Wetting Current/PLC Input
- Thread starter fifty60
- Start date
- Status
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
159108-1717 EST
You have asked similar questions in the past. If you really need 100 mA if that is the real problem, then put a shunt resistor across the PLC input to provide a 100 mA load.
.
You have asked similar questions in the past. If you really need 100 mA if that is the real problem, then put a shunt resistor across the PLC input to provide a 100 mA load.
.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
That assumes that the PLC is not the power source for that input.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
150108-1726 EST
Then add an additional power source to provide the current,
.
Then add an additional power source to provide the current,
.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
You would need to provide an additional 93 mA of wetting current.
Is this a 24VDC input? If so, A 258 Ohm resistor in parallel with the input would work. It would drop 2.3 Watts though when the input was on. Maybe a better relay is your answer.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
159108-2409 EST
fifty60:
You need to provide a better description of your circuit problems.
First, define the specifications and limitations of the PLC input.
Second, why do you get hung up on contact wetting current?
What is the contact material, and why use it? Why are you stuck with a particular relay? In the past I have suggested reed relays to you. What about gold contact, or even plain silver. Never use silver-cadmium-oxide in any low voltage circuit, even 24 V. Potter and Brumfield have available both contact materials in the KUP relay. There are mercury wetted contact relays, no contact bounce. Also consider an SSR of the correct type.
.
fifty60:
You need to provide a better description of your circuit problems.
First, define the specifications and limitations of the PLC input.
Second, why do you get hung up on contact wetting current?
What is the contact material, and why use it? Why are you stuck with a particular relay? In the past I have suggested reed relays to you. What about gold contact, or even plain silver. Never use silver-cadmium-oxide in any low voltage circuit, even 24 V. Potter and Brumfield have available both contact materials in the KUP relay. There are mercury wetted contact relays, no contact bounce. Also consider an SSR of the correct type.
.
Gar, this is a relay inside of a control system mounted on a motor. To change the relay, I would have to change the motor. I could run their relay into another relay with a lower wetting current...but it adds cost...but may have to be done.
The PLC has a 24VDC input that is rated for 7mA...the wetting current for the relay in the control box is 100mA...I need the solution to not mess anything else up in the circuit...sounds like a 5W resistor in parallel with the input is the way to go..
The PLC has a 24VDC input that is rated for 7mA...the wetting current for the relay in the control box is 100mA...I need the solution to not mess anything else up in the circuit...sounds like a 5W resistor in parallel with the input is the way to go..
If you don't want the constant power draw on the 24VDC supply while the contact is closed, you may be able to get by with an RC snubber across the contacts. The capacitor charges upon opening of the contacts, then discharges upon closing of the contacts... which could easily supply the wetting current yet eliminate the constant power draw that a parallel resistor would require.
Exactly what type of PLC input are you using? Digital in? What is the max voltage rating for the input? Does it expect you to limit the current to 7mA with an external resistor, or is it a DC voltage input?
He's asking about the PLC input. If it's just on/off, that's considered discrete input... not digital, not analog.
Is it really rated 24Vdc or is that just what your supply is? Many PLC's will operate over a voltage range... some equally well on either a 24 or 48Vdc supply.
And the input itself cannot be a set of contacts.
Perhaps it would be best if you just provided make/model of PLC or the input card thereto, or better yet, a link to the data sheet.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
150109-1025 EST
fifty60:
Your various questions could be better understood and responded to if you provided a more complete description of your problem and limitations, and you need a more fundamental understanding of how the different components you use operate.
Your present problem appears to be:
1. You have some sort of motor with an internal relay that has a contact wetting rating of 100 mA, but no specification at what voltage. At 10,000 V I can probably breakdown any contamination layer on normal contacts.
Suppose you really need 100 mA of current at 24 V DC for reliable operation of your system, then you have to provide that as a source. Thus, you need a 24 V supply that is switched by your relay and a load resistor of 240 ohms. The next smaller nominal is 220 ohms. Power dissipation at 24 V and 220 ohms is 2.6 W. Use a 5 W resistor.
2. What are the specifications for the PLC input. We have no clue except that you mentioned 7 mA. Is it AC or DC? Not mentioned.
With the additional information you supplied it is probably a nominal 24 V DC, but possibly not and is only a TTL input with a 714 ohm pull-up resistor (highly unlikely).
Another is an optically coupled input requiring a minimum of 7 mA, but at what voltage?
Or possibly it is an optical coupler that internally has a 7 mA pullup, and when you short the PLC input terminals 7 mA flows externally.
And there are other possibilities. Provide information. Exactly what this PLC input is will determine how you interface.
Some experiments on relays I have sitting around.
A Fluke 27 in ohms position has an open circuit voltage across the terminals of about 0.75 V, and a short circuit current of about 0.46 mA.
Measuring a normally closed contact pair on a Potter & Brumfield KUP relay (20 to 30 years old) reads about 0.3 ohms on either a 5 A or 10 A relay (silver vs silver-cadmium-oxide contact material). This is essentially the reading of the two test probes shorted together. Thus, not an accurate contact resistance measurement, but it is low.
On a very old sensitive relay (10,000 ohms coil, pull-in 12.5 V, 16 mW), thus low contact pressure, the normally closed contacts read open with the Fluke 27. Probably silver contacts.
When using each of the following source voltages, and a 100k series resistor the contact voltage drop was about: .
