Op Amp Circuit Course Design Analog Electronic Technology

There's nothing I can do ......

Another 200 points wasted .........

Don't be in a hurry, these books are available, go to the library to check the information, or find some exercise sets, analyze from similar topics, and soon be able to do a good job, your description involves debugging, its purpose is to let you do experiments, test data and waveforms, these need to go to the laboratory to do experiments, if you really can't find laboratory experiments, you can first get the value of the theory, and then make some modifications on some waveforms, because there is a deviation between theory and practice

I can't see **, there is a virus, please describe it in words.

This is what patent documents require.

Just add a buzzer with a bridge circuit.

Do your own research and research, you won't waste much time watching movies and playing Dota.

The two inputs of an ideal amplifier are short-circuited and disconnected, which means that the voltage between the two terminals is equal but no current flows.

In the circuit, the potential of both the positive and negative ends of the amplifier is 0, and all the current passing through R3 flows into the RF.

The current flowing through R1 is 3V (R1+R2 R3)=1mA, because R2=R3, the current distributed to R3 is.

Hence vo=

The inverting input of the OPA at the bottom is a virtual ground, and the midpoint of the four resistors is also a virtual ground. Therefore, the current of R1 and R4 is equal, and the OPA on the top is used as a voltage follower, and its output voltage is equal to UO1, and R6 has no current, which is also equal to UO2.

uo2=uo1=-（r4/r1)ui

Since the midpoint of the four resistors is imaginary ground, the resistor r2 has no voltage and no current, i2=0

The comparator uses positive feedback, which constitutes a hysteresis comparator.

First of all, the comparative functional analysis is to do a DC analysis and remove the capacitance.

Secondly, due to the particularity of the internal circuit structure, the 339 output needs a pull-up resistor to remove R16.

Let's analyze the circuit again:

When the comparator input is 0 (very low), the output level is high, and the voltage at the non-inverting input of the comparator becomes a voltage v1 between and (specifically the voltage divider resistor), that is, the flip threshold from low to high is v1;

When the comparator input is very high), the output level is high 0, and the voltage at the inverting input of the comparator becomes a voltage v2 between 0v and 0v (specifically the voltage divider resistor), that is, the flip threshold from high to low is v2;

Due to v1 > v2, the comparator is hysteresis.