transimpedance amplifier analysis - ijsmargan/articles/trans_z_amplifier.pdf · transimpedance...

Transimpedance Amplifier Erik Margan

- 1 -

Transimpedance Amplifier AnalysisErik Margan

System Description

In the transimpedance first order system is shown. It consists of anFig.1inverting amplifier accepting the input signal in form of a current from a highimpedance signal source, such as a photodiode or a semiconductor based detector forradiation particles, and converts it into an output voltage.

The transimpedance at DC and low frequencies is . However, the@ Î3 œ Vo i fhigh impedance signal source inevitably has a stray capacitance , which deprievesGithe amplifier from the feedback at high frequencies. Therefore the amplifier’sfeedback loop must be stabilized by a suitably chosen phase margin compensationcapacitance . Owed to the presence of these capacitances, and because of theGfamplifier’s own limitations, the system response at high frequencies will be reducedaccordingly.

The system analysis follows from the standard circuit theory in Laplace space.Upon the derived equations the system’s response can be optimized by a suitableselection of component values.

ii

Rf

Cf

oC i

A

1

( )s

Fig.1: Generalized transimpedance system schematic diagramme

Amplifier Description

The amplifier’s inverting open loop voltage gain is modeled as:@ =

@ = = = œ E = œ E œ E

o

ia b ! !

! !

! !

=

=(1)

where: is the complex frequency variable;= is the amplifier’s open loop DC gain;E!

is the amplifier’s real dominant pole, so that:=! , and is the open loop cutoff frequency= œ œ # 0 0 Þ! ! ! != 1

Note: the expression comes from the gain normalization of the function= Î = =! !a bJ = œ "Î = = J = œ J ! ÎJ =a b a b a b a b a b! , so that . This makes the low frequency gainNof unity and independent of , only its cutoff frequency depends on .J = = =Na b ! !


- 2 -

System Transfer Function

The sum of currents at the node is:@i

3 œ @ @ @" "

=G "

V =G

ii i o

i

ff

(2)

From (1) and (2) we have:

3 œ =G @@ @ " =G V

E E Vi i o

o o f f

fŠ ‹ (3)

Reordernig (3) gives:

3 V œ @ =G V " E " =G V"

Ei f o i f f fc da ba b (4)

The normalized transfer function is obtained by dividing by :@ 3 V! i f

@ "

3 V =G V " E " =G Vœ E

o

i f i f f fa ba b (5)

Since is a function (1) of :E =

@ "

3 V = œ E †

=G V " E " =G V=

o

i fi f f f

!!

!!

!

!

=

= =

=Œ a b (6)

We multiply the last term in the denominator:

@ "

3 V = œ E †

=G V " =G V E " =G V=

o

i f i f f f f f!

!

! !!

!

=

= =

=a b (7)

and multiply the numerator and the denominator by :a b= =!

@ E

3 Vœ

= G G V = " G V " E G V " E

o

i f i f f i f f f

! !#

! ! ! ! !

=

= = =a b a b a ba b (8)

We divide all the terms by the coefficient of the highest power of , which is=a bG G Vi f f , to obtain the canonical form of the normalized transfer function:

@

3 Vœ

†E " E

" E G G V

= = " G " E G V " E

G G V G G V

o

i f

i f f

i f f

i f f i f f

! ! !

!

# ! ! ! !

=

= =

a ba bc d a ba ba b a b(9)

The term is the system’s DC gain, and it is slightly lower thanE Î " E! !a bunity. The error is caused by the finite open loop gain. With being usually aboutE!

10 , the error is approximately , and is independent of frequency, so it can be& &"!neglected. The term represents the transition frequency (in rad s) of= =! !a b" E ÎTthe open loop amplifier, at which the amplifier has unit gain, . For highE œ "a b=T


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speed amplifiers, the open loop cutoff frequency is often between 1 and 10 kHz, so0!the transition frequency is usually between 100 MHz and 1 GHz, or slightly above.0T

To design the system, a set of limitations must be accounted for.

