gain bandwidth product - Understanding of Dominant Pole Compensation - Electrical Engineering Stack Exchange

according miller effect, within amplifying devices such transistors have inverting voltage gain higher one, there should increased input capacitance increased voltage gain of amplifier.

i has been said effect limiting amplifier @ higher frequencies. comes incorporation of method of increasing transistor's bandwidth. 1 of them dominant pole compensation.

according wikipedia:

"when capacitor introduced between input , output sides of amplifier intention of moving pole lowest in frequency (usually input pole) lower frequencies, pole splitting causes pole next in frequency (usually output pole) move higher frequency."

i part when states 2nd pole should moved higher in frequency (that 2nd roll-off), why heck want decrease 1st pole , consequently first gain roll-off begins @ lower frequencies before. what point of it?

i mean, shouldn't desired gain vs. frequency curve flat possible, long possible - higher frequencies? addition of capacitor between input , output node of voltage amplifying stage, 1st pole decreased in frequency , @ same time flat gain curve became shorter before.

i kind of upper plot trying achieve there - constant decrease of gain increase of frequency; straight dotted line 1st 2nd roll-off of gain curve. still, gain decreasing in area, decreasing! wouldn't desired flat long possible?

there still 2 or 3 questions unclear me (if comments added post can edited):

• referring phase margin: between 2 points phase shift occurs? within negative feedback? added basic 360° shift output of amplifier input of negative feedback?

also read somewhere phase margin negative feedback positive feedback factor. can represented in such manner?

i add comments phase shift.

the phase shift show here

is phase shift between input , output terminal of amplifier without feedback.

next should ask question -180° means?
means voltage @ output reversed (180° out of phase).

and if add feedback network , feed signal inverting input receive positive feedback amplifier ( output voltage start rise).

this unwanted phase shift inside amplifier caused parasitic component (capacitor , inductor). these 2 components provide inertia in circuit because electric field , magnetic field cannot change instantaneously, time needed. , why have phase shift in amplifier.

and here can see how "number" of "pole's" effect phase shift

here can see how phase margin effect's step respond

as can see phase margin , "circuit speed" inversely related.

edit

the simplest closed loop system

where:

\$ \$ open-loop gain ( forward gain )

\$\beta \$ feedback factor ( feedback network gain )

in case \$ \beta = \frac{v_c}{v_d} = \frac{r_y}{r_x+r_y}\$

base on can write equation because want find closed loop gain

\$a_{cl} = \frac{v_{out}}{v_{in}} = \frac{v_d}{v_a}\$

so have

\$v_d = v_b \cdot \$ (1)

\$v_b = v_a – v_c \$ (2)

\$v_c = v_d \cdot β \$ (3)

and can calculate closed loop gain, substitute 1 2

\$ v_d = (v_a – v_c) \cdot = v_a \textrm{a} - v_c \textrm{a}\$ (4)

now take 3 (vc = vd*β) , substitute 4

\$v_d =v_a \textrm{a} - v_c \textrm{a}= v_a\textrm{a} - v_d β \:\textrm{a} \$

\$ v_d + v_d \cdot β \cdot = \cdot v_a \$

\$ v_d (1 + aβ) = a\cdot v_a\$

and

$$a_{cl} = \frac{v_{out}}{v_{in}} = \frac{v_d}{v_a} = \frac{a}{1 + aβ}$$

this important equation. if divid get:

$$ a_{cl} = \frac{1}{(1/a) + \beta} $$

we can see closed loop gain equal

\$\large \frac{1}{\beta} = 1 +\frac{r_x}{r_y}\$ recognizing equation?

if open loop gain \$a\$ large (ideally \$a = \infty \$) approaching infinity.

but let form

$$a_{cl} =\frac{a}{1 + aβ}$$

this equation true negative feedback circuits.

and let see happens if \$a\beta = 1\$ (at frequency)

$$a_{cl} =\frac{a}{1 + aβ} = \frac{1}{1 + 1} = 0.5$$

for negative feedback circuit.

but let see happen if add delay equal 180 degrees phase shift @ frequency. 180 degrees reversing sign of sinewave.

hence, negative feedback became positive feedback.

therefore \$aβ\$ became \$-1\$

and denominator of \$a/(1+aβ)\$ changes sum difference.

so closed loop gain \$a_{cl} =\frac{a}{1 - aβ}\$ (positive feedback).

and again if @ frequency have \$a\beta = 1\$

the closed loop gain becoming

$$a_{cl} =\frac{a}{1 - aβ} = \frac{1}{ 1 - 1 } = \frac{1}{0}$$

wow, create amplifier infinite gain though open-loop gain not infinity. means can have output 0 volts @ input.

in real world, amplifier (with negative feedback) oscillate @ frequency 2 contrition met.

the magnitude of loop gain \$a\beta(j\omega) = 1\ \$ (0db) , additional phase shift reaches value of -180deg.

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also, noticed have troubles recognizing type of feedback in amplifier circuits.

the negative feedback vs positive feedback @ "dc".

1 - rise in voltage @ point x cause voltage @ point z rise.so if connect these 2 points (x , z) have positive feedback on circuit.

in case (dc) positive feedback dos not automatically means osculations. instead of oscillation, circuit can latch @ positive or negative rail.

2 - time rise in voltage @ point a corresponds voltage drop @ point d. so, if connect 2 points (a d) end negative feedback circuit.

try yourself.

as homework try recognize type of feedback in 2 circuits

the first circuit

and second circuit

the feedback resistor \$r_{fb}\$