Load Cell Signal Boosting

I’m using a variety of load cells, the highest output is ~3mV/V and I’d like to use an excitation of 10V. thus the output range of the loadcell will be -30mV to 30mV. Any ideas how to boost that voltage swing to a -5 to 5 range with minimal noise?

I’m currently using an INA333 and successfully boosting the signal to 0 to 5V but due to noise in the signal I want to utilize the full -5 to 5.

I’m currently suffering from +/-7.2mV of jitter on the output of the current setup, changing the gain didn’t change the amount of jitter so in theory if I can double the useful range of the signal while keeping the jitter at 7mV the noise will effectively be halved.

Which Instrumentation Amplifier is up to this task?

Hi Funglee,

So it appears that you are limited with the INA333A to a 5.5V supply, which is why you can’t get the -5V to +5V output. It looks like the AD8220ARMZ-R7CT-ND should take care of that problem.

It can do a dual supply from ±2.25 V to ±18 V or a single supply from 4.5 V to 36 V. Most of it’s relevant specs appear to be similar or better that the INA333, with the exception of the quiescent current, which would increase from about 50uA to as high as 750uA.

You could still get by with a single supply if you change your reference voltage to +5V and power it with 10V. If you do so, note that the output will only reach within about 3% of the supply rail, so if you use a 10V supply, Vout max would be approximately 9.7V. Note that this arrangement will not provide a -5V to +5V swing, but rather a +0.3V to +9.7V output range. If you could provide a 12V supply, it would give you the full 10V swing easily.

If you use a dual supply, you have the same limitations described above with regard to the output rails, so you would want a -5.25V to +5.25V supply or wider to get the full +/-5V output swing. If you go this route, you can just connect Vref to ground.

One other note is that the formula for calculating the gain changes, too, so your R2 value would change. In the AD8220, you calculate gain as the following:

G = 1 + 49.4K/Rg

(where Rg is your R2 resistor)

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Hey David,

That looks like it’ll work perfect. Building on the dual supply idea, if I use something like LM2776DBVR to invert my 5v supply (Powered via USB, I’ve seen 4.85 to 5.1v depending on the pc) and connect it to E- of the load cell and V- of the amplifier everything should be within operating limits, correct?

Also, I believe a 300ohm resistor should be used to set the gain to 166 thus amplifying the 30mV to 5V. I’m willing to take the loss of the last 0.2V on both ends to reduce the component count.

Thanks for the assist.

Hi funglee,

Yes, using 300 Ohms will give you a gain of 166 and should give you about +5V with a 30mV signal.

Using the LM2776 might be a good idea, but it depends a bit on your application. Switched capacitor inverters are going to produce some output ripple, and if you connect the output of the LM2776 directly to E-, that will cause your load cell output to ripple proportionately. Looking at the graph in Fig.1 of the datasheet, that ripple could be as high as 2.2% (ripple <= 120mV with Vin = 5.5V). Now, since the ripple is only on the negative half of the total excitation voltage of the load cell, the percent ripple seen at the output is only going to be 1/2 of this value (assuming no ripple on the positive rail).


So, depending on your system requirements, there are a few different things you can do. If you not need high precision measurements, then you could probably go with this directly and not do anything more to improve your results. If you used only a 1% tolerance gain resistor, that alone would give you about as much error as the ripple noise.

However, if you needed more precision, here are a few things to consider which might improve your results:

  1. If you only need to take readings at a very low frequency (say once per second, for example), you could simply oversample the output by taking many readings (say at a rate of 256Hz, or once every 3.9ms) and average them out. The average value should converge toward the true value, the more samples you take.

  2. If you notice from Fig.1, the ripple drops significantly if the load current exceeds about 70mA, down to about 30mV, or about 0.55%. Based on your circuit, it looks like the nominal load for the LM2776 would be no more than a few milliamps, so the ripple is going to be in the higher range. If you can afford drawing some extra current in your system, you could increase the load on the LM2776 by placing an appropriate value resistor in parallel with its load to ground. The output ripple drops significantly when the load current exceeds about 70mA, so if you used a resistor in the range of 60 to 70 Ohms, you can reduce the ripple by more than half, down to about 0.55%.

  3. By using larger value output caps on the LM2776, one can also reduce ripple. However, if you do so, you may have to change values of the other capacitors as well. They don’t recommend using anything other than ceramic capacitors because of the requirement of low ESR in the capacitor, but if you need higher output capacitance than you can reasonably get with a ceramic cap, you can use a tantalum, aluminum-polymer, or tantalum-polymer capacitor and place a ceramic cap in parallel with it. This will give you the higher capacitance while maintaining the low ESR. Just make sure that their rated voltage is sufficiently higher than your output voltage. Ceramic capacitor’s capacitance values are voltage dependent, so as the applied voltage is increased, their capacitance decreases. The phenomenon is worse for lower voltage rated parts and for smaller package sizes. Therefore, use a ceramic capacitor rated for at least 10V and in the largest physical size you can. See here for more information on this:

    Voltage coefficient of capacitance

Hope this helps!



Good call on the dummy load, I checked the ripple in Figure 12 but didn’t even think to check the minimal-load ripple. I’m going to use a 0.5Watt 68ohm resistor to maintain the 70mA minimum current like you suggested.

does using a higher precision resistor decrease error? I’m under the impression that a 5% and 1% resistor will offer the same precision as long as the calibration is done after assembly to account for the variance. Of course the gain will be slightly different but other than that the overall accuracy doesn’t depend on the resistor tolerance, correct? Or does the tolerance also specify how much the resistor will vary over time as well?

Thanks again


Resistor tolerance values don’t actually tell a person very much; they describe variations in resistance between devices due to variabilities in manufacturing only. The influence of factors such as temperature, humidity, drift, and applied voltage on the apparent resistance of any given device are not included in that figure. The effect of those influences can/will change over time and with applied stresses, and in a really sloppy resistor they might stack up to something in the territory of 20% of nominal value.

However, that very sloppy resistor would still be considered (for example) a 1% resistor so long as the difference between devices under standard test conditions at the end of the production line is held to 1% or less. Beyond that, thermocouple effects have potential to mess things up in really high-precision applications, and chemical degradation from atmospheric moisture or sulfur have potential to make a resistor go completely open-circuit over time. Long story short, a decent thin film or metal film (different names for the same basic thing) resistor would be a common choice for an application like this, and metal foil would be the place to look if you want the absolute best that (lots) of money can buy.

All that said, your initial post mentions several mV of noise that reportedly didn’t change when the gain was changed, suggesting that it’s not anything connected to the Vin terminals of the inamp that’s responsible. It’s totally possible for that noise to be coming from your measurement apparatus rather than the circuit in question, or is being created by whatever is being used to make the 2.5v bias.

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That lines up with what I was expecting, but I wasn’t aware of how much the values may vary over time!

Thanks for the insight, sounds like the 5% resistors on hand should work fine then.


I think Rick’s point was that the type of resistor used is as important as its initial tolerance. If you want to make a system that is stable, long term, and over varying temperature and humidity conditions, then you should go with either a thin film/metal film type, or, if money is no object, a metal foil type.

Most 5% tolerance resistors are not one of those types, though there are a few out there. Typically, the more stable types are specified as 1% or better. However, if this is not for a long-term product, and environmental conditions are reasonably well controlled, then a carbon film or thick film resistor might be just fine.