In the majority of cases we are given a signal source producing current inresponse to irradiation (consisting of either photons or particles). The source has aconversion sensitivity defined as a ratio of the produced instantaneous current byW 3the irradiation power , or (in A W). We would like to have some standardT W œ 3ÎT Îr rvoltage value for a standard amount of input power, say V µW, or similar,@ ÎT œ " Îo rso we need to select a suitable value of the feedback resistance to satisfy theVfrelation , so that (for constant input).@ ÎT œ V W @ œ V 3o r f o f

The source also has a stray capacitance (proportional to the detector’s activeGiarea, and the dielectric constant of the detector material, and inversely proportional toits thickness). This capacitance will cause a reduction of the feedback signal at highfrequencies, so the feedback loop must be phase compensated by a suitably chosenfeedback capacitance . The amplifier is chosen on the basis of its noise performanceGfand with enough bandwidth to cover the frequency range of interest, so once theamplifier has been selected, the only element by which we can optimize the systemwill be the feedback capacitance .Gf

However, even cannot be chosen at will. With a value too large the systemGfwill respond slowly, and with a value too small the response may exhibit a largeovershoot and long ringing, or even sustained oscillations. A system with a lowestsettling time has the poles in conform with a Bessel system. The Bessel system familyis optimized for a maximally flat envelope delay up to the system cutoff frequency,therefore all the relevant frequencies will pass through the system with equal delay,and the response will exhibit the fastest possible risetime with minimal overshoot.The optimal component values can be calculated from the system poles, which arethen compared to the normalized Bessel poles and scaled accordingly by the systemcutoff frequency. By comparing the system transfer function (9) with the generalcanonical normalized form (10) we obtain two equations from which the poles can becalculated. The general form of the transfer function with only poles is:

(10)œ K œ KJ == = = =

= = = = = = = = = =a b a ba ba ba b a b! !

" # " #

" # " # " ##

So we can find the system poles from the following two equations:

(11)œ= =" G " E G V

G G V" #

! != c da ba bi f f

i f f

(12)œ= =" E

G G V" #

! != a ba bi f f

For the second order Bessel system, normalized to the unity group delay, thevalues of the poles are:

= œ „ 4 œ " „ 4$ $ $ $

# # # $" #,

È È (13)


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To tune the system for the desired response we need to find the system polesfrom (11) and (12), and with (13) as the guide for achieving the necessary imaginaryto real ratio of the pole components.

We start by expressing from (12):="

œ=" E

= G G V"

! !

#

= a ba bi f f(14)

With this we return to (11):

œ =" E " G " E G V

= G G V G G V

= =! ! ! !

##

a b c da b a ba bi f f i f f

i f f (15)

We multiply all by and put everything on the left had side of the equation:=#

= = œ !" G " E G V " E

G G V G G V##

#! ! ! != =c d a ba ba b a bi f f

i f f i f f(16)

This is a second order polynomial, and it is solved using the standard textbookexpression of a general form:

+B ,B - œ !#

which has the roots:

B œ œ " „ 4 " , „ , %+- , %+-

#+ #+ ," #

#

#,È Ê

In our case (16) the coefficient , so we have:+ œ "

œ " „ 4 "=" G " E G V

# G G V

%" #

! !"E

G G V

" G "E G VG G V

#,i f f

i f f

= c da ba bÎ ÑÐ ÓÐ ÓÏ Ò

ÍÍÍÍÌ ’ “=

=

! !

! !

a ba bc da ba b

i f f

i f f

i f f

œ " „ 4 "" G " E G V % " E G G V

# G G V " G " E G V

= =

=

! ! ! !

! !#

c d a ba ba ba b Ëe fc da bi f f i f f

i f f i f f

(17)

It is difficult to obtain the required value of from this relation, since there isGfno simple way to solve it analytically. But we can always solve it numerically byentering the known component values and varying until the correct value (13) ofGfthe determinand (under the square root) of (17) is obtained:

% " E G G V "

" G " E G V " œ

$

=

=

! !

! !#

a ba be fc da bi f f

i f f (18)

Another possible way would be to calculate the envelope delay of the systemfor a range of frequencies and varying until the envelope delay is essentially flat upGfto almost the system’s cutoff frequency. We will see the relation for the envelopedelay a little later.


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In we have plotted the poles (17) as a function of changing fromFig.2 = G" #, f1 pF to 0.1 pF in steps of 0.01 pF. Initially both poles are real and the one closer to thecoordinate system’s origin is the dominant pole. As decreases the poles moveGftowards each other until they meet, at which point the system is critically damped. Byfurther decreasing the poles form a complex conjugate pair and travel along aGfcircle centered at the origin with a radius equal to the critically damped system values.Fig.3 shows the transfer function magnitude above the complex plane (Bessel poles).

