The problem of mutual influence between the operational amplifier voltage follower circuit and the amplifier circuit in parallel

flyriz

The problem of mutual influence between the operational amplifier voltage follower circuit and the amplifier circuit in parallel [Copy link]

 

Hello everyone, as shown below:

An AC input signal is input to the input end of a voltage follower circuit and an amplifier circuit. The purpose is to select through the optocoupler according to the amplitude of the input signal. The problem now is: when the frequency of the input signal is relatively low and the amplitude is relatively large (such as 4.4V, 4HZ), the optocoupler selects the voltage follower circuit below. At this time, the signal on the right side of C129 will be abnormal. At this time, if:
1. Short-circuit C129, the signal is normal;
2. Disconnect R87 or change R85 to 3K, the signal is normal;
3. Reduce the input signal, such as from 4.4V to 1V, the signal is normal
4. Increase the frequency of the input signal, such as from 4HZ to 20HZ, the signal is normal
In any of the above situations, the signal will return to normal. See the following pictures for normal and abnormal signals:

Please ask everyone, what is the cause of this?

@gmchen Mr. Chen, could you please take a look too?

flyriz

This phenomenon can also be restored by simulation.

maychang

I guess C129 in the picture is damaged. Try another one.

flyriz

maychang posted on 2020-4-24 08:27 I guess C129 in the picture is damaged. Try another one.

There is no damage, because the simulation also has this phenomenon.

maychang

flyriz posted on 2020-4-24 08:22 This phenomenon can also be restored by simulation.

Previously, we considered the possibility that D20 was asymmetric in two directions, but when we removed D20 during simulation, this phenomenon still occurred, so we ruled out this possibility.

maychang

"Reduce the input signal, for example, from 4.4V to 1V, the signal is normal"

The input is 1V, and the upper op amp is not saturated (gain 10 times), but the upper op amp is saturated when the input signal amplitude is larger. The waveform distortion may be related to the saturation of the upper op amp.

But this reason cannot explain why "the signal becomes normal when the frequency of the input signal is increased, for example, from 4HZ to 20HZ."

flyriz

maychang posted on 2020-4-24 12:11 "Reduce the input signal, for example, from 4.4V to 1V, the signal is normal" The input is 1V, and the above op amp does not enter saturation (gain 10 times) ...

Yes, there is also this phenomenon that cannot be explained:

1. Short-circuit C129, the signal is normal;

maychang

flyriz posted on 2020-4-24 12:43 Yes, there is also this phenomenon that cannot be explained: 1. Short-circuit C129, the signal is normal;

"Short-circuit C129, the signal is normal" This can be explained. The op amp enters saturation, so that the op amp's non-inverting input is no longer in a state of very large input resistance, and there is current at the non-inverting input during part of a cycle of the signal. If the signal is connected in series with C129, then the non-inverting input current during "part of a cycle" will charge and discharge C129. But if there is no C129 (short circuit), these currents will be completely absorbed or provided by the signal source.

flyriz

Comparing the five phenomena mentioned above, I think the reasonable explanation is: in the deep saturation state, the op amp cannot form negative feedback, and the feedback coefficient F is basically 0. According to the input resistance expression of the series negative feedback amplifier circuit: (1+AF)Ri, the input impedance of the circuit at this time will be reduced to the input impedance Ri of the op amp itself. OP2177 does not provide the input impedance parameters in the official data. I estimate that it will not be very high, so I am embarrassed to write it out. Generally, the input impedance of the Bipolar-input op amp is only tens to hundreds of kΩ. So at this time, the input impedance of the circuit drops sharply to the level of hundreds of kΩ or even lower. The impedance of the 2.2uF capacitor at 4HZ is 18kΩ, so the input waveform is distorted. In this way, the following five phenomena are reasonable.
1. Short-circuit C129, the signal is normal;
after the short circuit, the resistance is always 0, and the signal will not be affected.
2. Disconnect R87 or change R85 to 3K, the signal is normal;
the amplification factor becomes smaller, and the circuit does not enter deep saturation.
3. Adjust the input signal to a smaller value, such as from 4.4V to 1V. The signal is normal. The
input signal becomes smaller and the circuit does not enter deep saturation.
4. Increase the frequency of the input signal, such as from 4HZ to 20HZ. The signal is normal.
After the frequency increases, the impedance of the capacitor decreases.
5. Replace the op amp
with OPA2134. OPA2134 is FET-Input, and the input impedance can be regarded as infinite.
I wonder if you have any other opinions.

