ISOLATING 4-20mA INPUT SIGNAL USING IL300 Linear Optocoupler

May 30, 2005 4 Replies

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alcuemar 21 years ago

Dear Friends,

I have read today that there is one possibility to isolate the 4-20mA input signal using the IL300 octocoupler, I have read the application note 54 from siemes but i have not too much clear the functioning of the design.

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My design require 8 analog inputs from 4-20mA, with the possibility of to detect the broken sensor or short-circuit sensor, and out of range sensor, so the option to do that without isolation is to put simple shunt resistor for 220 ohm, 4mA = 0.88V, and 20mA=4.4V, so we can detect the out of range easitly, the short-circuit will be 0V, ... and the broken sensor will be 0 too.

How to do the same using the IL300 linear octocoupler, in the application note 54 page 11, they equation to join the input and output is Vo/IL=(R3/R4)*R5*K3, what is the value for K3??.

I have some doubt about how to use this design.

If someone has make something similar I will appreciate the help.

Thanks in advance.

Alberto

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alcuemar 21 years ago

Thanks Dan for your feedback.

I will test this figure 19 schematic in a new prototype, just i have the doubt of how to calculate the Vo voltage depending the I input. I have see the equation:

Vo/IL=(R3/R4)*R5*K3

But I have not clear the value of K3.

Thanks in advance.

Alberto

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Rich Grise 21 years ago

The value of K3 will be determined by the implementation of your circuit.

In other words, let the math mumbo-jumbo go to the bit bucket, where all mumbo-jumbo belongs, and lash up the circuit in figure 19, page

11-204, and calibrate it by putting a known 4 mA through "line", and measure the voltage at "output", and put a known 20 mA through "line", and measure the voltage at "output".

In short, K3 = K2/K1.

Hope This Helps! Rich

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Spehro Pefhany 21 years ago

and (16), it

"The remaining variable is the IL300?s transfer gain, K3. The part to part variation of the transfer gain offers (sic) a range of 0.56 to

1.53"

K3 is a datasheet parameter. Nominally it's 1.0 (the ratio of the small currents you get out of the two photodiodes) but it could be anywhere between 0.56 and 1.65, according to the data sheet (0.56 and

1.53 sez the AN, but that's suspect).

Best regards, Spehro Pefhany

"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

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Dan Hollands 21 years ago

I think the circuit of figure 19 is what you want. It is self powered on the input meaning that up to 4 ma of current is used to power the input. The ma from 4 to 20 creates a 1 to 5v signal isolated from the input.

If the input is shorted, open or out of range the output will be outside the normal 1 to 5 volts.

Dan

Dan Hollands

1120 S Creek Dr Webster NY 14580 585-872-2606 snipped-for-privacy@USSailing.net

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