Floating voltage-controlled resistor with LM13700: How does it work?

_{Note: Thank you @LvW for your answer. It helped me to gain more insight in the circuit. I think I got it now, how it works. I'll explain these insights below in more detail.}

A VCR (Voltage Controlled Resistor) is a circuit that emulates a resistor. It must behave in all practical ways like a resistor to any "external circuit" connected to its terminals. Although there are several ways to build a VCR, we'll dig into the technique offered by the LM13700 component.

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1.  The grounded VCR

1.1  Overview

The easiest VCR's emulate only grounded resistors (one terminal is connected to Gnd). The following figure clarifies how that happens:

The VCR exposes 3 terminals:

1. V_dd: The positive supply pin.
2. Gnd: The negative supply pin.
3. R_x: The resistive input pin. This represents the emulated resistor terminal which is not tied to Gnd.

When the external circuit imposes a voltage V_x on the R_x pin, the VCR circuitry draws a current I_x equal to:

such that it behaves like a normal resistor. Watch out: the current paths clearly show that the VCR circuit must be tied to the same power supply as the external circuitry for this process to work!

 

1.2  How it works

The schematic below shows the VCR circuit fundamentals. The opamp-alike-symbol in the middle is an OTA (Operational Transconductance Amplifier, like the LM13700). It converts the voltage difference between its V_in+ and V_in- into a current I_out

with g_m being the transconductance parameter. Supposing an ideal component, it won't be affected by the voltage on the output pin. It just forces the current I_out to flow no matter what.

But how can this component be used to build a VCR? Take a look at the (emulated) resistive terminal of the VCR. A positive voltage V_x is applied on that terminal. The (darlington) transistor on the right is connected as an emitter follower, duplicating voltage V_x at its emitter output. Next, the voltage is trimmed down with the resistive divider R_B and R_A to V₁.

The output current of the OTA is now:

The minus '-' sign is due to the fact that we tie V₁ to the negative input.

Now look again at the schematic below. The green arrow represents I_out flowing out of the OTA. The blue arrow I_x is what you'd expect to flow. That's right, you'd expect a current flowing into the OTA, as we applied a positive voltage V_x on the resistive terminal. So I_x = -I_out and we can say that:

We're almost there! Knowing that V₁ results from a resistive divider, we can substitute:

and reformat into:

And voila! We have proved a linear relationship between the applied V_x and the resulting flow of I_x. The linear factor is the term on the right: the (emulated) resistance value. With V_CONTROL you can adapt the g_m transconductance parameter, which in turn affects the resistance value.

 

1.3  Intuitive

After doing the maths, let us just analyze the circuit on "gut feeling". You impose the voltage V_x on the VCR terminal. The transistor on the right duplicates that voltage to its output. It behaves as a buffer, drawing almost no current on the base (remember: in the actual circuit, it's a darlington).

The OTA component turns a voltage into a current, just like a resistor does. So all we need to do is feed back the (buffered) V_x voltage to the input of the OTA. We don't tie V_x directly to the input of the OTA though, but first divide it with R_A and R_B into V₁.

Suppose we'd tie V₁ to the positive input of the OTA. Then we'd get a negative resistance: the higher the voltage V_x applied, the more current this VCR pushes into the external circuit. Therefore, we tie V₁ to the negative input to create a normal resistor. The higher the voltage V_x applied, the more current the VCR pulls from the external circuit.

 

1.4  Notes

• We suppose that the darlington transistor on the right only draws a negligible current on its base pin.
• The applied voltage V_x must stay withing the power supply range. It should keep some distance from the rails.
• Both devices - the external circuit and the VCR circuit - must be tied to the same power supply. Otherwise the current flows needed for the system to work are blocked.

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2.  The floating VCR

2.1  Overview

We'll simulate a resistor again, but this time with both terminals floating (none of them tied to Gnd). An external circuit connects to the two (emulated) resistor terminals R_x1 and R_x2, applying respectively the voltages V_x1 and V_x2 (assuming V_x1 > V_x2). It expects a current I_x to flow into terminal R_x1 and coming back out of terminal R_x2.

To emulate that, we use two OTA devices. The first one emulates the first terminal R_x1, pulling in a current I_x1. The second OTA emulates the second terminal R_x2, pushing out a current I_x2.

Now this setup is quite dangerous. What if I_x1 differs from I_x2? Then the external circuit will certainly think: "Hmm... weird resistor. It's losing/gaining current on the way through." Therefore it's vital that these two currents are perfectly matched!

 

2.2  How it works

Investigate the following circuit:

The external circuit applies voltages V_x1 and V_x2 directly on the bases of the darlingtons T1 and T2. Being connected as emitter followers, they duplicate (and effectively buffer) these voltages. Next, we see again a resistive divider appearing. Let us assume that:

ΔV_x = V_x1 - V_x2
ΔV = V₁ - V₂

with V_x1 > V_x2.

From this resistive divider, we can calculate that:

This differential voltage ΔV is applied on the inputs of both OTA devices, be it reversed on the one to the right, and straight on the one to the left. Consequently, these are the current outputs of both OTA's:

The output currents are equal and opposite. At the terminal where the highest voltage V_x1 is applied, the current flows into the OTA. At the other terminal, it flows out of the OTA. Just as you would expect when you apply these voltages to an everyday resistor.

The emulated resistive value is:

You can tune the transconductance parameter g_m with the control voltage V_CONTROL.

