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I have a three wire RTD 100 probe and a BD-300 Isolating Signal Converter. I wired up the BD-300 like the datasheet said to, and the unit powers powers on and indicates that it has power (red LED is on). However I do not get a stable voltage from the output pins. I have my multimeter hooked up black to output pin 7 and red to output pin 8. The voltage that I get cycles from 0mv to about 400mv and then repeats.

If I read the center lead of the RTD (while hooked up and powered on) and an outside lead, I get about 110ohms which looking against this table tells me I'm in the ball park for the right temperature. So it seems like my probe is good to go.

The converter did come with a funny looking yellow resistor looking item with markings that I can't make out, bridging to the 4 and 6 input pins.

Any help with this or pointing me in the direction to do more reading would be greatly appreciated, my google searches have turned up nothing other than advertisements (I probably don't know the correct search terminology).

If it matters my end goal is to be able to read the temp value, accurate to 0.1 degree, and use it with a PID loop to control mash temperatures for homebrewing. So if there is an easier way to interface this probe, I'm all ears!

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  • \$\begingroup\$ Is the "funny looking yellow resistor looking item" still in place, or did you remove it? \$\endgroup\$ – HikeOnPast Jan 31 '13 at 1:18
  • \$\begingroup\$ I removed it, and left in while taking measurements. It seemed to make no difference. \$\endgroup\$ – Michael Dillon Jan 31 '13 at 12:16
  • \$\begingroup\$ The datasheet is about the worst I've ever seen for a product of this type. Can you confirm that you ordered a voltage output model and not a current output model? \$\endgroup\$ – HikeOnPast Jan 31 '13 at 17:07
  • \$\begingroup\$ Oh wonderful... I definitely got a current output model, 4-20ma. Didn't read the ebay post very well. \$\endgroup\$ – Michael Dillon Jan 31 '13 at 17:29
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If you were expecting a 1-5V output and you accidentally purchased a current output model isolator, simply connect a high-precision 250 ohm resistor across the output terminals (or even better, across the input terminals of whatever is connected to your signal isolator). Resistors of this type are commonly available for exactly this purpose.

Ohm's law:

V = I * R

V = [4mA - 20mA] * (250 ohms)

V = 1V - 5V

While it may seem like a pain, a 4-20mA signal is preferable to a voltage output in many applications:

  • It allows the detection of a broken sensor wire (it goes to 0mA) in addition to sensing the process variable (sensor output).
  • Some sensors can be 'loop powered' - they require less than 4mA to operate, and can therefore be used as a two-wire device without need for a separate power supply wire. The savings in wiring cost alone can be significant.
  • A current signal eliminates error due to resistive losses in the conductors between sensor and readout/ADC. (Though it trades this for errors related to the 250 ohm resistor)
  • Noise immunity is substantially better due to the inherent common mode rejection. This is a huge win in industrial settings when tend to be very electrically noisy.
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  • \$\begingroup\$ And then just read the voltage across the resistor? \$\endgroup\$ – Michael Dillon Jan 31 '13 at 19:36
  • \$\begingroup\$ Edited to add more detail. \$\endgroup\$ – HikeOnPast Jan 31 '13 at 19:53
  • \$\begingroup\$ Thanks so much @HikeOnPast! Great stuff! I'm no EE, but I should be able to get this going! (I'm a CS guy) \$\endgroup\$ – Michael Dillon Jan 31 '13 at 19:59
  • \$\begingroup\$ Anything for the betterment of beer. :) \$\endgroup\$ – HikeOnPast Jan 31 '13 at 22:59

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