I know this is going to sound a basic to some of you Wizards but....
I am building a large hydraulic press. The central ram had a 4-20ms stroke position sensor built in and I have connected that to a 24v dc Siemens Logo8 PLC. As the ram moved I can display its position on a screen.That was the easy bit.
Next I want to set a digital limit for the ram travel using a 10 turn, 200 K potentiometer.
The PLC has both 0-10v and 4-20ma inputs available, either could be used to provide a suitable human adjustable input.
However my first attempt at building a simple voltage divider didn't go well, the resistor changed colour , the wires melted and all the smoke leaked out of the circuit. I have trawled through all the topics on this forum, but can't find quite the right answer.
Any help and a simple diagram would be much appreciated.
Thank you

 

Apparently you used too small resistor values which generated dissipation higher than there rating.
Power = V squared divided by the Resistance.

So what voltage were you connecting the voltage divider too?
It's not clear what voltage you are trying to reduce.
What will you do with this adjustable output?
Please try to explain clearly exactly what you want to do.

 

A 200K potentiometer connected end to end across 24 VDC has but .003 watts of power in it, hardly enough to do any damage.

What could do damage is connecting it with the center tap shorted to one end, or driving a low impedance load. I believe the 4-20 ma inout is such a load, though the 0-10V input should not be such.

So... Exactly how and where are you putting this potentiometer?

And welcome to the forums!

 

If the pot were wired incorrectly and adjusted all the way to either end it could appear to the system as a short and cause short-type things to happen. Connection diagram / photo of the installation?

ak

 

Apparently you used too small resistor values which generated dissipation higher than there rating.
Power = V squared divided by the Resistance.

So what voltage were you connecting the voltage divider too?
It's not clear what voltage you are trying to reduce.
What will you do with this adjustable output?
Please try to explain clearly exactly what you want to do.

The PLC uses a 24v power supply, that's were I am drawing a power feed from.
I want to take that 24v supply, pass it through the ten turn potentiometer reducing it down to either a 4-20ma or 0-10v signal.
Then feed that adjustable output into one of the PLC's input channels. then I can take that signal and program the PLC to use an analogue comparator function block to compare the actual position of my hydraulic ram relative to the soft limit I have set on the potentiometer.

I don't mind if that input signal is 4-20ma or 0-10v in theory either should work just fine.
Likewise I don't mind changing the pot for something other than 200 Ohm. I picked 200 Ohm as I understood it would give a reasonable fine level of adjustment, but I you any of you guys recommend something else then I will change it.

 

If the pot were wired incorrectly and adjusted all the way to either end it could appear to the system as a short and cause short-type things to happen. Connection diagram / photo of the installation?

ak

Yes the pot was adjusted all the way to the end when all the smoke escaped.
I haven't posted a diagram as clearly I my design was flawed and it would be safer for everyone to not put it in the public domain.
I was rather hoping one of you wizards would be able to post up a better one that actually works.

 

I picked 200 Ohm

In post #1 you said 200K. Which is it? It makes a big difference to how the pot could be used.

 

A 200K potentiometer connected end to end across 24 VDC has but .003 watts of power in it, hardly enough to do any damage.

What could do damage is connecting it with the center tap shorted to one end, or driving a low impedance load. I believe the 4-20 ma inout is such a load, though the 0-10V input should not be such.

So... Exactly how and where are you putting this potentiometer?

And welcome to the forums!

The pot is going to draw power from the 24v PLC power supply, reduce it to an adjustable 0-10v signal and feed that into one of the PLC's inputs so I can compare it to the actual position of the ram.

How can I stop the potentiometer from shoring out when its turned to one end ?

 

Until you post a schematic we can only guess.

Bob

 

There are a number of ways things could've gone wrong and burned your parts up. Without more information from you, we can only guess. If it was a 200 ohm pot instead of a 200k pot, that would explain a lot! The "k" stands for kilo, in this case kiloohm. So, a 200k pot would present 1000 times as much resistance as a 200 ohm pot.

Anyway, assuming it's wired up properly and fed into a high impedance input (which the 0-10V input should be, but the 4-20mA input probably isn't,) I think the following pot/resistor arrangement should meet your needs, converting 24V into a 0-10V signal. Technically, it appears to top out at 10.04V, but I'm guessing that's close enough, and it was the best I could do with standard values and low part counts.

In case it's not clear in the drawing, the two resistors in the middle are meant to simulate a standard 10k trim pot. Connect a 6.8k and a 150ohm resistor in series between the 24V supply and one side of the pot. Connect the other side of the pot to ground. Connect a 10k resistor in parallel with the "lower" side of the pot, joining the wiper and ground. The output on the wiper will sweep from 0 to just over 10V.

