dear all,

I am new on the forum.

I would like to use my logo 6ED1052-1MD08-0BA0 to show on screen the fluid level that is measured by a probe with variable resistance from 33 ohm to 240 ohm. I also have the AM2 RTD module (for PT100/1000 readings) and the AM2 module for 0-10V/0-20mA/4-20mA readings. I do not request high level of precision

What do you suggest as the best connection? If I want to use the 24VDC of the PLC, I will have to use a resistance in serie of 350ohm so that I can read between 2 and 10V. The max current to the PLC will be 62mA. is it too much for the PLC? I will still consume 1.5W which is quite high for a permanent connection.

I could also use a small 5VDC to feed my probe with a 220ohm resistance in serie and read mA in the PLC. Then I will consume only 0.1W. Is this possible? If I go even further, I could use a AA rechargeble battery to feed my probe with 1.2V and with a 27ohm resistance in serie, my consumption will drop to 0.025W.

What do you suggest?

I hope my request makes sense.

thanks

Kito

Kikikito

I would like to use my logo 6ED1052-1MD08-0BA0 to show on screen the fluid level that is measured by a probe with variable resistance from 33 ohm to 240 ohm. I also have the AM2 RTD module (for PT100/1000 readings) and the AM2 module for 0-10V/0-20mA/4-20mA readings. I do not request high level of precision

What do you suggest as the best connection? If I want to use the 24VDC of the PLC, I will have to use a resistance in serie of 350ohm so that I can read between 2 and 10V. The max current to the PLC will be 62mA. is it too much for the PLC? I will still consume 1.5W which is quite high for a permanent connection.

 

Generally you can use this option, but i don't understand your calculation of the current to the PLC. The current will flow mainly through the external resistors. The analog inputs of the base modul have a impedance of ~ 70kOhm. If you have a maximum voltage of 24V, you have only a current of 0.34mA.

Let us have a look to the external resistors, the 350 Ohm resistor is connected to +24V, the 33-240 Ohm resistor is connected to 0V. The connection between the resistors are connected to the analog input (you can ignore the 70kOhm impedance for calculations).

Max current is 24/(350+33)=62.7mA, min. current is 24/(350+240)=40.7mA

P=I²xR 0,0627²x33=0.13W 0.0407²x240=0.4W => Umax=10V needed with 0.4W

Check the specifications of the variable resistor, if okay, then it works. Check the fixed resistor the same way for power consumption. Also think about that resistors have a temperature drift.

Regards, Scorp

Scorp

Let us have a look to the external resistors, the 350 Ohm resistor is connected to +24V, the 33-240 Ohm resistor is connected to 0V. The connection between the resistors are connected to the analog input (you can ignore the 70kOhm impedance for calculations).

Max current is 24/(350+33)=62.7mA, min. current is 24/(350+240)=40.7mA

P=I²xR 0,0627²x33=0.13W 0.0407²x240=0.4W => Umax=10V needed with 0.4W

Check the specifications of the variable resistor, if okay, then it works. Check the fixed resistor the same way for power consumption. Also think about that resistors have a temperature drift.

Regards, Scorp

 thanks Scorp for your message.

Good that your calculation is about the same as mine. The power you calculated is the one dissipated inside my probe. If I add the power dissipated in the added resistor of 350 Ohm, it is I²xR = 0.0627² x 350 = 1,37W. So in total, the power required is about 1.5W.

How would you connect all this to the Logo?

What do you think about using a 5VDC (instead of the 24V of the PLC) or even a 1.2V battery and to connect all that to the 4-20mA analog input?

best regards and stay safe to all

Kito

Kikikito

 Good that your calculation is about the same as mine. The power you calculated is the one dissipated inside my probe. If I add the power dissipated in the added resistor of 350 Ohm, it is I²xR = 0.0627² x 350 = 1,37W. So in total, the power required is about 1.5W.

How would you connect all this to the Logo?

What do you think about using a 5VDC (instead of the 24V of the PLC) or even a 1.2V battery and to connect all that to the 4-20mA analog input?

best regards and stay safe to all

Kito

I have already described, how to connect the resistors. From +24V down to 0V first the 350 Ohm, then the 33-240 Ohm as voltage divider, the point between the resistors e.g. to I7(AI1).

I see no advantage to use lower voltage except for the power consumption. If you have problems with the temperature of your probe resistor, then you can increase the 350 Ohm resistor as needed and loose the full resolution.

Which relation is between resistor value and fluid level, linear or anything else ? You will have problems with non linear relation or the non linear voltage divider. Maybe it is a good idea to create an excel sheet to see the relations. There are only base arithmetics in the Logo to determine the fluid level or you must work with complicated polynoms to get an approximation.

Regards, Scorp

thanks Scorp.

I see how to connect and how to reduce power by increasing resistor with consequence of losing full resolution.

I will close the topic

regards to you

Kito

it is better to use a special interface with its probe to measure a liquid level......

1. a dc current will result in electrical corrosion from the probe.

2. You also need to install a input protection from the PLC to avoid that a high voltage discharge is damaging the input.