I'm building a VCR (Voltage Controlled Resistor) using the LM13700 component. You can read more details on that here:
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:
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:
From which we can deduct that:
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:
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).
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 ΔVIN ∼ I2.
I've also calculated this relationship, and I get the exact same value for RTOT as before, namely:
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:
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:
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.
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:
According to my calculations, the relation RTOT ∼ RVCR is now:
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:
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:
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.