We are currently using AD7795 to get ratiometric reading of 4 load cells. It works, but the chip stocks are rather volatile. So, I am thinking on using much cheaper and easier to come by MCP3424/MCP3428, with added benefit of smaller PCB footprint.

Since MCP chips use internal reference I needed a stable excitation voltage. So I stumbled upon LT1019 precision reference which is just about perfect for our needs. The very first schematics in datasheet is somewhat puzzling.

enter image description here

I understand that they use LT1019 to get stable +5V source, and then invert it with LT1637 to get [supposedly] stable -5V. I think I also understand a second comment, which says that gain amplifier LT1001 practically does not affect bridge in any meaningful way.

I think I do understand the role of 357 Ohm resistors, which, according to the first comment "reduce reference and amplifier loading to ~0". It seems they work as voltage divider to apply slightly under 10V to the 350 Ohm bridge, so the reference and inverter only have to supply tiny current to get it to full 10V.

What I do not understand is why they included "active element" into feedback path for the inverter. Certainly that should make the -5V fluctuate with the load, which could have been avoided by connecting to the opposite side of the bridge. The comment "not critical, because A2 acts as differential amplifier" is not a good reason to not get stable excitation, IMHO, especially if you call the result "ultralinear".

Note, that I am not trying to replicate the design, as we only need positive source for cells. But I am curious about the schematics above and also the trick with resistor supplying majority of current allows the use of much cheaper voltage references.

  • 1
    \$\begingroup\$ Why not simulate the circuit? \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2021 at 7:42
  • 2
    \$\begingroup\$ @Andyaka I am pretty sure the simulation will show it works just fine, but it won't answer the question "why". Also, I admit, I never used simulation before. With my day having 14 hrs work and 4 hrs sleep it is doubtful I will ever learn how. \$\endgroup\$
    – Maple
    Commented Oct 1, 2021 at 8:00

3 Answers 3


The purpose of A1 is not to produce -5V, a mirrored version of the +5V reference. It is to maintain the voltage at the top of "active element" at exactly 0V. This can be understood by knowing that negative feedback causes the opamp to equalise its two input potentials (one of which is held at ground), by adjusting its output to make that so.

In the case where the "active element" is exactly 350Ω, this also happens to cause A1 to output -5V, since the left-hand path of the bridge is now dividing by exactly 2.

As the active element changes its resistance, A1 modulates the voltage at the bottom of the bridge so that the top node of the element is kept at 0V.

I've never seen this done before, and it intrigued me, so I analysed it, to test the claim of linearity. What we have is this bridge:


simulate this circuit – Schematic created using CircuitLab

The junction of the left path (containing the lower, active, element \$R_E\$) is point A, held at 0V by the opamp. At the top of the bridge, point X, is the exitation potential \$V_X\$. The opamp is controlling \$V_Y\$, the voltage at Y, the bottom node of the bridge. Also, \$R_1 = R_2 = R_3\$.

We wish to know the voltage across the bridge \$V_B\$.

The potential \$V_Y\$ is whatever voltage is needed to produce 0V at A. This is stated here:

$$ V_A = V_Y + (V_X - V_Y) \frac{R_E}{R_E + R_1} = 0V $$

Rearrange to make \$V_Y\$ the subject:

$$\begin{aligned} V_Y - V_Y (\frac{R_E}{R_E+R_1}) &= -V_X (\frac{R_E}{R_E+R_1}) \newline \newline V_Y (1 - \frac{R_E}{R_E+R_1}) &= -V_X \frac{R_E}{R_E+R_1} \newline \newline V_Y &= -V_X \frac{\frac{R_E}{R_E+R_1}}{1-\frac{R_E}{R_E+R_1}} \newline \newline &= -V_X \frac{R_E}{R_1} \end{aligned}$$

Since \$R_2 = R_3\$, \$V_B\$ is half way between \$V_X\$ and \$V_Y\$:

$$\begin{aligned} V_B &= \frac{V_X + V_Y}{2} \newline \newline &= \frac{V_X - V_X \frac{R_E}{R_1}}{2} \newline \newline &= V_X \frac{1}{2} (1 - \frac{R_E}{R_1}) \end{aligned}$$

Since \$V_X\$ is constant at 5V, this does actually produce a voltage across the bridge which varies linearly with \$R_E\$, and which is zero when \$R_E = R_1\$. Amazing, at least to me.

