also energize a relay to switch the resistor in and out.
 

Ian0

Joined Aug 7, 2020
6,318
As we are dealing with AC, a snubber circuit will be equally as effective as a resistor, without generating so much heat.
A capacitor might do, but I’d be a little bit wary of adding any completely capacitive loads.
 

Thread Starter

cdebackere

Joined Nov 2, 2022
10
That would be 10W, then. But there is no reason to leave the resistor connected when the contactor is on. Whatever is turning on the contractor can also energize a relay to switch the resistor in and out.
Hi,

the reason to leave the resistor connected is that this is a monitoring 'circuit'/input from a power distribution board. So I cannot easily switch off the resistor. It needs to be passive.

Putting it all together it ssems like a class X1 capacitor is what I need. I have also read the article on the site here about X and Y capas, which was very helpfull.
Largest X1 capa I can easily order is 4,7nF. Such capa has an impedance of 700k.
Is this OK according to you guys?
 

Ian0

Joined Aug 7, 2020
6,318
Hi,
Ω
the reason to leave the resistor connected is that this is a monitoring 'circuit'/input from a power distribution board. So I cannot easily switch off the resistor. It needs to be passive.

Putting it all together it ssems like a class X1 capacitor is what I need. I have also read the article on the site here about X and Y capas, which was very helpfull.
Largest X1 capa I can easily order is 4,7nF. Such capa has an impedance of 700k.
Is this OK according to you guys?
Maybe.
4.7nF would be the largest Class-Y. You should be able to get class-X up to 1uF.
I‘d suggest 47nF in series with 47Ω.
 

Thread Starter

cdebackere

Joined Nov 2, 2022
10
OK, cristal.
I found indeed up to 50nF, but the three main suppliers is checked (digi key, RS, Farnell) only have up till 10nF available. 47nF is months waiting time.
For the R, what power rating is needed? For 220V I could calculate P=220^2/47 ~1kW .. but that can't be right :)
 

Ya’akov

Joined Jan 27, 2019
6,610
Hi,

the reason to leave the resistor connected is that this is a monitoring 'circuit'/input from a power distribution board. So I cannot easily switch off the resistor. It needs to be passive.

Putting it all together it ssems like a class X1 capacitor is what I need. I have also read the article on the site here about X and Y capas, which was very helpfull.
Largest X1 capa I can easily order is 4,7nF. Such capa has an impedance of 700k.
Is this OK according to you guys?
It isn't needed, but for the record the idea was to bypass the resistor when the contractor is energized. The sensor would be connected with the resistor when the contactor is off, and without when the contactor is is energized and the ghost voltages aren't a problem.

The snubber is a better idea for sure.
 

Ian0

Joined Aug 7, 2020
6,318
OK, cristal.
I found indeed up to 50nF, but the three main suppliers is checked (digi key, RS, Farnell) only have up till 10nF available. 47nF is months waiting time.

:)
https://uk.rs-online.com/web/p/film-capacitors/1262231
https://uk.farnell.com/wurth-elektronik/890324025009cs/cap-0-047-f-275vac-10-pp/dp/2450012
https://www.rapidonline.com/kemet-r46ki24700001m-47nf-radial-poly-capacitor-20-275v-x2-15mm-63-1136
https://www.digikey.co.uk/en/products/detail/würth-elektronik/890324025009CS/5038919
For the R, what power rating is needed? For 220V I could calculate P=220^2/47 ~1kW .. but that can't be right
Z = √(R^2)+1/(2πfC)^2))
I = V/Z
P = I^2 R
 
Last edited:

Thread Starter

cdebackere

Joined Nov 2, 2022
10
Thanks Ian, I indeed overlooked the impedance of the C which dramatically reduced I, hence P over the R.

Odered components, I'll confirm once installed
 

MrAl

Joined Jun 17, 2014
9,553
Thanks Ian, I indeed overlooked the impedance of the C which dramatically reduced I, hence P over the R.

Odered components, I'll confirm once installed
Are you talking about a 0.01uf cap in series with a 47 Ohm resistor?
Then yes the power in the resistor is almost completely moot.
Since the current will be so low i wonder if 0.01uf will be large enough.
 

Ian0

Joined Aug 7, 2020
6,318
Make completely sure it is off, and put your meter on AC Current and measure across the sense points. Then you will know how much leakage current there is. (If you measure when the power is switched on, you will need a new meter)
 
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