Analog electronic technology, instantaneous polarity method to judge positive and negative feedback

When discussing instantaneous polarity, the DC component of the signal is not considered! You think vcc=0 is fine, although it's not!

in the transistor. The collector applies Kirchhoff's current law.

Current in the C-E direction of the triode branch = current from the RC branch VCC (in this case, ground) to node N (the node common to R2, RC, C2, and the triode collector) + the ground to node current of the RLC2 branch.

Due to the triode base.

The voltage is positive, so the collector-emitter current of the triode is also positive, then the two currents on the right side of the above equation are positive, since the current flows from ground to node n, then the potential of node n is lower than ground, and the voltage uo is also lower than 0.

As for "why the instantaneous polarity of the UO to the ground is -, the polarity of the UO feeding back to the input via R2 will also be -" is almost impossible to explain, isn't it logical? If you have to explain, the current II will be separated by the R2 branch, which is the if, which causes the current IID to decrease, because the size of the IID current is proportional to the ICE, so the ICE current also decreases, which is called negative feedback.

Finish. However, the above "insist on explaining" and the latter statement are not very scientific and cannot be put on the table. In the final analysis, the invention of instantaneous polarity analysis is only to establish an intuitive method for judging negative feedback, and to master the intuitive method, it is necessary to have an intuitive understanding of the circuit, which is a thing that practice makes perfect.

It's like how many tea leaves are put in tea, everyone relies on feelings, and they don't weigh the scale. Engineering originally needs to be rigorous, but for the sake of convenience, some intuitive cognitive methods are often established, sometimes it is difficult to explain thoroughly, of course, it is not impossible to explain, the above instantaneous polarity method is strictly proven, but the discussion is complex and not suitable for amateur solutions!

As can be seen from the figure, R2 is the feedback line. Let the polarity of the left end of R1 be positive, then after passing through the triode, the polarity of the collector pole of the triode will be negative (the triode can be regarded as co-ejection, and the input and output are inverted). ）。

Through R2, the polarity of Fan Kuo at point C1 is negative, so this feedback is voltage negative. Personally, I think that judgment feedback can be judged directly, and try not to remember the rules.

The instantaneous voltage polarity of the C and B poles of the triode is opposite, and the resistance does not change the polarity.

For the inductance with the middle tap, the instantaneous polarity judgment method: if either end of the head and tail is AC grounded, the polarity of the middle tap and the other end is the same; If the middle tap is AC grounded, the polarity of the head and tail ends is opposite.

For this problem, the upper end of the inductor L is AC grounded (CB is considered to be AC shorted), where the polarity of the middle tap is the same as the lower end.

It is mainly distinguished by the polarity of the feedback signal, if the polarity is the same or the phase is the same, it is positive feedback, otherwise it is negative feedback. The so-called instantaneous polarity method can be understood as the polarity at a certain moment in time, for example: an op amp is connected to the reverse input, and the feedback signal is fed back from the output to the reverse input.

It can be analyzed as follows: assuming that the input signal is positive at some point, the phase is reversed through the op amp, the output is negative, and the feedback to the input counteracts the positive input signal, and this feedback is negative.

Negative feedback sampling is generally done using current sampling or voltage sampling. Because negative feedback has its own unique advantages, it is widely used in actual amplifiers, and it changes the performance of amplifiers. The use of negative feedback stabilizes the closed-loop gain of the amplifier and eliminates the effect of open-loop gain.

This bai is simple, this is just an analysis of the du method, don't care about it, just know that it can be used to judge the positive and negative feedback; Teach you a simple way.

Regardless of what type of feedback it is, just remember that the feedback object and the original signal plus object are added together, and the result is larger than the original signal, then it is positive feedback. On the other hand, if the sum is smaller, it is a negative feedback.

For example; When the input signal voltage continues to increase, the output signal voltage is also increasing, and the increased signal voltage is introduced (returned) to the input terminal, then the result must be greater than the original signal, because the input signal is not only the original signal but also the amplified output signal voltage.

Negative feedback is still this example, if the output voltage is not increased but decreased, then the increased signal voltage is introduced (returned) to the input terminal, then the result must be smaller than the original signal, which means that the amplifier circuit can not be amplified as required, then.

The input signal in this case is the difference between the original input signal and the feedback signal.

For a single-stage arithmetic amplification circuit, the feedback circuit from the output back to the inverting input is negative feedback, and the feedback circuit from the output back to the non-inverting input is positive feedback.

The relationship between the base voltage vb and the emitter voltage ve is:

vb=vbe-ve

Among them, VBE is constant, so when the emitter voltage Ve increases, the base voltage Vb decreases, and when the emitter voltage Ve decreases, the base voltage Vb increases, so it is a negative feedback.

The first two equations (i.e., f and af) are actually definitions. The feedback coefficient f is the output signal (xf) of the feedback network compared to the continuous input signal (xo); The closed-loop gain (i.e., the closed-loop amplification) is the output signal (xo) of the entire closed-loop circuit (i.e., the negative feedback circuit) compared to its input signal (xi). Under the condition of deep negative feedback, since af=1 f, xi=xf can be obtained after substitution, that is, the essence of deep negative feedback is that the input signal and the feedback signal are equal (that is, the net input signal after superposition is 0).

If you still can't understand it, you should compare it with the diagram of the negative feedback circuit in the textbook.

Hope it helps!

No matter how you look at this circuit, it is a negative feedback circuit.

Method 1, the feedback of the circuit is connected from the output to the reverse input, and such a circuit is negative feedback;

Method 2: instantaneous polarity method, because the input terminal selects the reverse input terminal, when the input signal at the input terminal is instantaneous +, the output is -, and the feedback signal is also -, because the feedback signal is connected in parallel with the input signal, the feedback signal is inverted with the input signal, so the feedback is negative feedback.

STEP1 Set the instantaneous polarity of the UI signal to +

STEP2 Since the UI acts on the inverting input of the integrated op amp, the polarity of the output signal is - [At the same time, answer that this is an inverting amplifier, if you don't understand this, then you have to look at the differential amplification circuit].

Step3 The feedback signal and the input signal are superimposed at the inverting input terminal, that is, acting on the same node Step4 The signals with different polarities at the same node are superimposed as negative feedback [it can also be understood that under the action of positive UO, the IF current to the right is generated, and the III current generated under the original positive polarity is shunt, so that the net input signal IID becomes smaller, so it is negative feedback].