Mode voltage for op amp items

OP amp Pee

(a) we must first understand the virtual short, virtual break. This is what constitutes the op amp always

①

Virtual Break : 2, 32 points between the open circuit. No current (how much bit)

Virtual Short : 2, 32 point voltage, no pressure difference (somewhat)

② Why is there a virtual break in the op amp, short?

We take the CA3130 that we use often.

This is CA3130 's internal circuitry.

Chip internal circuit can be divided into 3 parts

1. The upper left part is a high input impedance current source (level 2)

2. The differential amplification circuit composed of the lower left field FET and the mirror current source composed of the BJT

3. The right part of the OCL composed of FET circuit (I think this chip in use, the output should be added capacitance ^_^)

We need to analyze the virtual break, virtual short simply look at the second part of it

Virtual Short : The Open-loop voltage amplification of general-purpose OP amps is more than DB. The output voltage of the OP amp is limited, typically at V~14 v.

So

Wd=%e8%bf%90%e6%94%be&hl_tag=textlink&tn=se_hldp01350_v6v6zkg6 "rel=" nofollow "> Op amp differential mode input voltage less than 1 MV. Two-input approximation

Wd=%e7%ad%89%e7%94%b5%e4%bd%8d&hl_tag=textlink&tn=se_hldp01350_v6v6zkg6 "rel=" nofollow "> Equipotential, equivalent to" short circuit ".

Open loop

Wd=%e7%94%b5%e5%8e%8b%e6%94%be%e5%a4%a7%e5%80%8d%e6%95%b0&hl_tag=textlink&tn=se_hldp01350_v6v6zkg6 "rel = "nofollow" > The higher the voltage magnification, the closer the potential of the two inputs is to equal. But it's not really a short circuit. (Baidu knows)

I think it is not possible to rely on the input voltage is small to come to a short conclusion, for the differential amplifier, single-ended input, and double-ended input, input resistor, the same voltage gain, perhaps this is one of the reasons

virtual break : Because op amp differential mode input resistance is very large, general-purpose

Wd=%e8%bf%90%e7%ae%97%e6%94%be%e5%a4%a7%e5%99%a8&hl_tag=textlink&tn=se_hldp01350_v6v6zkg6 "Rel=" The input resistors of the nofollow "> op amp are above 1mω. Therefore, the current flowing into the op-amp input is often less than 1uA, much smaller than the current of the circuit outside the input terminal.

Therefore, the two inputs of the OP amp can usually be considered as open circuit, the equivalent of a "circuit breaker" and the greater the input resistance, the closer the two inputs to the open circuit. But it's not really a disconnect.

Own previous notes, to be able to read the lower differential circuit

(ii) OP amp Circuit

When we know the virtual short, the virtual break can be composed of several operational circuits and the role in the overall circuit

① Linear, nonlinear circuit

For the first half of the common-shot common-base, we'll talk about

Linear nonlinearity In the graph we can understand this simply

1, linear: negative feedback

2, non-linearity: No negative feedback (for circuits with positive feedback that is the voltage comparison device.) or hysteresis voltage comparator called Schmitt trigger)

Negative feedback: There is a device connection to the output end of the reverse input terminal

Several arithmetic circuits of ②

Remember: The principle of calculation is:

This two point voltage is equal (for this circuit voltage is 0) virtual short

This two-point current is zero and the virtual break

Operation:

So gain: Au =-rf/r1

The same can be based on the properties of the capacitor to form integral differential circuit

The following is the integral differential circuit

^_^ not thoughtful place to leave a message ah

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Mode voltage for op amp items