How does this current sink work (Figure 15 LM675)

Question

Here is figure 15 of the LM675 datasheet. I was interested in it because it shows an opamp voltage-controlled current device that doesn't include a FET. The LM675 is a power opamp that can source and sink 3A.

I've been trying to figure out how to derive the figures in the bottom right (I_out = 2.5 Amp/Volt * V_in), and how to select the potentiometer value for R_out.

I tried to do some nodal analysis:

Shortcut constants:
R_out is the resistance between the opamp output and GND.
It is equal to 4 + [(10k + 200 potentiometer + 1k) || R_load]
k is the ratio between V+ and V_out, since  is being connected through a voltage divider.

V_out (the output of the opamp) = A(V+ - V-)
V+ = k * V_out
KCL: (V_in (the input) - V-) / 1k = (V_out - V-) / (10.1k).
I isolated V- to be equal to (10.1 / 9.1) V_in - V_out

After solving and taking the limit of A to infinity, I got:
V_in = ( (1k / 10.1k)(1-k) + k ) V_out

Here's where I got stuck. This is supposed to be a constant current device, but the output voltage depends on k, which in turn depends on R_load. I can't find a way to make R_load disappear, and I don't know whether it's supposed to since there's a potentiometer to tune it.

How can I derive / modify the current to input voltage ratio?

Answer 1

I don't fancy doing a complete precise-to-the-ohm analysis of this, and it's not really necessary. Instead, we can make some observations which will really simplify things. Start with a suitably labelled schematic:

^{simulate this circuit – Schematic created using CircuitLab}

For example, we can say that negative feedback will cause the op-amp to adjust its output \$V_X\$ to whatever potential is necessary to equalise the potentials of its two inputs, \$V_P\$ and \$V_Q\$. That fact would eventually emerge from a complete analysis, as a consequence of taking the limit as \$A\rightarrow \infty\$, but it's much easier to start with this equality:

$$ V_P = V_Q $$

A second observation is that current in \$R_4\$ and \$R_3\$ is three orders of magnitude smaller than load current in \$R_L\$. I'll make the statement that:

$$ I_L = I_S $$

This allows us to avoid the ugly arithmetic associated with the parallel combination of \$(R_3+R_4) \parallel R_L\$. Then we have this simple expression for \$V_Y\$:

$$ \begin{aligned} V_Y &= V_X - I_SR_S \\ \\ &= V_X - I_LR_S \end{aligned} $$

Perhaps you already see the implication that any algebra from this point on is independent of \$R_L\$. Perhaps, also, you begin to see the role that \$R_S\$ is playing here, somehow involved in setting transconductance, the ratio \$\frac{I_L}{V_{IN}}\$.

Let's continue, though:

$$ \begin{aligned} V_P &= V_Y \frac{R_3}{R_3+R_4} \\ \\ &= \frac{10}{111}V_Y \\ \\ &= \frac{10}{111}(V_X - I_LR_S) \\ \\ \end{aligned} $$

Also:

$$ \begin{aligned} V_Q &= V_{IN} + (V_X-V_{IN})\frac{R_1}{R_1+R_2} \\ \\ &= V_{IN} + (V_X-V_{IN}) \frac{10}{111} \\ \\ &= \left(1 - \frac{10}{111}\right)V_{IN} + \frac{10}{111} V_X \\ \\ &= \frac{101}{111} V_{IN} + \frac{10}{111} V_X \\ \\ \end{aligned} $$

Since \$V_P=V_Q\$, we can equate these two expressions, and rearrange to find \$I_L\$ as the subject:

$$ \begin{aligned} \frac{101}{111} V_{IN} + \frac{10}{111} V_X &= \frac{10}{111}(V_X - I_LR_S) \\ \\ 10.1V_{IN} + V_X &= V_X - I_LR_S \\ \\ I_L &= -10.1\frac{V_{IN}}{R_S} \end{aligned} $$

It seems that you must adjust \$R_S\$ to set your desired relationship between \$I_L\$ and \$V_{IN}\$. Also, as long as \$R_3\$ and \$R_4\$ are large compared to \$R_S\$. this relationship is independent of load resistance \$R_L\$.

The last thing to address is the role of the potentiometer in the original circuit. It allows you to change the transconductance (closed-loop gain) very slightly, above and below the center value of −10.1 calculated above.

Essentially, it makes \$R_4\$ variable between 10.0kΩ and 10.2kΩ, permitting transconductance to be adjusted to exactly \$-2.5AV^{-1}\$. Other than this "trimming" role, it has no significant influence over transconductance, which is squarely in the hands of \$R_S\$.

Answer 2

Simon has done an analysis of the way this circuit (a Howland current pump) works.

It would be good to recognize that this circuit, though it avoids the pass transistor, both requires the op-amp to supply all the load current and is very sensitive to resistor matching and drift.

It has one advantage- it has a bipolar output (it can drive current in either direction), and that's about it.

That's why you need the 10-turn trimpot (and manual adjustment) to get decent performance even with 1% resistors. And it will as easily drift out of calibration, resulting in lower than expected output resistance (an ideal current source has infinite output resistance \$\text R_{out}\$). A circuit with a pass transistor typically will always have high output resistance (without trimming) and the sensitivity of the current to input voltage will vary proportional to resistor errors (as does the LM675 circuit, of course).

There's also a Zobel network on the output, presumably because the load is expected to have inductance, being a servo.

Also keep in mind that datasheet application circuit are concocted (in this case, about 25 years ago) to sell chips and may not be the ideal solution for your problem.