Gain of operational amplifier...

[Externet]

Hi.
Is the gain in this case, Rf/Ri = R1/R2 = 10K/100 = 100. Is that correct ?
The coil itself has 1000Ω DC resistance.
Is the gain still 100 or is it Rf/Ri = 10K/ 1100 = 9 ?
If R2 is substituted by 0Ω, the coil being 1KΩ. What is the gain ?

The point is if the coil can be considered as a {0Ω coil plus a 1KΩ resistor} in series to pin 2.

 

[Arouse1973]

The gain is 1+(R1/R2) for a non-inverting stage. The coil is on the input so the gain remains the same. If R2 is zero, well it can't be zero but will be very low then the gain will be very high, too high I suspect for the circuit to work correctly.

Thanks
Adam

 

[Audioguru]

You posted a very fuzzy schematic saved as a Jpeg file type instead of a very clear PNG so I cannot read the opamp part number. The datasheet of every opamp shows a graph of its gain from DC to a high frequency where its gain is less than 1. When you short R2 then you are removing all negative feedback so its DC gain is millions then drops slowly as frequencies are higher.

 

[Ratch]

Hi.
Is the gain in this case, Rf/Ri = R1/R2 = 10K/100 = 100. Is that correct ?

No.

The coil itself has 1000Ω DC resistance.
Is the gain still 100 or is it Rf/Ri = 10K/ 1100 = 9 ?

No.

If R2 is substituted by 0Ω, the coil being 1KΩ. What is the gain ?

AC or DC gain?

The point is if the coil can be considered as a {0Ω coil plus a 1KΩ resistor} in series to pin 2.
View attachment 38943

Assuming DC gain is of interest, the input resistance at point 2 is in the multi megohm range. Therefore the relatively small resistance of 1 kilohm is not going to affect the voltage from point 1 very much. So, 1 mv applied at point 1 will be 1 mv at both points 2 and 3. That means 10 ua will exist in R2. All that current has come from R1, so the voltage across R1 will be 100 mv. Add to the 1 mv at point 3 gives 101 mv at point 6. Therefore the DC gain is 101 mv/1mv = 101.

Ratch

 

[hevans1944]

It appears (upon further magnification and eye squinting with tongue precisely positioned just so...) that the op-amp is an Analog Devices AD8628 rail-to-rail device. @Ratch's unusually non-pedantic analysis of the DC gain is a spot-on first-order analysis of a typical non-inverting gain. How could @Externet not know about this already?

 

[AnalogKid]

If R2 is decreased to zero, the circuit becomes a comparator, not a linear amplifier.

To determine if the opamp has enough bandwidth for your application:

1. Determine the highest frequency of interest. This might be a higher harmonic of a complex signal waveform.

2. Determine the maximum circuit gain needed.

3. On the opamp datasheet, find the chart of open loop gain versus frequency.

4. Locate your circuit gain and signal frequency operating point on the chart.

5. A good rule of thumb is for your operating to be at least 20 dB below the curve. This is enough headroom for the feedback loop to "close" properly. It also reduces phase distortions caused by the opamp's internal frequency compensation.

ak

 

[Ratch]

If R2 is decreased to zero, the circuit becomes a comparator, not a linear amplifier.

To determine if the opamp has enough bandwidth for your application:

1. Determine the highest frequency of interest. This might be a higher harmonic of a complex signal waveform.

2. Determine the maximum circuit gain needed.

3. On the opamp datasheet, find the chart of open loop gain versus frequency.

4. Locate your circuit gain and signal frequency operating point on the chart.

5. A good rule of thumb is for your operating to be at least 20 dB below the curve. This is enough headroom for the feedback loop to "close" properly. It also reduces phase distortions caused by the opamp's internal frequency compensation.

ak

If R2 is very low or close to zero, the large current supplied by R1 will lock up the circuit to the positive rail voltage as fast as the slew rate can take it there.

Ratch

 

[AnalogKid]

the large current supplied by R1

0.3 mA is less than 1% of the short circuit current limit.

ak

 

[Ratch]

0.3 mA is less than 1% of the short circuit current limit.

ak

The ability of the opamp to supply the current is not the problem. If 1 mv is applied to the input and R2 is 0.1 ohm, then 10 ma of current will have to be supplied by the opamp via R1. That makes 100 volts across R1. Of course the opamp will hit and stick at the voltage rail long before that happens.

Ratch

 

[AnalogKid]

As above, for low values of R2, the circuit behavior is that of a comparator with a 10 K load, not a linear amplifier. We still do not know the input signal voltage range at the non-inv input.

ak

 

[hevans1944]

We still do not know the input signal voltage range at the non-inv input.

Well, the back-to-back diodes across the "Search coil winding" sort of places a maximum limit of ±0.7 V DC (if silicon diodes). However, since it is a "search coil winding" it is likely that the anticipated signal is much less. With a closed-loop gain of 101 and a negative supply rail of -3 V, this rail-to-rail op-amp would swing to the negative rail with anything more than a mere 3/101 V input signal, or about 30 mV. We don't know what the positive supply rail is and of course @Externet hasn't bothered to inform us as to WTF he is trying to DO here, so any further speculation is a FWOT.

 

[Ratch]

As above, for low values of R2, the circuit behavior is that of a comparator with a 10 K load, not a linear amplifier. We still do not know the input signal voltage range at the non-inv input.

ak

Is this what you are thinking about?

Ratch

 

[AnalogKid]

Is this what you are thinking about?