10 V ---- 30 to 80 mV
1 V ----- 2 to 50 mV
0.1 V --- 1 to 15 mV
0.01 V -- 4 to 7 mV
You can calculate the current in each case. The environment can have a major affect on contact resistance.
.
fifty60:
Your various questions could be better understood and responded to if you provided a more complete description of your problem and limitations, and you need a more fundamental understanding of how the different components you use operate.
Your present problem appears to be:
1. You have some sort of motor with an internal relay that has a contact wetting rating of 100 mA, but no specification at what voltage. At 10,000 V I can probably breakdown any contamination layer on normal contacts.
Suppose you really need 100 mA of current at 24 V DC for reliable operation of your system, then you have to provide that as a source. Thus, you need a 24 V supply that is switched by your relay and a load resistor of 240 ohms. The next smaller nominal is 220 ohms. Power dissipation at 24 V and 220 ohms is 2.6 W. Use a 5 W resistor.
2. What are the specifications for the PLC input. We have no clue except that you mentioned 7 mA. Is it AC or DC? Not mentioned.
With the additional information you supplied it is probably a nominal 24 V DC, but possibly not and is only a TTL input with a 714 ohm pull-up resistor (highly unlikely).
Another is an optically coupled input requiring a minimum of 7 mA, but at what voltage?
Or possibly it is an optical coupler that internally has a 7 mA pullup, and when you short the PLC input terminals 7 mA flows externally.
And there are other possibilities. Provide information. Exactly what this PLC input is will determine how you interface.
Some experiments on relays I have sitting around.
A Fluke 27 in ohms position has an open circuit voltage across the terminals of about 0.75 V, and a short circuit current of about 0.46 mA.
Measuring a normally closed contact pair on a Potter & Brumfield KUP relay (20 to 30 years old) reads about 0.3 ohms on either a 5 A or 10 A relay (silver vs silver-cadmium-oxide contact material). This is essentially the reading of the two test probes shorted together. Thus, not an accurate contact resistance measurement, but it is low.
On a very old sensitive relay (10,000 ohms coil, pull-in 12.5 V, 16 mW), thus low contact pressure, the normally closed contacts read open with the Fluke 27. Probably silver contacts.
When using each of the following source voltages, and a 100k series resistor the contact voltage drop was about: .
10 V ---- 30 to 80 mV
1 V ----- 2 to 50 mV
0.1 V --- 1 to 15 mV
0.01 V -- 4 to 7 mV
You can calculate the current in each case. The environment can have a major affect on contact resistance.
.
Gar, the PLC is an Idec
| FC5A-D16RS1 here is a link to their website http://catalog.idec.com/australiaf/datasheet.asp?PN=FC5A-D16RS1&FAM=PLCs&P=2955,2980 |
Looking at the datasheet for that model I'm getting...
Power Supply 24Vdc nominal
(allowable 20.4 to 26.4Vdc including ripple)
Maximum input current 700mA (26.4Vdc)
Input impedance calculates to 37.7ohms @ 26.4Vdc
Current interpolated to 24Vdc is 636mA.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
So where did the 7ma figure come from? The minimum current needed to activate the input?
Sounds like the 100ma wetting current will not be s problem, but a series resistor would be s good idea to keep power dissipation down.
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
I suspect this is the power required to run the PLC itself, not the input current on an individual digital input.
Digital inputs rarely require much current when they turn on. Despite what the specs are on minimum wetting current, most contacts will work fine at lower levels than what the specs might indicate.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
150109-1347 EST
fifty60:
You make it very difficult to communicate with you. And very time consuming for us to get information.
Instead of pointing us directly to where the information is located you point to a very general starting website.
You should have pointed to http://www.idec.com/language/english/brochure/PLC_Brochure-au.pdf.pdf page 33. Then by deduction it appears that one of the circuits shown with a 3.3 k resistor is the likely input. At 24 V and 3.3 k the input current is 7.27 mA. But there is voltage drop in at least the optical coupler of at least 1 V. A voltage of 23 divided by 3.3 is 6.96 mA.
Now we probably know what your PLC input looks like. Pretty much you need to supply 24 V DC to the input. If the input includes a bridge rectifier, or a reversed biased shunt diode, then AC is OK as an input.
Now if you are hung up on 100 mA, then shunt the PLC input with 220 ohms or less. I really doubt that 100 mA is absolutely necessary.
.
fifty60:
You make it very difficult to communicate with you. And very time consuming for us to get information.
Instead of pointing us directly to where the information is located you point to a very general starting website.
You should have pointed to http://www.idec.com/language/english/brochure/PLC_Brochure-au.pdf.pdf page 33. Then by deduction it appears that one of the circuits shown with a 3.3 k resistor is the likely input. At 24 V and 3.3 k the input current is 7.27 mA. But there is voltage drop in at least the optical coupler of at least 1 V. A voltage of 23 divided by 3.3 is 6.96 mA.
Now we probably know what your PLC input looks like. Pretty much you need to supply 24 V DC to the input. If the input includes a bridge rectifier, or a reversed biased shunt diode, then AC is OK as an input.
Now if you are hung up on 100 mA, then shunt the PLC input with 220 ohms or less. I really doubt that 100 mA is absolutely necessary.
.
I believe you are correct. I knew 700mA was quite high for a typical input... but I've seen weirder things. Table row should have been labeled "Maximum Current" rather than "Maximum Input Current". I didn't see any [other] spec relating to I/O input. There's some spec's for analog voltage input... but going by the drawings, that's not what we're looking for.
- Status