Operation amplifier from narrow gain sheet pole amplifier is infinite gain approximately amplifier of sheet pole operation, derivation gives the gain of circuit of the amplifier that cross block, if pursue,1 is shown.
 

 

Graph 1: One looks be like simple circuit to have two parts of an apparatus only: Operation amplifier and feedback resistor.

 

Be informed from the first part, derivation gain crosses impedance to be namely:

 

 

 

The utmost is:

 

 

 

Amplifier gain makes we have an opportunity to use control at circuit theoretically. The importance when this example controls the specification character of trends of circuit of academic and reasonable solution and practical. Carry out stage by stage, is not whole lot of a head comes in, the hope can reach his to apply means to have thorough knowledge to controlling a technology so.

 

The utmost is right (2) multinomial expresses to be normally:

 

 

 

The τn = 1/ωn = 1/(2 X of syntonic time constant of amplifierπX Fn) and damp ζ are respectively:

 

 

 

When ζ<1, the utmost turns into plural the utmost is right, extremely horn is:

 

 

 

To solid the utmost, ζ >1 andφ= 0.

 

To constant group (or bag sth resembling a net) when delay (the biggest even defer that include sth resembling a net / MFED or Beisaier) answer, phasic reduce along with frequency linear, and happen inφOf = 30o extremely horny part. Of all frequency when delaying is identical, keep undee and changeless. Next:

 

 

 

Answer to crossing block amplifier MFED:

 

To critical damp (what did not rush too is the fastest rank jump answer) , ζ= 1 and τT = 4 X τi or FT = Fi/4. Two the utmost are Fi/2.

 

Greaten as RR, Fi is reduced, amplifier is shown in Vix bigger too strong. Going up somehow, this to Z-meter it is advantageous, because extremely hornφ= 45° , dampζ= Cos(φ) = Cos(45o)≈0.707, and frequency (or extent) answering is constant or even, be close to bandwidth frequency. This is the biggest even range (MFA) frequency is answered. To stable state (frequency domain) application, MFA is answered is first-rate. The transient state that answers to having ideal rank to jump (time domain) application, MFED is answered is first-rate. (in oscillograph in the design of perpendicular amplifier, optimizing two kinds of response standards is conflict. )

 

 

Slow carry is put have low FT and τT >>τi, bring about two solid the utmost to leave further. In limiting value:

 

 

This is origin and the utmost that Fi is in. FT must enough small in order to maintain FT <<Fi. However, reduce as FT, annulus road gain decreases, the likelihood can’t maintain error of concessional operation amplifier gain. Below this kind of circumstance, precision needs fixed rate.

 

Put the addition of FT as carry, the damp of Zm is reduced, stability is reduced. To given ς and Fi:

 

 

 

Be like FT = 1MHz and G0 = 105, criterion FG = 10Hz, and (of critical damp loopζThe Fi = 40Hz of = 1) . Hypothesis Ci = 10pF, so RR = 398MΩ , can maintain Fi to any lesser values so>40Hz.

 

Graph 2 showed closed circuit the limit as FT (rapidder movement is put) the circumstance that increase and moves. In origin and the depart that Fi(–1/τi) is in the utmost is in Fi/2(right nowπ= 1) place gathers together, turn into next plural the utmost is right. Increase as FT, extremely horn increases and ζ is reduced. Amplifier becomes not stable, answer more oscillation.

 

 

Graph 2: Closed circuit the utmost moves as the addition of FT.

 

The parameter that wants change only (graph 2 medium FT or τT) the S2 that appears in multinomial at the same time and S in, the position with mobile the utmost or contrail can show in the graph. Amplifier is when infinite FT damp is the smallest, when τT→Position of the maximum when 0s is in limiting value:

 

 

 

There are two values on Jxω axis, its answer be stability (is not oscillation) : Origin and ±jx ∞ are in. Both is infinite, (Zero(0) is the) of infinitesimal. When τT→When 0s, two in the S of the utmost multinomial adjacent 0, leave 1 constant, and do not suffer frequency effect. Below ultimate circumstance, the utmost is located in on Jxω axis, ζ= 0 (the condition of the oscillator) , but be in in the finite value of S, their extent is 0. The utmost frequency is very tall, damp no longer important. They and Fi apart too far and won’t affect annulus road trends. This is the condition of ideal operation amplifier. Accordingly, we can conclude, to very slow or very fast operation amplifier, the utmost is sufficient depart, answer stability in order to make. It is only inside the limits of FT, this fortune calculates amplifier and Ci the limit too stand by, damp is in enough part of small Fn of the utmost frequency is excessive reduce, the vibration with quite big range produces in amplifier at the same time.