No. There is no hysteresis in the post #1 schematic.

ak

 

[AnalogKid]

Well, the back-to-back diodes across the "Search coil winding"

Brain fade, forgot about the ground connection on the left side of the coil.

ak

 

[Ratch]

No. There is no hysteresis in the post #1 schematic.

ak

Yes, there is. A op amp cannot just switch from one voltage rail to the other instantaneously. See https://www.google.com/search?q=hys...j69i57j0l4.13407j1j7&sourceid=chrome&ie=UTF-8

Ratch

 

[AnalogKid]

Yes, there is. A op amp cannot just switch from one voltage rail to the other instantaneously.

OKaaaayyy...

True, but slew rate and intentional circuit hysteresis are not related.

There is no direct relationship between an opamp's output stage slew rate and it's input stage gain and performance as a level detector. In a circuit where an opamp is acting as a comparator (no feedback of any kind to either input), there probably are microvolts or a few millivolts difference between the effective voltage trip points of the upward and downward direction crossovers, probably due to base charge differences or something else related to the fact that the two transistors in the input differential pair have significantly different Vce's and collector currents as the input stage goes from overdriven in one direction to overdriven in the other direction. I think it is fair to say that that is *not* what is meant when the concept of circuit hysteresis is discussed on this forum.

Separate from that -
In the circuit in post #1, there is no external positive feedback explicitly designed and configured to shift intentionally the "trip point" of the circuit when acting as a conventional voltage comparator.

ak

 

[Ratch]

OKaaaayyy...

True, but slew rate and intentional circuit hysteresis are not related.

There is no direct relationship between an opamp's output stage slew rate and it's input stage gain and performance as a level detector. In a circuit where an opamp is acting as a comparator (no feedback of any kind to either input), there probably are microvolts or a few millivolts difference between the effective voltage trip points of the upward and downward direction crossovers, probably due to base charge differences or something else related to the fact that the two transistors in the input differential pair have significantly different Vce's and collector currents as the input stage goes from overdriven in one direction to overdriven in the other direction. I think it is fair to say that that is *not* what is meant when the concept of circuit hysteresis is discussed on this forum.

Separate from that -
In the circuit in post #1, there is no external positive feedback explicitly designed and configured to shift intentionally the "trip point" of the circuit when acting as a conventional voltage comparator.

ak

Nevertheless, the output will go from one rail to the other depending on the polarity of the input. Hysteresis determines when it starts to move and slew rate determines how long it takes to get somewhere. If you call an overdriven op amp a comparator fine, but I would call it a clipped amplifier.

Ratch

 

[AnalogKid]

Nevertheless, the output will go from one rail to the other depending on the polarity of the input. Hysteresis determines when it starts to move

That is not my understanding of what hysteresis is.

All digital and saturated analog circuits - opamps, comparators, diff amps, inamps, logic gates, single-ended inputs, differential inputs, whatever - are linear amplifiers that are overdriven to output saturation. (Exception: ECL) There are many internal circuit configurations, but they all operate the same way - as the input voltage transitions across a threshold value or transition level, the output voltage moves from saturated against one rail to saturated against the other rail. If, when the input moves back across what is essentially the same transition level in the opposite direction, the output responds with a similar movement back to the original rail, that is not an example of hysteretic action.

If the input moves across the transition level slowly enough, the output spends significant time in its non-saturated, linear active region. In this state the device is acting as a very high gain amplifier with nothing but circuit noise as its input, so the output can appear as a noise burst. A hysteresis circuit specifically prevents this by modifying the inout transition level based on the state of the output. There are several ways to accomplish this, but most of them involve some form of positive feedback.

The high gain linear state has uses. A section of a CD4049 CMOS hex inverter makes a nice electret microphone preamp.

ak

 

[Externet]

Thanks, gentlemen. That fuzzy schematic does use a AD8628. Other similar circuits with the sense coil and other instrumentation amplifier ICs show the coil directly to inputs, but have provisions for gain setting.

The intention is attempting to detect a 1Tesla magnet slowly spinning at less than 1 Hz from the farthest possible distance. Hopefully 50 metres. Coil is iron core over 10 thousand turns enameled #36. Output is to a plain zero-centered galvanometer. Can also be counterparalleled LEDs.
I was looking at different approaches for such high sensitivity needs and wondered what role would the coil DC resistance affect other than only raising R1 value for higher gain.

Environment is very quiet a mile or more away from dwellings, power lines. disturbances.

 

[Ratch]

That is not my understanding of what hysteresis is.

All digital and saturated analog circuits - opamps, comparators, diff amps, inamps, logic gates, single-ended inputs, differential inputs, whatever - are linear amplifiers that are overdriven to output saturation. (Exception: ECL) There are many internal circuit configurations, but they all operate the same way - as the input voltage transitions across a threshold value or transition level, the output voltage moves from saturated against one rail to saturated against the other rail. If, when the input moves back across what is essentially the same transition level in the opposite direction, the output responds with a similar movement back to the original rail, that is not an example of hysteretic action.

I said that hysteresis determines determines when something will move. I should have added that when it will move after a stimulus is applied. Operating a linear amplifier in saturation in not hysteresis, even though hysteresis my affect its characteristics.

If the input moves across the transition level slowly enough, the output spends significant time in its non-saturated, linear active region. In this state the device is acting as a very high gain amplifier with nothing but circuit noise as its input, so the output can appear as a noise burst. A hysteresis circuit specifically prevents this by modifying the inout transition level based on the state of the output. There are several ways to accomplish this, but most of them involve some form of positive feedback.

The high gain linear state has uses. A section of a CD4049 CMOS hex inverter makes a nice electret microphone preamp.

ak

All the above two paragraphs are applications of hysteresis. However, the definition of hysteresis is the lag of response to a stimulus.

Ratch