-8 -7 -6 -5 -4 -3 -2 -1 0×107

-4

-3

-2

-1

1

2

3

4 ×107

σ

ωj

s1

s2

0

θ

Cf 1pF ... 0.1pFstep 0.01pF

Fig.2: The position of the poles changes with decreasing Gf

-6-4

-20× 107

-6 -4 -2 0 2 4 6× 107

0

1

2

3

4

5

6

σ

j ω

|M(s

)|

[rad/s] [rad/s]

Fig.3: The transfer function magnitude (absolute value) over the complex plane, shown herefor the Bessel poles case. Above the poles the magnitude is infinitely high. The shape of thesurface cut along the axis is the system’s frequency response (in linear scale).4=


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In most cases the system’s high frequency cutoff , at which the0 œ Î#h h= 1system gain drops by 3 dB relative to its value at low frequencies, can be calculatedfrom the expression for the last characteristic polynomial term (9):

==

hi f f

# ! !œ

" E

G G V

a ba b (19)

so the cutoff frequency in Hz will be:

0 œ œ# # G G V

" " Eh

h

i f f

= =

1 1 Ë a ba b! ! (20)

Note that this holds for only a secon order system optimized to have amaximally flat amplitude, or the Butterworth system, where the normalized poles are= œ # Î# „ 4 # Î# œ # Î# " „ 4 œ" #, È È ÈŠ ‹a b, and the angle ±135° (as measured)

from the positive real axis). Because the poles of a Bessel system have a differentlayout (the angle ±150°), the cutoff frequency will be lower by , see (13).) œ $Î$È

0 œ 0$

$h(Bes ) h(But )# #

È(21)

Transfer Function Magnitude (‘Frequency Response’)

The usual meaning of the term ‘frequency response’ is the magnitude (absolutevalue) of the complex frequency transfer function. The magnitude can be calculated asthe square root of a product of the transfer function with its own complex conjugate:

¹ ¹a b a b a bÉJ = œ J = † J =* (22)

Ordinarily we are not interested in the shape of the transfer function magnitude overthe entire Laplace complex plane (as in ), but only in the shape of that surface cutFig.3along the imaginary axis with := œ 4 Ð œ !Ñ= 5

¹ ¹a b a b a bÈJ 4 œ J 4 † J 4= = = (23)

We thus rewrite (9) in response to :4=

@

3 Vœ

†E " E

" E G G V

4 4 " G " E G V " E

G G V G G V

o

i f

i f f

i f f

i f f i f f

! ! !

!

# ! ! ! !

=

= == =

a ba ba b c d a ba ba b a b

(24)

We need first to separate the real and imaginary terms and rationalize thedenominator, then multiply the relation by its own complex conjugate, as in (22), (23):

@

3 Vœ

†E " E

" E G G V

4 4" E " G " E G V

G G V G G V

o

i f

i f f

i f f i f f

i f f

! ! !

!

# ! ! ! !

=

= == =

a ba ba b a b c da b a ba b (25)


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To keep the equations short we shall use the general form (10), so (25) becomes:

@ K † = =

3 Vœ

4 = = 4 = =

o

i f

! " ##

" # " #a b a b= =(26)

The denominator is rationalized by multiplying both the numerator and thedenominator by the denominator’s complex conjugate:

(27)œ@

3 V

K † = = † 4 = = 4 = =

4 = = 4 = = 4 = = 4 = =

o

i f

! " # " # " ##

# #" # " # " # " #

‘a b a b ‘ ‘a b a b a b a b= =

= = = =

The following terms are present in the denominator after the multiplication:

=4

= =" ## #

= ==#" #

#a b = = = == =# #

" # " #

4 = = 4 = == =$ $" # " #a b a b

4 = = = = 4 = = = == =a b a b" # " # " # " # (28)

The last three lines contain terms with alternating signs and they cancel. So therationalized form is:

@

3 Vœ

K † = = † = = 4 = =

= = = =

o

i f 4! " # " # " #

#

" ## # #

" ##

‘a ba b

= =

= =(29)

The imaginary unit is contained only in the last term in the brackets of thenumerator, so the complex conjugate will only occur there. The common (squared)terms can be extracted from the square root. Then the absolute value is:

(30)œ@ K † = = = = 4 = = = = 4 = =

3 V = = = = º º Èc dc da b a ba bo

i f 4! " # " # " # " # " #

# #

" ## # #

" ##

= = = =

= =

After multiplying of the brackets under the root the imaginary terms will havealternate signs and cancel, so the system’s transfer function magnitude is:

º º Éa b a ba b@

3 Vœ

K † = = = = = =

= = = =

o

i f 4

! " # " # " ## ## #

" ## # #

" ##

= =

= =(31)