 

voltage dividers schematic and eq. are faound on many places by trying google

diag: Vin > R1 * Vout * R2 > Vin_GND

``````````````````````.---------------->
``````````````````````|
(+) --- Vin-R1-Vout-R2
`````````````````````|
(-) -----------------.------------->
GND

eq:
Vout == Vin (R2/(R1+R2))
Vin == I (R1+R2)
Vout == I R2
Vout == Vin (R2/(R1+R2))
I == Vin/(R1+R2)
Vout == Vin
H == Vout/Vin == R2/(R1+R2) == V div. ratio

i dislike these "web editing" boxes that feed webadmins with what you are typing before you send it, and as well, tamper with "deleting and adding spaces" (ruined my ascii art dang it)

 

^^ using more resistors "in a tree" wont help (unless they are providing R1 or R2), you only need two

or the variable output of a transistor can lower voltage (not so reliably or maybe to so many magaohms)

there are other kinds of electronic protection, such as spike protection, current protection, voltage regulation

all depends on if your ok with blowing parts or need to keep it clean

 

There are a number of ways things could've gone wrong and burned your parts up. Without more information from you, we can only guess. If it was a 200 ohm pot instead of a 200k pot, that would explain a lot! The "k" stands for kilo, in this case kiloohm. So, a 200k pot would present 1000 times as much resistance as a 200 ohm pot.

Anyway, assuming it's wired up properly and fed into a high impedance input (which the 0-10V input should be, but the 4-20mA input probably isn't,) I think the following pot/resistor arrangement should meet your needs, converting 24V into a 0-10V signal. Technically, it appears to top out at 10.04V, but I'm guessing that's close enough, and it was the best I could do with standard values and low part counts.
View attachment 100626
In case it's not clear in the drawing, the two resistors in the middle are meant to simulate a standard 10k trim pot. Connect a 6.8k and a 150ohm resistor in series between the 24V supply and one side of the pot. Connect the other side of the pot to ground. Connect a 10k resistor in parallel with the "lower" side of the pot, joining the wiper and ground. The output on the wiper will sweep from 0 to just over 10V.

 

Dear edeowulf17

Thanks for the elegantly simple solution.
I will order some new bit today and let you know the result next week.

Thank you very much

 

Dear edeowulf17

Thanks for the elegantly simple solution.
I will order some new bit today and let you know the result next week.

Thank you very much

No problem. Happy to help! Definitely do let us know how it works out.

Hopefully I've properly understood your situation and needs, and this should do exactly what you're looking for. If not, my apologies and we can take another crack at it.

As a side note, the two series resistors could of course just be one 6.95k resistor. I drew it up as two separate resistors because non-standard values can be rare or expensive depending on required mounting type, size, etc. If you happen to find a reasonably priced single resistor to substitute there, feel free to do so.

 

voltage dividers schematic and eq. are faound on many places by trying google

diag: Vin > R1 * Vout * R2 > Vin_GND

``````````````````````.---------------->
``````````````````````|
(+) --- Vin-R1-Vout-R2
`````````````````````|
(-) -----------------.------------->
GND

eq:
Vout == Vin (R2/(R1+R2))
Vin == I (R1+R2)
Vout == I R2
Vout == Vin (R2/(R1+R2))
I == Vin/(R1+R2)
Vout == Vin
H == Vout/Vin == R2/(R1+R2) == V div. ratio

i dislike these "web editing" boxes that feed webadmins with what you are typing before you send it, and as well, tamper with "deleting and adding spaces" (ruined my ascii art dang it)

I'm not certain, but I believe if you use the code tags ("CODE" and "/CODE" in square brackets) to block off your ascii art, it will be displayed correctly.

 

No problem. Happy to help! Definitely do let us know how it works out.

Hopefully I've properly understood your situation and needs, and this should do exactly what you're looking for. If not, my apologies and we can take another crack at it.

As a side note, the two series resistors could of course just be one 6.95k resistor. I drew it up as two separate resistors because non-standard values can be rare or expensive depending on required mounting type, size, etc. If you happen to find a reasonably priced single resistor to substitute there, feel free to do so.

Hi ebeowulf17,

Parts arrived today, your circuit works perfectly.
Thank you very much indeed.

Paul
Pump Parts Ltd

 

Hi ebeowulf17,

Parts arrived today, your circuit works perfectly.
Thank you very much indeed.

Paul
Pump Parts Ltd

Excellent! Glad to hear it. Congrats on getting your system working!

 

Just wanted to let you all know this circuit is still working hard every day, without any issues.

 

Sweet! Glad to hear it.

Yeah, there's not much to go wrong there. Eventually the wiper on the trim pot will probably get glitchy - they all wear out sooner or later due to mechanical wear and/or moisture.

The external resistors should ensure that there's no way for a pot failure to let more than ~10V hit your input, but you could get erratic behavior if a worn pot makes the signal start jumping around randomly.