 

2.3  Notes

• The circuit only works if the external circuit and the VCR circuit are connected to the same supply.
• We assumed that V_x1 > V_x2 while doing the maths. The circuit is perfectly symmetric, so you can do the same maths for the opposite case.
• The applied voltages V_x1 and V_x2 must stay within the power supply ranges, preferably keeping some distance from them.

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3.  Mathematical analysis

The (unbuffered) output stage of the LM13700 is very limited. Therefore, we've got severe capability constraints on this circuit. Let's dig into the details.

 

3.1  V_ABC as a function of I_ABC

Figure 10 in the datasheet plots the voltage V_ABC one has to apply on the amplifier bias input to cause current I_ABC to flow. This is important, because I_ABC eventually determines the transconductance g_m.

Figure 10 from the datasheet:

_{Note: It's the IABC current that eventually defines the transconductance parameter gm.}

The datasheet only provides this figure to show us the relation I_ABC ∼ V_ABC. No formulas. So I've deducted myself the following formulas from the figure:

In SI units:

And with these formulas, I've plotted the relation I_ABC ∼ V_ABC:

My plot resembles the plot from the datasheet, so I believe the formulas I've deducted are okay. Note: there is a logarithmic relationship I_ABC ∼ V_ABC, not a linear one!

 

3.2  Transconductance g_m as a function of I_ABC

The transconductance g_m is plotted agains I_abc in Figure 8 from the datasheet:

_{Note: From my understanding of the datasheet, the unit in this plot is uS (micro-siemens). A transconductance of 1uS means that 1uA of output current is generated per differential voltage at the inputs of the OTA.}

As both the x-axis and y-axis are logarithmic in this figure, the relation between g_m and I_ABC is linear:

and in SI units:

I've plotted the relation myself as well:

 

3.3  Transconductance g_m as a function of V_CONTROL

We have all intermediate formulas to get the final relationship between g_m and V_CONTROL. In the end, that's what you need to know: what transconductance do I get for a given control voltage?

Using the formula's above, I've plotted the relationship:

 

3.4  Peak output current I_peak

Before proceeding to calculate the emulated resistance, it's wise to take a look at the limited output current the OTA can deliver:

in SI units:

which I've plotted as:

Please compare this to Figure 4 in the datasheet. The plots should match.

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4.  Emulated resistance

As we got the transconductance g_m, we can proceed to the final step: define the emulated resistance R_x. We already know that:

The datasheet of the LM13700 proposes values R_A = 1k and R_B = 100k in its schematic for the VCR circuit. So let's take those values (as I suppose them to be safe).

4.1  Resistance in function of the transconductance

Using the equation above, we get:

LEGEND:

• Blue curve: resistive value R_x.
• Red curve: peak current I_peak that OTA is capable to deliver.
• Green curve: current I_30V drawn from OTA when you apply 30V on the terminals of the emulated R_x.

The LM13700 is usually fed by a 30V supply (or ±15V). But as you can see from the figure, the worst-case-scenario (applying 30V on the terminals of the VCR) leads to an overcurrent condition on the OTA!

So we can either put a constraint on the applied voltage to the terminals (no more than ΔV = 10V), or we choose other values for R_A and R_B

 

4.2  Safe values for R_A and R_B

I've calculated these safe values for R_A and R_B:

R_A = 1k R_B = 320k

That gives us the following curves:

LEGEND:

• Blue curve: resistive value R_x.
• Red curve: peak current I_peak that OTA is capable to deliver.
• Green curve: current I_30V drawn from OTA when you apply 30V on the terminals of the emulated R_x.

Please notice that the red and green curves are on top of each other. In other words, the current I_30V drawn from the OTA when you apply 30V on the terminals will never exceed the current I_peak that the OTA is capable to deliver.

 

4.3  Resistance in function of V_CONTROL

The resistance R_x in function of the control voltage is:

The values for R_A and R_B are still 1k and 320k respectively.

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5.  Output stage

The current capability of this circuit is very poor. Therefore, I need to add some kind of "output stage" to the terminals of the VCR. I've put that in another question, over here:

Power stage for VCR (Voltage Controlled Resistor)

Please have a look at it ^_^.

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_{Note: I have done lots of efforts to share correct insights and calculations on this VCR circuit with you. If you find any errors, please leave a comment. I would be happy to make corrections.}

^{simulate this circuit – Schematic created using CircuitLab}

Description edited:

OK - let me try an intuitive explanation. The circuit diagram shows the basic method for creating a grounded OTA resistor.

Using basic OTA relations - and with Vs as a test signal source - we can write (gm: OTA transconductance):

Iout=-gm*V1 with V1=VsRA/(RA+R)

Rin1=-Vs/Iout=(RA+R)/gm*RA.

(Here I have assumed that we have an ideal emitter follower)

The given circuit for a floating resistor is a symmetric extension of this shown basic OTA-resistor - however, with cross-coupled OTA input terminals. Hence we can expect a similar expression for Rin2.

Because in the given circuit the resistor is defined BETWEEN both base nodes of the transistor we have to use the sum of both expressions (Rin1 and Rin2 in series):

Rx=Rin2+Rin1=2(RA+R)/gm*RA.

Up to now, we have not yet considered the fact that the resistor RA is connected not to ground. Instead, it has a common connection to the inverting terminals of BOTH OTAs. Hence it has a "normal" (positive) influence to one OTA (as in the basic model) and a negative influence on the other OTA.

For this reason, the influence of RA will be cancelled out in the sum of both expressions - and we arrive at the given formula:

Rx=2R/gm*RA.