@maychang

maychang

flyriz posted on 2020-4-24 15:54 After comparing the five phenomena mentioned above, I think the reasonable explanation is: in the deep saturation state, the op amp cannot form negative feedback, the feedback coefficient F is basically 0, and the root...

In the deep saturation state, the op amp cannot form negative feedback, and the feedback coefficient F is basically 0. According to the input resistance expression of the series negative feedback amplifier circuit: (1+AF)Ri, the input impedance of the circuit at this time will be reduced to the input impedance Ri of the op amp itself.

I think it is a linear working state (not saturated), the voltage between the two input terminals of the op amp is very small, almost zero (virtual short). But after the op amp enters saturation, the voltage between the two input terminals increases, and the input impedance at this time is obviously different from that when it is not saturated.

maychang

flyriz published on 2020-4-24 15:54 After comparing the five phenomena mentioned above, I think the reasonable explanation is: in the deep saturation state, the op amp cannot form negative feedback, the feedback coefficient F is basically 0, and the root...

After changing to OPA2134, the two input tubes are FETs. When the voltage between the two input terminals is not zero (for example, 0.5V), the resistance between the two input terminals is certainly different from that of the bipolar tube.

For JFET, when the voltage between the two input terminals is larger (for example, 2V), it will definitely be different from when the voltage between the two input terminals is close to zero, and it will also change.

埋土书生

You have a power supply problem, I can reduce the simulation frequency to 1hz and increase the voltage to 9v without any problem

埋土书生

Maidu Shusheng published on 2020-4-25 13:07 You have a power supply problem, I can reduce the simulation frequency to 1hz and increase the voltage to 9v without any problem

If it is not the power supply, then it may be that the 2177 above is saturated, the op amp input characteristics have changed, and the input bias current has increased. You can consider reducing the input to ground resistance to 10k, or choose a mosfet input device.

埋土书生

Or you can consider adding an analog switch to the front, choose a follower for large signals and a 10x amplifier for small signals

maychang

flyriz posted on 2020-4-24 15:54 After comparing the five phenomena mentioned above, I think the reasonable explanation is: in the deep saturation state, the op amp cannot form negative feedback, the feedback coefficient F is basically 0, and the root...

After switching to OPA2134, this problem was solved.

However, there are other considerations for this circuit from the outset.

The purpose of this circuit is to amplify the signal by 10 times when the signal amplitude is small, and not to amplify the signal when the signal amplitude is large. Two op amps are used to achieve 10 times amplification and no amplification respectively, and an optocoupler is used to select the output of the two op amps.

If you want to control the gain of an amplifier, so that the input signal is not amplified when the amplitude is large, and is amplified 10 times when the signal amplitude is small, then only one op amp is needed. There are many variable gain amplifiers, even programmable variable gain amplifiers, on the market. Even if you use an ordinary op amp, you can also achieve gain control by using an analog switch to select whether the input signal is attenuated to 1/10 or not attenuated, and then amplifying it. In this way, you don't need to use an optocoupler, and only one op amp is enough.

maychang

flyriz posted on 2020-4-24 15:54 After comparing the five phenomena mentioned above, I think the reasonable explanation is: in the deep saturation state, the op amp cannot form negative feedback, the feedback coefficient F is basically 0, and the root...

Some op amps have two diodes (actually two transistors with collectors and bases short-circuited) connected in anti-parallel between their two input terminals, such as NE5532, as shown in the figure below. When the voltage between the two input terminals of the op amp is very small (such as a few mV), the resistance of the two anti-parallel diodes is very large (linear working state), but when the voltage between the two input terminals is slightly larger, the resistance decreases sharply (saturated working state).

There is no circuit diagram in the OP1117 manual. If the same is true for the input of OP1117, then the deformation of the input waveform after the op amp enters saturation can be easily explained.