Several issues with op amp circuits

shaorc

Several issues with op amp circuits [Copy link]

 

As shown in Figure 1, the input signal VIN is given by the DSP chip. In Figure 1, R3 and C1 as well as R2 and C3 form a two-order low-pass filter. What is the principle of this filtering? It is easy to understand that R3 and C1 filter to ground, but I don't understand the principle of the connection of R2 and C3. In Figure 2, the input is a changing DC voltage value. Is the LM158, which was originally an op amp, used as a comparator in the figure? If the op amp is used as a comparator, then the op amp should be working in a positive feedback state, but it is still negative feedback in the figure.

maychang

"In Figure 2, the input is a changing DC voltage value. Is the LM158, which is originally an op amp, being used as a comparator?"

Figure 2 is obviously not a comparator, but an inverting amplifier.

maychang

"I don't understand the principle of connecting R2 and C3"

If the upper end of C3 in Figure 1 is grounded, it is a simple two-section RC low-pass filter. Now the upper end of C3 is connected to the output of the amplifier, which can increase the Q value and form an active filter (Sallen-Key filter).

maychang

Qualitatively speaking, when the signal frequency is much lower than ωo, the capacitive reactance of C3 is very large and there is not much positive feedback. When the signal frequency is much higher than ωo, the capacitive reactance of C2 is very small and the signal at the op amp's in-phase input is also very small. Only when the signal frequency is close to ωo does positive feedback play a relatively large role.

shaorc

maychang posted on 2020-9-15 17:45 "In Figure 2, the input is a changing DC voltage value. Is the LM158, which was originally an op amp, used as a comparator in the figure?" Figure 2 shows...

In Figure 2, if LM158 is still used as a reverse amplifier

What signal is amplified?

Is it to directly amplify the voltage divided by R2 on the input Vin?

Or amplify the difference between the voltage on R2 and the positive terminal (5V)?

The topology of Figure 2 is neither like a typical inverting amplifier circuit nor like a subtractor circuit.

maychang

shaorc posted on 2020-9-16 08:32 In Figure 2, if LM158 is still used as an inverting amplifier, what signal is amplified? Is it directly amplifying the voltage divider value of R2 on the input Vin? ...

"In Figure 2, if the LM158 is still used as an inverting amplifier, what signal is amplified?"

What is amplified is the change of Vin divided by R1R2. At this time, the input resistance of the inverting amplifier is the parallel value of R1 and R2. The low-frequency amplification factor is R3/(R1//R2).

maychang

shaorc posted on 2020-9-16 08:32 In Figure 2, if LM158 is still used as an inverting amplifier, what signal is amplified? Is it directly amplifying the voltage divider value of R2 on the input Vin? ...

"Does it directly amplify the voltage divided by R2 on the input Vin? Or does it amplify the difference between the voltage on R2 and the positive terminal (5V)?"

In fact, the voltage between the two input terminals of the op amp is the difference between the Vin voltage divider and the 5V voltage.

Therefore, the op amp can only work in the linear region when the difference between Vin after voltage division and 5V is very small. If Vin is slightly larger or smaller, the op amp will enter saturation.

shaorc

maychang published on 2020-9-16 08:48 "Does it directly amplify the voltage divider value of R2 on the input Vin? Or does it amplify the difference between the voltage on R2 and the positive end (5V)?" In fact, the operation...

【1】“What is amplified is the change in Vin divided by R1R2. At this time, the input resistance of the inverting amplifier is the parallel value of R1 and R2. The amplification factor in the low-frequency band is R3/(R1//R2).”

What is amplified is: the voltage divided by the change in VIN on R2? Why is the resistor at the reverse input end the parallel connection of R1 and R2?

【2】“Therefore, the op amp can only operate in the linear region when the difference between Vin after voltage division and 5V is very small. If Vin is slightly larger or smaller, the op amp will enter saturation.”

Does it mean that if the reverse input is much larger or much smaller than the 5V at the forward end, the op amp will enter the saturation region and act as a comparator? If both ends are 5V, what will the op amp output be?

maychang

shaorc posted on 2020-9-16 08:32 In Figure 2, if LM158 is still used as an inverting amplifier, what signal is amplified? Is it directly amplifying the voltage divider value of R2 on the input Vin? ...

"The topology of Figure 2 is neither like a typical inverting amplifier circuit nor a subtractor circuit"

Figure 2 is an inverting amplifier circuit, but a 5V bias is added to the non-inverting end.

maychang

shaorc posted on 2020-9-16 08:32 In Figure 2, if LM158 is still used as an inverting amplifier, what signal is amplified? Is it directly amplifying the voltage divider value of R2 on the input Vin? ...

The output of the op amp in Figure 2 varies around 5V. The ratio of this variation to the variation in Vin is the voltage gain of the amplifier circuit.

maychang

shaorc posted on 2020-9-16 09:40 【1】“What is amplified is the change in Vin divided by R1R2. At this time, the input resistance of the inverting amplifier is the parallel value of R1 and R2. The low-frequency amplification factor...

"Why is the resistance of the inverting input terminal the parallel combination of R1 and R2?"

It should be said that "the internal resistance of the signal source when the inverting input terminal looks toward the signal source" is the parallel value of R1 and R2.

For this problem, take R1 and R2 separately, and the internal resistance of the signal source looking to the left at the junction of the two should be the ratio of the signal voltage to the current to the left when an infinitesimal signal is applied to this junction. This ratio is exactly the parallel value of R1 and R2.

maychang

shaorc posted on 2020-9-16 09:40 【1】“What is amplified is the change in Vin divided by R1R2. At this time, the input resistance of the inverting amplifier is the parallel value of R1 and R2. The low-frequency amplification factor...

"It means that if the reverse input is much larger or much smaller than the 5V at the forward end, the op amp will enter the saturation region and act as a comparator?"

It is easy to see if you treat 5V as zero potential and "ground" as negative 5V (12V power supply as 7V).

maychang

shaorc posted on 2020-9-16 09:40 【1】“What is amplified is the change in Vin divided by R1R2. At this time, the input resistance of the inverting amplifier is the parallel value of R1 and R2. The low-frequency amplification factor...

"If both ends are 5V, then what is the output of the op amp?"

According to the previous reply, both input terminals of the op amp are zero, and the ideal op amp output (feedback through R3) is also zero. But this zero is after the potential moves, and before the potential moves ("ground" is zero), it is 5V.

maychang

shaorc posted on 2020-9-16 09:40 【1】“What is amplified is the change in Vin divided by R1R2. At this time, the input resistance of the inverting amplifier is the parallel value of R1 and R2. The low-frequency amplification factor...

"So the op amp goes into saturation and acts as a comparator?"

It cannot be said to "act as a comparator". Even if it is considered as a comparator, it is in saturation state, which is no different from the saturation state of an op amp.

shaorc

maychang posted on 2020-9-16 10:11 "So when the op amp enters the saturation region, does it act as a comparator?" It can't be said to "act as a comparator". Even if it is regarded as...

I understand the problem. Thank you.