Need help understanding schematics in LT1019 datasheet

Question

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.

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.

Answer 1

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.

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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\$.

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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:

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.

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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.

Answer 2

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:

Answer 3

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.