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I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

(OP)
I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

This is to monitor the level of charge in some UPS batteries on site, my IQ3 can "see" between 0-10V DC but my batts will be between 0-24V DC

Any Thoughts

RE: I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

Your batteries will probably float at about 28V or so, depending a little on the charger, rather than 24V. I would use a 30V full scale value to give yourself a little margin. You could do this using a resistive divider chain but that option will depend on whether the battery is earthed or not and whether the analog input is earth-referenced. Or you could use an isolating transducer to convert levels and provide isolation, but that will be more expensive.

RE: I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

Disclosure: I'm a consultant for Kele, and was employed there for many years. This product is available from other folks (BAPI, ACI) too, so shop if you want.



It can be field calibrated, or you can make amends for any inaccuracy in your dropping resistor inside your IQ3 panel.

DT-13 isolator

See the attahed chicken-scratched diagram for a voltage-divider input considering ScottyUK's 0-30V suggestion.

Voltage Divider for 0-30V in = 0-10V out

Best to you,

Goober Dave

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RE: I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

(OP)
On that diagram, could I use a 1k and a 2k resistor instead of the 27 and the 13.3 ?

RE: I would like to be able to be able to convert a 0-24v dc signal into a 0-10v dc signal

The 13.3 kΩ resistor is the internal impedance of the product, it cannot be changed.

You can do the 1 kΩ and 2 kΩ divider, but be aware that your 1 kΩ must be put in parallel with that 13.3 kΩ internal impedance (across the input terminals). Effective will be 930Ω. That's not off by much, but you have to get two resistors instead of just one.

Best to you,

Goober Dave

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