 

Return the amplifier that cross block again, if operation amplifier is ideal almost, that is to say, speed gets on for τT≈0s, criterion the utmost multinomial is about 1. To enough fast operation amplifier, FT >>Fi, and the utmost is apart, can have stable annulus way. To offer additional damp, make operation amplifier FT (with annulus road gain) too won’t low, capacitor Cf needs to pass RR billabong. Use the circuit algebra calculation that includes Cf next:

 

 

 

The utmost is to parameter:

 

 

 

The action of Cf is in 2 coefficient lieutenant general τf adds τi, more important is to add linear medium τT, this meeting increases damp. Because of τi = τT, so:

 

 

 

To critical damp, setπ= 1; so τT = (3 + 2 X √2) X τi≈3.414 X τi and τn≈1.848 X τi. If do not have Cf (Cf = 0pF) , be like previous place calculative, τT = 4 X τi. If have Cf, answer in identical trends below, operation amplifier is OK faster, have taller G0 to realize taller precision namely.

 

Frequency answers extent and phasic be:

 

 

The fast operation amplifier to ideal (τT = 0s) and when Cf = Ci(τf = τi) , have in frequency Fg(or ωg) place answer:

 

 

 

If Fi = 10 X Fg, so ％ of extent error ≈0.5. Because of Fi = 10 X Fg, phasic error≈6o. Phasic error is compared to frequency effect more sensitive to extent error. This is in design of impedance plan circuit is very important, be sometimes in photoelectricity exploration amplifier is very important also, because photoelectricity exploration weaveform wants as undee as a few other synchronism.

 

 

The Z-meter that crosses block amplifier to a few belts (ZM) design, RR wants enough big, namely 10MΩ or bigger. When RR becomes very big, want to get the damp of expectation, billabong Cf must very small, and capacitance of resistor billabong parasitism returns a likelihood pass big. To avoid this problem, can use the following circuit to replace.

 

Graph 3: Use this circuit to avoid capacitance of resistor billabong parasitism too big.

 

Should let operation amplifier make tall gain amplifier of sheet pole operation, G≈–1/s X τT (refer to this series article the derivation of concerned G of the first part) . Feedback scale down implement transfer function is:

 

 

 

And τf = RR X Cf. Use Rp = R1 when circuit | | When R2 seeks solution:

 

 

Ideal operation amplifier (the Zm of τT = 0s) reduces:

 

 

 

To Rp = 0Ω , cross block to reduce further to:

 

 

If fast ×1 buffer amplifier is inserted between output and RR and Cf, criterion R1 and resistor of output of R2 voltage divider do not need too small (Rp <<RR) . So when Rp = 0Ω and when operation amplifier has τT:

 

 

 

This circuit and did not output scale down implement the situation has two different: RR and τT increased 1/Hdiv effectively.

 

 

 

Through the article of two parts elaborate can see, even if only the simple circuit of two parts of an apparatus involves complex dynamic derivation possibly also. Design personnel can avoid to use these derivation to reduce mathematical calculative to bother sometimes, but use these formula to be able to understand the property performance that gives circuit to fall in all sorts of conditions better. The analysis of the amplifier that cross block that we introduce can offer a pattern plate for such circuit design, offer the guiding give a demonstration that how analyses amplifier trends character.

 

Because,do not want cubic or taller second polynomial and reject to use S region algebra to solve circuit trends problem. We encountered a cubic metre in this example, but not necessary go solving it, because can fall multinomial through simplifying,be quadratic equation, convenient the following analysis is calculated. This kind of circumstance is very common, because circuit is in design phase often by modular, each other keep apart their or, or undertakes through controlled port impedance controlled and alternant. Pattern plate program can use in the design, but normally S of be confined to inside the country quadratic equation.