Phase Angle

We calculate the phase angle of the transfer function of the order system: 8th

as the arctangent of the ratio of the imaginary to real part of the transfer function,which is equivalent to finding the individual phase shift of each pole: =5a b= œ „ 45 5 55 = and then summing them:

: = : == =

5a b a be fa be fa b " "œ œ œ

e J = …

d J =arctan arctan

5œ" 5œ"

8 8

55

5

(32)


- 8 -

Our frequency response function (32) has two complex conjugate poles,therefore the phase response is:

: == = = =

5 5Ð Ñ œ

arctan arctan

(33)" "

" "

Envelope Delay

We obtain the envelope delay as the phase derivative against frequency:

7: =

=e œ

.

.

a b(34)

Because the phase response (33) is a sum of individual phase shifts for eachpole, the same is true for the envelope delay. Each pole contributes a delay:

. . …

. .œ œ

…

: = = = 5

= = 5 5 = =

a b ” • a barctan

(35)i i

i i i# #

and the total envelope delay is the sum of the contributions of each pole.

For the 2-pole case we have:

75 5

5 = = 5 = =e œ

" "

" "# #

" "# #a b a b (36)

It is important to note that because the poles of stable systems are on the leftside of the complex plane, their real part must be negative. In (36) the denominators5are sums of squares and thus positive. So the envelope delay is a negative function,the negative sign indicating a time delay. Since this function is a ‘delay’, we mighthave neglected the negative sign. But there is a deeper meaning in this sign: it reflectsthe sense of rotation of he phase angle with frequency, and for real stable systems thephase always decreases with frequency. Whenever we see the phase increasing weshould watch for the possible source of instability within the system.

In the system’s transfer function magnitude is plotted, along with theFig.4phase angle and the envelope delay (phase derivative against frequency). For this plotthe system components have been chosen to conform with a Bessel second orderresponse (constant group delay almost up to the cutoff frequency).

The component values resulting in the Bessel system response are:

Amplifier Source Feedback

rad s kpF

pFE œ "! G œ !Þ&)

œ # † "! Î V œ "!!G œ (!

!&

!%= 1 H

if

f

With these components the poles (17) have the following values:

× × rad s= œ #Þ&(' "! 4 "Þ&!%" "! Î"( (

× × rad s= œ #Þ&(' "! 4 "Þ&!%" "! Î#( (

The imaginary to real part ratio is 1.5041 2.576 , which is wellÎ œ !Þ&)$*

within the component tolerances from the ideal value of 0.5774.È$ Î$ ¸


- 9 -

10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

-210

-180

-150

-120

-90

-60

-30

0

-80

-70

-60

-50

-40

-30

-20

-10

0

f [Hz]

F (s)| |

ϕ (s)dsd

A0 = 10 5

ω π2 ⋅10 40 = rad s/

Ci = 70 pF

Cf = 0.58 pFRf = 100 kΩ

[dB]

[°]

[ns]

Cf

Rf

A

i i

o

i

Ci

ϕ (s) = ℑ ℜ

F (s) F (s)

arctan

τ =

Fig.4: System transfer function (absolute value in dB), phase (in °) and group delay (in ns)

Input Impedance Analysis

The amplifier’s differential input resistance is assumed to be vey high (modernamplifiers having jFET or MOSFET input transistors have their input resistancewithin the range 10 – 10 ), and its input capacitance ( 2 pF) can be considered"# "% H as being a small part of . Then the system’s input impedance is:Gi

^ œ œ œ@ @ @

3 E † 3 E 3=

ii o o

i i i!!

!

=

=

(37)

We need again the relation between the input current and the output voltage (4):3 @i o

3 V œ @ =G V " E " =G V"

Ei f o i f f fc da ba b (4)

which we divide by :Vf

3 œ @ =G V " E " =G V"

EVi o i f f f

fc da ba b (38)

By inserting (38) into (37) we have:

^ œ@

E † =G V " E " =G V@

EV

io

o

fi f f fc da ba b (39)

We cancel the common terms in the numerator and the denominator:

^ œV

=G V " E " =G Vi

f

i f f fa ba b (40)


- 10 -

Replace with its full expression from (1):E

^ œV

=G V " E " =G V=

if

i f f fŒ a b!!

!

=

=

(41)

We multiply the numerator and the denominator by the term and regroup the= =!

coefficients having the same power of :=

^ œV =

= G G V = " G V G V " E " Ei

f

i f f i f f f

a ba b c d a ba b=

= = =

!#

! ! ! ! !