Edit 1 - On active element current

I noticed that A1 and the 350Ω resistor provide a constant current through the active element. It's easier to see if I redraw those parts like this:


simulate this circuit

From this perspective it's easy to see that with a constant current through \$R_E\$, obviously the opamp output will be indirectly proportional to the resistance of \$R_E\$.

Edit 2 - On the 357Ω resistors

You are right about the 357Ω assisting the LT1019 regulator by raising point X nearer to +5V, thus relieving the regulator of the burden of providing all of the current required to do that. However, I wanted to find out the exact voltage at X that this resistor would create, with no regulator there at all. That's not trivial, and I admit I cheated here, using the simulator to derive the voltage at X, rather than doing the algebra:


simulate this circuit

To bring \$V_X\$ up to 5V from 4.93V, the regulator clearly has to source a small current. There's a "Load Regulation" graph on page 5 of the LT1019 datasheet you linked to:

enter image description here

This chip is quite capable of sourcing and sinking 10mA, but it's better at sourcing than sinking. Also the regulation error is linear from 0mA and up, but if ever the output crosses over between sourcing and sinking, there's a kink in the curve there which will introduce a subtle non-linearity. It makes sense to keep the device sourcing, well away from that region of crossover distortion.

In my previous edit I showed that the current in the left path of the bridge will be a constant 14mA, but the current in the right path will vary to some degree, centered at 14mA (because it has the same resistance of 2 × 350Ω between X and Y) but rising or falling slightly as \$V_Y\$ changes.

You should normally calculate the maximum excursions of \$V_Y\$ (corresponding to the expected extremes of \$R_E\$) above and below -5V to figure out the upper and lower bounds of current in the right path. I can take a guess at this though, given that the differential amplifier stage (A2) has a gain of 100.

I'll assume that an output swing of ±15V corresponds to an input of ±0.15V at B, in turn corresponding to twice as much at \$V_Y\$ (due to the halving by the 350Ω+350Ω divider).

$$ V_Y = -5V ±0.3V $$

Current in the right path will vary between: $$ \frac{5V - (-5V \pm 0.3V)}{2 \times 350\Omega} = 14.3mA \pm 0.5mA $$

Add the left path current, and we get a total current to be drawn from point X (and returned to Y, of course) of about \$28.6mA \pm 0.5mA\$

So, the designer wants the LT1019 to source a small amount of current (significantly less than 10mA), but not so small that the variance of ±0.5mA will cause it to approach zero.

Edit 3 - On the differential amplifier

I finally got around to looking at A2, and the claim that it doesn't load the bridge. That part of the circuit initially looks like this:


simulate this circuit

There are some approximations that enable us to whittle that down into something much simpler. For starters, we know that \$V_A = 0V\$. We also can say that \$V_P = V_Q\$, due to opamp action under negative feedback. \$R_4\$ and \$R_5\$ have 0V across them, draw no current, are dividing 0V by something, to yield \$V_Q = 0V\$. All this allows us to make this statement:

$$ V_P = V_Q = V_A = 0V $$

The whole thing reduces to:


simulate this circuit

Under those assumptions, this is just a regular inverting amplifier with output:

$$ V_Z = -V_B \frac{R_3}{R_2} $$

Currents are easy to calculate, given the usual idealisation that the opamp inputs have infinite resistance, drawing no current:

$$ I_B = \frac{V_B}{R_2} $$ $$\begin{aligned} I_A &= -\frac{V_Z}{R_6} \newline \newline &= V_B \frac{R_3}{R_2 R_6} \end{aligned}$$

As the author states, we should have \$R_3 = R_6\$, so that simplifies to:

$$ I_B = V_B \frac{1}{R_2} = I_A $$

What this means is that the inputs of the differential amplifier draw the same current from each side of the bridge. Technically this isn't "no load", but it does prevent the bridge from becoming unbalanced.

We made the assumption that \$V_A = 0V\$, but, strictly speaking, it will be slightly non-zero, by an amount equal to the input offset voltage of A1. Its effect on the condition \$I_B = I_A\$ will be negligible, since the current drawn by \$R_4\$ and \$R_5\$ due to mere millivolts at A will be tiny compared to the many milliamps in the bridge elements.

