I'm building a VCR (Voltage Controlled Resistor) using the LM13700 component. You can read more details on that here:

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

The problem I'm facing is its severe current constraints: below 1 mA (*). Therefore, I'm building my own "output stage" for the VCR.

(*) Note: The output current of the darlingtons in the LM13700 is up to 20mA. However, it is the output of the OTA's (Operational Transconductance Amplifier) that define the current capabilities of the VCR (Voltage Controlled Resistor) circuit.


1.  The circuit

I've designed the following circuit to create a more powerful (read: more current capable) output stage for my VCR:

enter image description here

Let us analyse the circuit. Note that the circuit behaves differently for VIN1 > VIN2 versus VIN2 > VIN1. So we'll handle both situations.


1.1  VIN1 > VIN2

Thanks to the schottky diodes D1 and D2, only the mosfet on the left T1 can conduct current if VIN1 > VIN2.

The positive input of the opamp VIN+ is the result of a resistive division. The higher the voltage difference ΔVIN on the terminals, the higher VIN+ rises. We want the current I2 (going through the mosfet) to rise accordingly. Will that happen?
If I2 is "too low", the feedback to the negative input of the opamp VIN- is lower than VIN+, causing the opamp output to increase. This makes mosfet T1 more conductive (gate rises), so I2 goes up. This keeps happening until VIN- reaches the level of VIN+.

I've made following calculations:

enter image description here

From which we can deduct that:

enter image description here

We have proven here a linear relation between ΔVIN and I2. Sure, that's not the only current flowing between the terminals. But we assume that I2 >> I1, such that:

enter image description here

The green box encircles the total (perceived) resistance from one terminal to the other.


1.2  VIN2 > VIN1

The currents I1 and I2 now flow in the opposite direction. Only mosfet T2 on the right can conduct (thanks to the schottky's).

enter image description here

The positive input of the opamp VIN+ is of course still the result of a resistive division. But this time its value goes down with increasing ΔVIN. Well, at least if you keep VIN2 fixed and only pull at VIN1 to increase the differential ΔVIN.
For the feedback to work correctly, we hope that I2 increases in magnitude as ΔVIN grows. Does that happen?

Yes, it does. As VIN+ decreases, it slides below the value of VIN-. The opamp output takes a dive. The PMOS T2 gains more VSG voltage and consequently conducts more. Beware, this PMOS is upside down, so perhaps you've got to turn your head to see that happening.

In short, we've intuitively deducted a linear relationship ΔVINI2.

I've also calculated this relationship, and I get the exact same value for RTOT as before, namely:

enter image description here


1.3  Resistive value

Let us choose some decent values for R1 and RC:

R1 = 20kΩ
RC = 100Ω

With these values, plotting RTOT against RVCR gives:

enter image description here

Let us now plot RTOT against the control voltage VCONTROL that drives the VCR resistance on the left. I'm using a VCR circuit based on the LM13700, as explained here: Floating voltage-controlled resistor with LM13700: How does it work? Based on those formulas relating VCONTROL to RVCR combined with the previous formula relating RVCR to RTOT, I get:

enter image description here


1.4  Notes

Perhaps you've noted the resistors RA and RB around the mosfets. The purpose of RA is to protect the output of the opamp. Opamps don't like capacitive loads that much - depending on one type to another of course.
Mosfets have a "linear region" in which they act as a resistor. However, if you drive the gate just a little too high, you risk to leave that region in an instant. I know that the feedback mechanism should prevent this from happening. But I thought: let's give those opamps some margin. It doesn't hurt to increase artificially that linear region with the resistive divider RA and RB.


1.5  Components

These are the parts I'm going to use:

  • OPAMP  →  LTC6090 (Digikey: `LTC6090HS8E-5#PBF-ND‎`)
  • NMOS   →  IRF740 (Digikey: `IRF740LCPBF-ND‎`)
                        or STP3LN80K5 (Digikey: `497-17293-ND`)
  • PMOS   →  IRF6218 (Digikey: ` IRF6218PBF-ND‎`)
                        or FQP3P20 (Digikey: `FQP3P20-ND`)
  • Schottky  →  BAT46 (Digikey: `BAT46-TR`) Note: Some in parallel for higher current


2.  Increase the range

The example in the previous paragraph provided a range from 100Ω to 140Ω. Not so spectacular. This low-range problem has to do with the resistive division at R1 and RVCR. Even tuning their values doesn't really solve the low-range problem (I believe).

I've introduced an instrumentation amplifier to solve the problem:

enter image description here

According to my calculations, the relation RTOTRVCR is now:

enter image description here

with G the gain of the amplifier set by resistor RG. This gain can range from 1 to 1000.


2.1  RTOT plotted against VCONTROL

We'll choose the same values for R1 and RC as before:

R1 = 20kΩ
RC = 100Ω

For now, we keep the gain at G = 1. This is now RTOT in function of the control voltage:

enter image description here

You can see a big range on the resistance value: from 100Ω to 225Ω. The range almost doubled, and we didn't even use the potential offered by the gain factor in the instrumentation amplifier.


2.2  Using the potential of the gain

Using the "potential offered by the gain factor" from the instrumentation amplifier could intuitively extend the range. However, that is not really the case.

Let us increase the gain a little to G = 1.5. This is what we get:

enter image description here

Just applying the formula would gives the dashed lines. But they are not representative anymore once RTOT dives under the value of RC. RTOT can never go below that resistance.

The range doesn't extend. Instead, it shifts downwards and gets tighter. So we have to keep the gain G = 1!


3.  My question

I've got a couple of questions concerning this circuit:

  • Is the way I'm using the PMOS okay? It's upside down...
  • Do you see another option (instead of inserting an instrumentation amplifier) to increase the range of RTOT?
  • Do you see any other issues in the circuit?

I'm going to build the circuit in a few days on a PCB, and would like to catch errors early before spending a lot.

  • \$\begingroup\$ What specs for Vcr vs R? And tolerances or acceptance criteria? \$\endgroup\$ Commented Aug 11, 2018 at 21:38
  • \$\begingroup\$ Hi @TonyEErocketscientist, you can find the specs of the R_VCR (the voltage controlled resistor on the left) here: electronics.stackexchange.com/questions/389347/…. \$\endgroup\$
    – K.Mulier
    Commented Aug 11, 2018 at 22:13
  • \$\begingroup\$ Simulate before building it. \$\endgroup\$
    – Andy aka
    Commented Aug 12, 2018 at 10:34
  • \$\begingroup\$ Hi @Andyaka, I would like to do that. But currently, I have two main drawbacks when it's about simulating electronic circuits: (1) Costs: For the moment, I cannot afford expensive simulation software. (2) Time investment: learning to use simulation software requires time and efforts. The past few years, I've been focusing mostly on embedded software design, and didn't need to build many PCB's. Nonetheless, I'd like to take on more circuit-building-projects over the next years, so investing time in learning software might be a good idea. What software do you advise? \$\endgroup\$
    – K.Mulier
    Commented Aug 12, 2018 at 10:42
  • \$\begingroup\$ Micro cap student edition is free. Yes there’s a learning curve but it is 100% worth it. \$\endgroup\$
    – Andy aka
    Commented Aug 12, 2018 at 14:42

1 Answer 1


Is the upside down PMOS okay?

Yes, it act as a reverse-polarity "diode": https://electronics.stackexchange.com/a/525249/59133

Another option (instead of inserting an instrumentation amplifier) to increase the range of RTOT?

I don't see a biasing method that you haven't already graphed and explored.

Any other issues in the circuit?

You tell us. I'm assuming you built the circuit in the few days since you said, and it's been almost three years since.


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