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DaveX

Member Since: November 10, 2015

Country: United States

  • For better sensitivity, you can connect four of the three-terminal half-bridge cells in a ring by the excitation terminals, matching the colors, making a full wheatstone bridge consisting of two strain gauges per leg. You then excite with two of the sense terminals from opposite sides of the bridge, and sense on the two intervening sense terminals. See http://electronics.stackexchange.com/a/199470/30711 for a schematic.

  • You don't need the combinator. The load sensors are half-bridges and you'd hook the center taps directly to the S+/S- on the HX711. The end taps you would hook to the E+/E-, and depending on whether you want to add or difference the loads on the two sensors, you'd hook the end terminals up with opposite colors or matching colors. Opposite colors (black-to-white) will add loads, since increasing loads will pull one sensor(half of the wheatstone bridge) down while increasing load on the other sensor will pull that side up. If you match end-terminal colors with two 3-wire half bridges, then similar loads on each will move the sensor in the same direction.

    The combinator helps wire the 8 resistances of 4 half bridges into a full wheatstone bridge with 2 resistances in each arm, trying to make sure loads on each half bridge add up constructively to to unbalance the bridge.

  • No, the center tap should not go to 'W'. To turn the 8 resistances from the 4 three-wire sensors into 4 2-resistance legs of a wheatstone bridge per the combinator schematic the center taps should be on the 'R' terminals.

  • The three-wire sensor is a half-bridge--essentially two of the variable resistances in a wheatstone. You could make the other half of the full wheatstone bridge with another half-bridge sensor, or with a variable resistor with its center tap set to match the voltage at the sensor's center tap.

  • It is like the diagram in this stack exchange answer -- essentially, the eight resistances in the 4 half-bridge load cells become two resistances per arm of the wheatstone bridge. The combinator tries to hook the center taps of the cells to the E+/E-/S+/S- of the HX711, and wires up the end-taps so that the 4 tension resistances are on opposite diagonals, and the 4 compression resistances are on the remaining diagonals. This way, the wheatstone bridge adds up loads on the cells, becoming more unbalanced as the increasing load increases the tensions (increasing their resistances) and increases the compressions (decreasing their resistances).

  • With two half-bridge, three-wire load cells, you don't need the combinator. Identify the center taps and hook them up as S+/S-, and wire the remaining wires to their opposite colors (e.g. white to black, and black to white) and hook them up to E+/E-. The color-flipping puts the tension resistances on opposite diagonals of the bridge, so adding tension to them unbalances the bridge constructively. If you wire it up with matching colors, you get it backwards, and two half-bridges with the similar loads will balance each other out.

    The four-wire load cell is two separate gauges, you can make a four-wire cell into a three wire half-bridge by hooking one wire from each pair together to make a common terminal.

  • You are right. The trick in hooking the 8 variable resistances of 4 half-bridge load cells (like the ones in bathroom scales, and the ones Sparkfun sells) up as a 4-resistance wheatstone bridge is to identify the center taps, wire the non-center-taps into a big ring with matching colors, and then excite with two opposite center taps and sense on the other two center taps. The wheatstone bridge then has two tension (or compression) resistances on each leg, the two tension legs (four tension resistances) are on opposite diagonals, and the two compression legs (four compression resistances) are on the other opposite diagonal. The taps of the wheatstone bridge are the center-taps of the 3-wire load cells.

  • The voltage on the red will be approximately 1/2 the white-black voltage at no load, and will shift proportionally by 1mV/V of the white-black voltage by load/50kg.

  • Qualib, if you swap the excitation of one of the cells and you can use the bridge to add the loads.

  • Yes, about 10lb of load should shift the output about 0.454 mV from no-load, and you are only reporting measurements to a precision of 10mV. Try a 5 digit DMM to see a direct reading.

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