The last claim is that A1's input offset voltage and drift are cancelled by the differential amplifier. Whether this is true or not is not immediately obvious to me, but I'll cautiously trust the claim, and leave the explanation to someone else.

  • \$\begingroup\$ That was the one part I thought I understood... and you thoroughly dismantled it. So, this simple circuit is not that simple, after all. \$\endgroup\$
    – Maple
    Commented Oct 1, 2021 at 16:12
  • 3
    \$\begingroup\$ @Maple The thing is, somebody had to come up with this in the first place. It's just humbling. \$\endgroup\$ Commented Oct 1, 2021 at 16:34
  • \$\begingroup\$ At least, is my guess about the role of 357Ohm resistors correct? Can I use one to reduce the current requirements for the voltage reference? We have four 350 Ohm cells on 5V excitation, so about 60mA total. Such reference is hard to find, but there are plenty of cheap 10 - 20mA chips in SOT-23 \$\endgroup\$
    – Maple
    Commented Oct 1, 2021 at 22:26
  • 1
    \$\begingroup\$ I've added a section (edit 2) where I show the rational for the 357Ω resistor. Your intuition is good, but there's a little more to grok. \$\endgroup\$ Commented Oct 2, 2021 at 4:56
  • \$\begingroup\$ @Maple If you're still interested, for completeness I managed to dig into the differential amp part, to some extent. It's in the section "Edit 3". I'll leave it all alone now, unless someone points out any errors. \$\endgroup\$ Commented Oct 2, 2021 at 9:00

If you consider only the 4 bridge resistors, with a single variable one, the differential voltage variation will be non-linear (regardless if the the bridge is powered by a current or voltage source).

What the op. amp. tries to achieve is that it keeps the Vaa node (below) close to 0V, meaning that the R8 resistor will have a constant current. With that, the voltage across R10 will vary linearly with the resistance variation.

If the current through R8 and R10 remains practically constant, and the current through R7 also doesn't change (considering that the +15V has a good load regulation), the current through R9 and R11 changes linearly.

So, ideally, the differential voltage becomes linearly dependent on the variable resistor value.

The idealized circuit below tries to show that by exaggerating the resistance variation:

enter image description here

enter image description here

  • \$\begingroup\$ Amazing. So, all those linearity claims by load cell manufacturers are misleading since the bridge is inherently non-linear. Unless, of course, the cell has four active elements. It never occurred to me to analyze wheatstone this way. Reminds me of a case when a client asked me to rewrite lift control software to use position pot instead of end switches. They forgot to mention that they still have pull-ups on those pins and complained that the motion was non-linear. \$\endgroup\$
    – Maple
    Commented Oct 1, 2021 at 16:25
  • \$\begingroup\$ As you said, there are arrangements with 4 (and also with 2) active elements which are inherently linear. Also, for load cells, since the resistance variation is much smaller than the one shown here, the error due to non-linearity may be less relevant than others. \$\endgroup\$
    – devnull
    Commented Oct 1, 2021 at 16:33
  • \$\begingroup\$ Thank you very much for the clear and visual explanation of the bridge behavior. I wish I could accept more than one answer. In this case I have to accept Simon's version because he gave detailed analysis of other parts of the circuit as well. I did upvote yours, of course \$\endgroup\$
    – Maple
    Commented Oct 2, 2021 at 6:46

I feel like the other answers are too long.

As for why A1 is there. Keeping one node of the bridge at virtual ground is explained in this app note.

A bridge is calibrated so that the differential output voltage is around zero. It is of less importance that the common-mode output voltage be exactly half the supply voltage. That means if you drive the top of the bridge to exactly 5V and the bottom to exactly -5V, the outputs will probably be off from ground to the same direction, which needs the amplifier to have more CMRR than would be required if one output node is kept at ground.

As for linearity, if only one leg changes, then the voltage will be most linear in respect to its resistance if the current through it doesn't change. And this circuit does exactly that.

As for the resistor in parallel with the LT1019. Aside from dissipation problems, the LT1019 will keep the node labeled 5V at 5V, whether or not the resistor is there. During operation there is about 5V across each resistor of the bridge. So the voltage reference would have to supply 28mA. To reduce load on the reference, we can supply the 28mA with a resistor whose value is (15V - 5V) ÷ 28mA = 350Ω.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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