Opamp ciruit

The buffer is to prevent loading of the voltage created across the 100 ohm.

The 100 ohm is the most important part of the circuit, as it does the current to voltage conversion.

The pot is for setting the reference for the output.

Short answer is yes you could probably get rid of buffer.

It all depends on accuracy specs of the circuit goals.

The buffer probably could get dropped as its load, 10K + 31.25K, would
have negligible impact on the 100 ohm, in fact its affect ~ .25% error in
the 100 ohm, the non inverting path G, assuming all R's perfect values.

The transfer f() of the converter is basically a straight line, y = mX + b,
and m is the slope/scale factor, b is the 1.2V offset controlled by pot.
One could even look into replacing the 10K, 31.25K, and buffer with just
the computed scale factor formed by the 31.25K and 10K, eg. the value
of the I to V conversion R, the 100 ohm. Assuming one uses a low Ibias
OpAmp for the diff amp portion, so that offset due to bias current of
OpAmp does not contribute significant error.

Using worst case analysis, and desired accuracy, one could eliminate the
pot if the +15V was a precision Vref. Typically its a 5% type supply, so
worst case analysis again would determine if it can be eliminated, and also
determines precision of R's needed. Alternatively use a precision 1.2V Vref
part,


Also OpAmp bias, Voffset, PSRR, all contribute to error, so again worst case
analysis to see if circuit meets target accuracy/precision specs is prudent.

Note the 100 ohm AND the ratio of the 31.25K/10K, and the 1.2V Vref are
the circuits most important considerations in determining G (slope) and
V offset in y = mX + b.

There are many more considerations in 4 20 ma current loops -




Regards, Dana.

Thank you for your clarifying answers Nigel and Dana! I am still having trouble with getting a deep understanding of the circuit though...

Could someone help me reflect on the questions below?
1. Is the pot only used for precision tuning?
2. Do we always need an additional voltage source when we know that there will be an offset?
3. Isn't the 1150ohm enough to set the inverting input to -1.2V?
4. Do we have a buffer to make sure that the current doesn't "travel further"?
5. Are the transfer functions of most opamp circuits a straight line y = mx + b? How would you figure this out?

Keep in mind typically you have a sensor whose output is conditioned
into a current to transmit over a long cable/wire/twisted pair. Pressure,
Temperature, Force.....and you want an accurate reading of that sensor.
So thats a linear function that you need the 4 - 20 mA current loop to
conform to. You could transform in a non linear fashion, and then
linearize at the receive end, but that is much more complicated to get
accurate results.

1. Used to shift the straight line transfer function up or down to get
desired response at receiver with its Rload (converting I back to V).
"Normally" one wants @ 4 mA the receiver to see 0 V for input to ADC.

2. Yes, you need a method to insure the 4 - 20 mA range produces a
known V range at receiver, which is controlled by the receiver Rload
and the current. Otherwise you would have to inform the receiver,
somehow, what the offset is so you can conform to the transfer
function, straight line.

3. Yes if and only if the -15 is accurate. Otherwise you need to trim it
or use a V reference device.

4. Not sure what you are asking here.

5. Many circuits are amplifiers, so primarily used as linear functions.
Many circuits are non linear, used for other work. You figure out the T(s) by
analysis, DC using Thevenin and Norton, and AC using LaPlace or time domain
analysis using differential equations. Non linear analysis much more complicated,
many papers in IEEE and modern textbooks on methods to analyze non linear
circuits.







Regards, Dana.

dandedandelion said:

1. Is the pot only used for precision tuning?

Click to expand...

Yes, it is there so that the -1.2v can be set accurately to -1.200 volts, not 1.1?? or 1.2??.

dandedandelion said:

3. Isn't the 1150ohm enough to set the inverting input to -1.2V?

Click to expand...

No, the 1150 Ohm fixed resistor and the 100 Ohm potentiometer form a voltage divider.

dandedandelion said:

2. Do we always need an additional voltage source when we know that there will be an offset?

Click to expand...

If we have an offset, we need to get that offset from somewhere.
Either on of the op-amp supplies, if they are stable enough, or a separate dedicated supply.

dandedandelion said:

4. Do we have a buffer to make sure that the current doesn't "travel further"?

Click to expand...

A very simplistic naive way of thinking about it, as already stated it provides isolation between the summing amplifier and the 100 Ohm input conditioning resistor.
The circuit may work quite well without it, but as op-amps often come two or four in a pack, there is no great cost to incorporating the extra amplifier.

dandedandelion said:

5. Are the transfer functions of most opamp circuits a straight line y = mx + b? How would you figure this out?

Click to expand...

Many are.
To work out what is going on, you need to analyse and understand the individual circuits.

JimB