(42)

Divide by the coefficient at the highest power of :=

^ œ

V = "

G G V

= = " G V G V " E " E

G G V G G V

i

fi f f

i f f f

i f f i f f

a b a ba b a ba b a b=

= = =

!

# ! ! ! ! !(43)

Make the numerator’s frequency dependent term same as the last denominator term:

^ œ

V " E

" E G G V=

= = " G V G V " E " E

G G V G G V

i

f

i f f

i f f f

i f f i f f

a b a ba b a ba b a ba b a b

!!

!

# ! ! ! ! !

=

= = =(44)

From (44) we can extract three terms of the input impedance. The first one is afrequency independent term, which multiplies the two frequency dependent terms:

^ œV

" E"

!

fa b (45)

The two frequency dependent terms are: the unity gain normalized band-pass term:

^ œ

=" E

G G V

= = " G V G V " E " E

G G V G G V

#

!

# ! ! ! ! !

a ba ba b a ba b a bi f f

i f f f

i f f i f f

= = =(46)

and the unity gain normalized low-pass term:

^ œ

" E

G G V

= = " G V G V " E " E

G G V G G V

$

! !

# ! ! ! ! !

=

= = =

a ba ba b a ba b a bi f f

i f f f

i f f i f f

(47)

So the total input impedance (44) is:

^ œ ^ ^ ^i " # $a b (48)

It is clear from (45) that at low frequencies the input impedance must be verylow. Likewise, is unity when :^ = ¥$ =h


- 11 -

= ¥" E

G G VË a ba b=! !

i f f(49)

and then falls with frequency, owed exclusively to (because is in series with theG Gi famplifier’s output impedance, which increases at high frequencies, but which we haveneglected in this discussion). In contrast, increases with frequency and reaches a^#

maximum when and this maximum is proportional to the capacitance ratio:= ¸ =h

^ ¸ ^ "G

G# "max

i

fŒ (50)

after which falls off with frequency.^#

The absolute values (in ) of the input imedance and its components areHplotted in .Fig.5

10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 910 -5

10 -4

10 -3

10 -2

10 -1

10 0

10 1

10 2

10 3

f [Hz]

Z1

Z 2

Z 3

Z i

Z i = Z 1 Z 2 Z 3+( )

||

Z(

) s

Cf

Rf

A

i i

o

i

CiZ i = ii i

[Ω]

A0 = 10 5

ω π2 ⋅10 40 = rad s/

Ci = 70 pF

Cf = 0.58 pFRf = 100 kΩ

Fig.5: Input impedance with its components (absolute values in )H

Time Domain Calculation of the Impulse Response

The system’s impulse response can be calculated from the complex transferfunction by using the Laplace transform inversion via the Cauchy residue theory. Theprocedure has been introduced into circuit theory by Oliver Heaviside, who developedthe method independently of the existing mathematical knowledge.

The residue of a pole is found from the transfer function by canceling that poleand perform a limiting process in which approaches the value of that same pole; by=repeating the process for all the poles we obtain all the residues. The time domainresponse is the sum of all the residues.


- 12 -

A general order system has poles. For a system with simple poles (non-8 8th

repeating), with no zeros in the numerator, and if the impulse response is required, thegeneralized expression for the residue of the pole is:5 =th

5

< œ = == =

=

= =5 5

5

3œ"

8

3

3œ"

8

3

= > e (51)

limp

a ba ba b

$$

5

So our two-pole system with the pole values as in (13) or (17) wil have the residues:

< œ = = œ= =

= = = =

= = = = = =" "

"

" # " #

" # " #

= > = > e e (52)

limp

a b a ba ba ba b a b" "

< œ = = œ= =

= = = =

= = = = = =# #

#

" # " #

" # # "

= > = > e e (53)

limp

a b a ba ba ba b a b# #

In these equations we first cancel the corresponding and terms, anda b a b= = = =" #

then let assume the value of the particular pole. The system’s impulse response is=then equal to the sum of the residues:

œ < < œ C >= = = =

= = = =a b a b a b" #

" # " #

" # # "

= > = >

e e (54)" #

We can extract the common term:

œ C >= =

= =a b a b ˆ ‰" #

" #

= > = >

e e (55)" #

We can now write the poles in terms of their real and imaginary part, ,= œ „ 4" # " ", 5 =thus the time domain response (55) can be written as:

œ C > 4 4

4 4a b a ba ba b ‘5 = 5 =

5 = 5 =" " " "

" " " "

4 > 4 >

e e (56)a b a b5 = 5 =" " " "

We factor out e from both exponentials, rearrange the denominator and multiply the5">

numerator to obtain:

œ C >

#4a b ˆ ‰5 =

=" " > 4 > 4 ># #

"e e e (57)5 = =" " "

Since from Euler’s expressions of trigonometric functions follows:

e e4 > 4 >

"

= =" "

#4œ >sin=

we will have:

œ >C >a b 5 =

==" " >

# #

""e (58)5" sin

Note that the time domain response of any realizable function is alwayscompletely real (the imaginary components cancel)!


