HX711 on 3.3V

Hello,

I'm using a board that operates at 3.3V and am connecting it to an HX711 for reading a load cell. Connected as follows:

GND --> GND
DT --> 6
SCK --> 5
VCC --> 3.3V

The HX711 reads well around the calibration point, but seems less responsive the farther I get from from it. Would operating the board at 5V increase its accurate range?

Thanks!

Can't you run the board on 5volt, and drop the data/clock lines to 3.3volt with a voltage divider.
Leo..

The HX711 specs says it should work no problem with 3.3V, and I don’t recall any mention about losing accuracy at lower voltages.

I can only assume that might happen when load applied is much lower than cell’s nominal load.

Michael_Link:
The HX711 reads well around the calibration point, but seems less responsive the farther I get from from it.

This sounds like a calibration problem, not a "clipping" problem, but how can anyone know without seeing some data?

blimpyway:
The HX711 specs says it should work no problem with 3.3V, and I don't recall any mention about losing accuracy at lower voltages.

The chip can, a breakout board can't shouldn't.
The boards I have seen have a 4.3volt regulator for the exitation voltage that needs 5volt to work.
Lower/unstable exitation voltage is lower/unstable load cell output.

I don't see why two 1k:2k dividers on the output (to drop 5volt logic to 3.3volt logic) wouldn't work.
Leo..

Wawa:
The chip can, a breakout board can't shouldn't.
The boards I have seen have a 4.3volt regulator for the excitation voltage that needs 5volt to work.

~~In the datasheet's typical breakout application schematic, the chip's analog supply regulator controls the base of an external transistor, and it shows 2.7 to 5.5v on the breakout Vcc (which goes straight to the transistor collector...and the emitter is connected to E+). So it seems a breakout ought to work with 3.3v (and the 0.7v drop we see at 5v is probably the drop across the C-E junction.) ~~
But, like you say, it would be straightforward to supply 5v and drop the logic to 3.3v, if wider range/better resolution is really necessary (and it is not clear that it is). Oops - see below.

DaveEvans:
In the datasheet’s typical breakout application schematic, the chip’s analog supply regulator controls the base of an external transistor, and it shows 2.7 to 5.5v on the breakout Vcc (which goes straight to the transistor collector…and the emitter is connected to E+). So it seems a breakout ought to work with 3.3v (and the 0.7v drop we see at 5v is probably the drop across the C-E junction.)

Look at the diagram again. This is a typical LDO circuit.
The chip sinks base current. The PNP transistor could saturate (<0.5volt or even <0.2volt between CE).
The chip measures collector voltage with R1,R2,VFB pin (Voltage_Feed_Back).
And backs off base current untill set voltage is reached.
VFB pin threshold is 1.25volt, so if you know R1,R2 you can work out the voltage of the regulator.
Leo…

Hmmm... yep, you're right. The datasheet says the output voltage is Vavdd=Vbg*(R1+R2)/R2, and Vbg=1.25v, so Vavdd=4.3v with R1=20k and R2=8.2k as shown in the prototype. It also says Vavdd should be at least 100mV below the supply voltage....so yes, that would be a problem if the supply was 3.3v.

Maybe this can help: Modifying the HX711 Breakout Board for 3.3V operation

T

Lowering excitation voltage to 3.3volt is a bad idea if you have 5volt available.
Then ist's better to split the analogue and digital supply of the HX711 (see HX711 datasheet).

Sparkfun has already done that for you with this board.
VCC (5volt) for the analogue parts of the chip (excitation voltage and instrumentation amp),
and a separate VDD pin (3.3volt or 5volt) for logic.
Leo..