Adding voltage follower problem

Hi,
I designed and implemented a log amplifier that requires fixed voltage shift (bias) of its output. For that I added an adding voltage follower, but I am unable to get the voltage to be lineary shifted as required.
The expected voltage shift is significantly lower than expected of an adding circuit and is not linear.
Can anyone help with a working solution?
See attached schematics.

Common problem, I think. You have to learn the hard way. The summers etc. don;t work unless fed with a low Z source.
I think you need to buffer the reference. Lift a leg of R13 toward R12 and drive R13 with a power supply and it will likely work.

Another thing, I might do, is put a low value resistor between pin 6 of your buffer and the output connector. It's not really necessary, but it helps isolate capacitive loads and provides current limiting. Only works with buffers.

I suspect it would work much better using the opamp in inverting mode (with a second on the output to re-invert it if needed) as a virtual earth mixer. I'm also pretty dubious about the 4V reference source?.

This would be easier if *all* components had reference designators.

U1's 10 K output resistor, R12, R13, and R14 form a variable attenuator. The percentage of U1's output that appears at the U2 input varies with the adjustment of R12. Likewise, the offset voltage source at the R12 wiper is attenuated by over 50% at the U2 input (assuming the pot is not at its extreme rotation).

If you want to stick with this arrangement (as opposed to adding a true (inverting) summing stage before an inverting output buffer), then change the voltage reference to an adjustable type. This will present a consistant zero-ish ohm source impedance for the offset summing resistor that is independent of the wiper position. There still will be a 50% attenuation of the U1 output. National, ADI, and Linear Tech make adjustable references.

NOTE: Thinking this further through, this circuit has a problem that has no simple fix:

If the U1 output is 0 V and the offset is 4 V, what appears at the U2 input is 2 V (50% of the offset).
If the U1 output is 2 V and the offset is 4 V, what appears at the U2 input is 3 V, not 6 V.
If the U1 output is 4 V and the offset is 4 V, what appears at the U2 input is 4 V, not 8 V.

Because the U2 input is not a virtual ground, it allows interaction between the two currents; either current can go either direction through either resistor. The more narrow the U1 output voltage range, the less the interaction will be noticeable; but it never will go away and the voltage seen by U2 never will be a simple offset of the voltge at the output of U1. The error curve will be logarithmic.

ak

To get a linear shift, you can use an opamp configured as a differential amplifier with gain=1.
Connect the positive input to the signal you want to offset and the negative to the offset voltage.

You can add or subtract a voltage like that while keeping the same "span".

For unity gain, just use four equal value resistors:

Your offset reference circuit needs to be low impedance compared to the resistors in this part of the circuit.

rjenkinsgb said:

Your offset reference circuit needs to be low impedance compared to the resistors in this part of the circuit.

Click to expand...

And as with the original circuit, a 100 ohm pot introduces a 1% error with 10 K resistors. The alternative is an adjustable reference with a consistently low output impedance.

But - with a true diff amp, the offset voltage has to be negative (with respect to GND) to create a positive offset at the output.

ak

AnalogKid said:

But - with a true diff amp, the offset voltage has to be negative (with respect to GND) to create a positive offset at the output.

Click to expand...

Yep, I'm confusing myself...

Connect the negative input of the circuit to ground and reconnect the existing ground (resistor from amp + in) to the reference offset voltage.
That should add in a positive value.

Hi all,
Being new to this forum (although old in age...), I am overwhelmed with the many quick replies, all of which seem to be relevant. I believe I joined the perfect forum. Thanks.
As I thought to begin with, I'll try the one using an inverting adder (rather than the non-inverting) and will add an inverting v follower to get the positive output.
Yet, one question to rjenkinsgb: I am not sure I quite understand your idea that seems to be too simple. You suggest to connect the amp negative input to ground (?). The amplifier negative input is part of the unity feedback, connected to the amp. output.
Can you please elaborate ?

samtal said:

Hi all,
Being new to this forum (although old in age...), I am overwhelmed with the many quick replies, all of which seem to be relevant. I believe I joined the perfect forum. Thanks.
As I thought to begin with, I'll try the one using an inverting adder (rather than the non-inverting) and will add an inverting v follower to get the positive output.

Click to expand...

As I see it that will do everything you wish, and as it's a virtual earth mixer there's no interaction between the two inputs - it's the classic adder circuit from analogue computers.


Although I do think the formulas shown don't really do much except confuse

samtal said:

You suggest to connect the amp negative input to ground (?). The amplifier negative input is part of the unity feedback, connected to the amp. output.
Can you please elaborate ?

Click to expand...

Sorry for the confusion; not the opamp IC input, but the input of the overall "differential amp" circuit with the four resistors.

Using four equal value resistors, that translates the difference between the input terminals to an equal voltage between output and (normally) ground or 0V.
If you reconnect the "ground" point on that circuit to your voltage ref to be added, the output will then be relative to that instead.

In other words, add two equal resistors in your existing circuit - from pins 2 to 6 and from 2 to 0V on the OP07; that should give you the correct configuration.

That's all true, but the problem with any of these circuits is that the offset must be variable, and any arrangement based on a simple pot will introduce a variable error. The old school way is a fixed reference and a pot driving a unity gain inverter, that drives the inverting input of a true diff amp. Lotsa parts, but zero interactions.

ak

AnalogKid said:

That's all true, but the problem with any of these circuits is that the offset must be variable, and any arrangement based on a simple pot will introduce a variable error.

Click to expand...

Indeed.
As someone stated earlier, around 1% max with 10K resistors and the 100 Ohm pot.
That is a low input current amp, so using 100K resistors may be OK and improve accuracy somewhat.

Or, add a 100 ohm pot in series with the negative feedback resistor to balance out the slight gain change.

Well, based on my knowledge and help from members, I completed now the log electrometer amplifier.
I originally made it to replace a bad vintage vacuum Ion Gauge controller electrometer, but it can be used as a general purpose logarithmic amplifier to cover up to 7 decades of input current / voltage.
In my case I only use it for 5 decades, 1nAmp (10E-9A) and up to 100uA (10E-4A).
The offset voltage is just to match it with the original instrument and is not required otherwise.
Thanks all contributors.

Would it be an Varian 843 Ion gauge by chance?

The log thing was an Analog Devices part and I think, mostly unlabled.

samtal said:

1nAmp (10E-9A) and up to 100uA (10E-4A).

Click to expand...

1 nA = 1 x 10e-9
100 uA = 100 x 10e-6

ak

AnalogKid Good catch!

KeepItSimpleStupid said:

Would it be an Varian 843 Ion gauge by chance?

The log thing was an Analog Devices part and I think, mostly unlabled.

Click to expand...

No, it was Arun Microelectronics PGC2 vacuum gauge controller that can use any type vacuum tube. It was from the 90s, originally using discrete components log amplifier. Excellent piece, but went bad and I couldn't fix it, so I used a TI log IC instead.

AnalogKid said:

1 nA = 1 x 10e-9
100 uA = 100 x 10e-6

ak

Click to expand...

What do you mean? Was there anything wrong in what I wrote?

I think so. In the E-notation subset of scientific notation, the letter E is the symbol for 10 and what follows is the value of the base-ten exponent. In this format, 10E-9A means ten times (ten raised to the -9 power), or 10 uA.


In my post #15, the two instances of e should be capitalized. This prevents confusion with the natural logarithm.

ak