3
\$\begingroup\$

How do I connect/use a 3 Wire Digital Scale Strain Gauge Weight Sensor?

I took it from an old weight meant for people. :)

The wires are red, black and white.

I found this, http://www.nerdkits.com/videos/weighscale/ ,but they use a 4 wire.

Thanks

\$\endgroup\$
2
\$\begingroup\$

Does it look like this sensor? If so, put the power (maybe try 3 to 5V for starters) across black and white; signal is on red. Even if you apply the maximum force (50kg), you can expect the signal to change by an incredibly small amount. Make sure your voltmeter is up to the task; the sensor sold by SparkFun will vary 1mV per 1V of input, when the max force is applied. Additionally, it'll handle up to 10V of input, so you'll want to get as close to that as you can.

\$\endgroup\$
  • \$\begingroup\$ This is a pretty old question (July 10') -- not that I haven't made the same mistake myself! ;) \$\endgroup\$ – tyblu Dec 24 '10 at 1:59
  • 1
    \$\begingroup\$ I knew how old it was, but Lars still hadn't accepted an answer, and the other two answers had guessed the wrong pin out (understandably), so it still seemed very much worthwhile to answer. \$\endgroup\$ – Jay Kominek Dec 24 '10 at 16:57
3
\$\begingroup\$

I'd bet that the wires are:

  1. Red: power
  2. Black: ground
  3. White: signal

But it's hard to guess what the "power" voltage is. If you knew the power voltage, you could connect that voltage across the red and black wires, and check to see what signal comes out of the white wire.

If you can't figure out a way to determine the proper voltage, you could use an adjustable power supply. Start with a low voltage (3 V is probably safe), and then slowly raise the voltage while looking for a signal with an oscilloscope. There's a decent chance you'll go too high and burn the sensor, but it may be your only option.

Are there any identifying marks on the sensor? Did it come from a battery powered scale?

\$\endgroup\$
  • \$\begingroup\$ Perhaps I was not clear, sorry, but it is not a chip, it's an object from two metals and a white rubber where the 3 wires come out from. The scale works on 6V and putting this on the red and black and measuring the white (with black) on an oscilloscope shows no change when pressure is applied. Red - Black -> resistance: 1k White - Black -> resistance: 2k \$\endgroup\$ – Lars Jul 7 '10 at 9:39
  • \$\begingroup\$ Pressure has no effect on the resistance, very strange. \$\endgroup\$ – Lars Jul 7 '10 at 9:40
  • \$\begingroup\$ Oh and there are 4 of these metal objects (with each 3 wires) on each corner that come together on a separate circuit and than connect with 4 pins to the main circuit board. \$\endgroup\$ – Lars Jul 7 '10 at 9:58
  • \$\begingroup\$ I suspect what you have is a resistive strain gauge. I know you said that pressure has no effect on the resistance, but maybe it's just relatively insensitive. If it's from a scale, you probably need a few pounds of pressure on the sensor to move the signal. \$\endgroup\$ – pingswept Jul 7 '10 at 12:36
  • \$\begingroup\$ Ah, here's another thought-- maybe it's a capacitive sensor. You'd feed it a square wave on one side of the capacitor and measure how much the corners roll off on the other side. That theory doesn't work so well with the relatively low resistances you measured earlier, but it sure would be a cheap way to make a sensor for a consumer-grade scale. \$\endgroup\$ – pingswept Jul 7 '10 at 12:38
1
\$\begingroup\$

If the scale had a battery in it, the voltage for that type of battery is probably what you want to put across the red(+) and black(-) leads. A digital scale that I opened up had one of those 2032 'coin' cells in it, which is 3V. The signal will come back on the white wire, referenced to ground (i.e., the black lead), but the protocol may be tricky to sort out.

\$\endgroup\$
1
\$\begingroup\$

I am working on the same 3-wire-strain gauge. I am using 2 sets of the sensors to get a better results because I was experiencing same problem with you. Try to reverse the power polarity if you still cant get any values.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.