| Electronics > Beginners |
| How to capture 4-20mA current loop from a sensor |
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| roogadget:
I have an industrial sensor (4-20mA) and want to capture this using my STM32 F1 ADC pin. Do I need a "receiver" to do this? Or can I simply use a 250 ohm resistor?? ( 4mA * 250 = 1V & 20mA * 250 = 5V ) (Precision 4mA to 20mA CURRENT LOOP RECEIVER) http://www.ti.com/lit/ds/symlink/rcv420.pdf |O The second image is an update of my setup (passive).. |
| Benta:
Is it an active or a passive 4...20 mA sensor? In the first case, you just need a resistor to ground from the output. In the second case, you need to supply a voltage to one end of the sensor (normally 12...24 V). The other end of the sensor goes to the ADC input, with a resistor to ground. 250 ohms is a good value in both cases. |
| Zero999:
Yes, a resistor in series with the 0V side is all that's needed. It's possible to put it on the positive side but it will complicate matters, unless the 0V rail on the mictrocontroller circuit is isolated from the 12V to 24V rail. A 250R resistor would be dissipating 0.022*250 = 0.1W, which is too much for a 0402 surface mount part. Although there are 0402 parts rated to 100mW, I wouldn't recommend it. Use 0805 or larger parts. It would be better to use two smaller parts in series, such as 120R and 130R and spread the power dissipation between them. |
| roogadget:
Danke sehr! I have updated my original post with a new image. Is this what you described?? |
| Zero999:
There's still the issue of the 0402 surface mount resistor. Did you just use the default symbol and are going to use a through hole resistor when you build the circuit? If so please make that clear, otherwise it's confusing. Attached is a circuit showing what I meant. The 250R resistor can be 120R and 130R or 100R and 150R in series, if you can't find 250R. |
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