Very simple 24Vdc loop powered 4mA and 20mA trip amplifier

Hi

I am currently testing 4-20mA Analogue output cards of PLC’s to an LED/Resistor circuit. As you can imagine, this is simple but can be sometimes difficult to determine by the brightness of the LED for 4mA and 20mA.

Therefore I am looking for a simple trip amp solution that when a PLC output is between 4 to 5mA a Red LED is on, and from 19 to 20mA a Green LED is on.

In between 5mA and 18mA then both LED’s should be off.

Any ideas?

Thanks
BR
Gray

hi,
A quad LM339 comparator would give you 4 ' threshold, window' levels, the LM339 output is o/c and it can sink ~6mA, enough to power the LED indicators.
E

esp1 said:

hi,
A quad LM339 comparator would give you 4 ' threshold, window' levels, the LM339 output is o/c and it can sink ~6mA, enough to power the LED indicators.
E

Click to expand...

Hi,

LM339 is a right choice.Before given to the opamp the current (4-20mA) should be converted into voltage .is it right?

hi,
Using a load resistor of 250 Ohms in the current loop, will give an output voltage of 1V thru 5V for 4 thru 20mA, thru the resistor.

Connect the high side of the 250R via a 10K , into each +Inp of the LM339.
Then construct a resistive divider network from a +5V supply to 0V, choose the resistors in the divider to give the 4 threshold voltages, then connect the individual junctions of the divider to the -Inps of the LM339, via 10K.

Use a 1K or 2K2 from +5V to each LM339 output.

Do you follow that OK.?
Eric

Thanks guys.. interesting solution and I take the comment of using the LM339 onboard however the circuit is looking more complicated than I would like with Vcc requirements of the comparator being 36Vdc (or +/ - 18Vdc) and so a simple solution could be a bulky box than I would like with PP9's.

There are 4-20mA Trip Amps on the market which are loop powered and small (DIN rail) so maybe I will end up buying a few of them and ripping the bits out to fit in a box

Thanks again, not unless there are any more ideas?
Gray

hi GD,
The LM339 will work OK with single supplies as low as +5V.
[just ensure that any voltage on a LM339 input pin is less than Vsupply -1.5V, in order avoid latch up]

It really depends upon the LM339 Vout switching levels you want.??
A single 9V PP3 battery could power the LM339.
E

The way I would attempt to solve this would be to get a 12 V zener with a 200 ohm resistor in series with it, I think in the -ve end. So now with a loop current of 4 mA, you get .8V dropped across the resistor and 1V when the loop current is 5 mA. So now return the emitters of a NPN to the -Ve and the base to the zener/load current resistor, via a 10K and fiddle about with the collector load (returned to the +ve end of the zener) until you get a nice big transition , say 2-> 10V. repeat the circuit with another transistor, but its emitter has to be biased to + .2V, so this transistor comes on at 5 mA. Put a potential divider across the current sensing resistor and use a third transistor, to provide the transition at 18mA.
So now you have three signals >4mA, > 5 mA and >18 mA. The >18 mA can drive its LED direct via a current limiting resistor. Now you have to gate the signals to turn off the >4mA which will light its LED, so feed the LED via an emitter follower from the collector load of the >5mA, so when the signal is >4mA, the collector of the >5mA is high so the LED gets its supply, when the signal is >5mA, its collector drops so the voltage feed to the LED goes off, and stay off as the input signal is increased beyond it up to 20 mA.
Frank