- 13 -

Step Response Calculation

The system’s step response can be calculated in two ways:

1) by the convolution integration of the product of the impulse response (59) and theinput unit step; this procedure is easy to execute numerically on a computer, butcan be very difficult and often impossible to do it analytically;

2) by multiplying the system’s transfer function with the Laplace transform of theunit step operator and performing the Laplace transform inversion via residuetheory; this process may sometimes be lenghty but is always easily managable.

We are going to follow the second procedure. The Laplace transform of theunit step function is . If we multiply the system’s transfer function by we"Î= "Î=obtain a three pole function, with the new pole at the complx plane origin ( ).! 4!

K = œ J = œ"

=a b a b a ba ba ba b= =

= = = = =" #

" # (59)

This function has three poles and therefore three residues. We find the residues of thecomplex conjugate pole pair in the same way as we did for the impulse response:

< œ = = œ œ=p=

= = = = =

= = = = = = = = = =" "

"

" # " # #

" # " " # " #

= > = > = > e e e

lim a b a ba ba ba b a b" " " (60)

< œ = = œ œ=p=

= = = = =

= = = = = = = = = =# #

#

" # " # "

" # # # " # "

= > = > = > e e e

lim a b a ba ba ba b a b# # # (61)

The residue for the third pole at will be:= œ !

< œ = ! œ œ œ "=p!

= = = = = =

= = = = = ! = ! = = =$

" # " # " #

" # " # " #

=> !> e e

lim a b a ba ba ba b a ba b (62)

So our step response will be the sum of these residues:

ga b> œ < < < œ " = =

= = = =$ # "

" #

# " " #

= > = >

e e (63)# "

As before, we can extract the common term from the last two terms:

ga b ˆ ‰> œ " = ="

= =" ## "

= > = >

e e (64)" #

and by writing the poles by their real and imaginary components:

ga b a b a b ‘> œ " 4 4"

4 45 = 5 =5 = 5 =

" " "" " " "

4 > 4 >e ea b a b5 = 5 =" " " " (65)

Again we cancel the terms with alternate signs and reorder the expression to obtain:

ga b ‘> œ " 4 4#4

ee e e e (66)

5= = = =

"

" " " "

>

"" " " "

4 > 4 > 4 > 4 >

=5 = 5 =

We regroup the real and imaginary part within the brackets:


- 14 -

ga b ‘ˆ ‰ ˆ ‰> œ " 4 #4

ee e e e (67)

5= = = =

"

" " " "

>

"" "

4 > 4 > 4 > 4 >

=5 =

We multiply the brackets by the external exponential term:

ga b ˆ ‰ ˆ ‰> œ " 4 #4 #4

e ee e e e (68)

5 5= = = =

" "

" " " "

> >

" "" "

4 > 4 > 4 > 4 >

= =5 =

By moving the denominators to the exponentials with imaginary exponents we get:

ga b> œ "

#4 #

5

="

"

> >4 > 4 > 4 > 4 >

e e (69)e e e e5 5

= = = =" "

" " " "

By again employing the Euler’s trigonometric idenitities we obtain:

ga b> œ " > >5

== =

"

"

> >" "e e (70)5 5" "sin cos

And again the resulting step response (70) is a completely real function.The normalized impulse and step responses are plotted in .Fig.6

0 50 100 150 200 250 300 350 400 450 500

0

0.2

0.4

0.6

0.8

1.0

Cf

Rf

A

i i

o

i

Ci

A0 = 10 5

ω π2 ⋅10 40 = rad s/

Ci = 70 pF

Cf = 0.58 pFRf = 100 kΩ

t [ns]

y (t )

g(t )

Fig.6: Impulse response and step response . The ideal second order impulse responseC > >a b a bgrises abruptly from zero; in reality, the presence of a distant non-dominant pole in the amplifier(beyond ) will round up and delay the initial impulse response rising. Because of this non-E! !=dominat pole, the step response will also exhibit a slightly increased delay.

Noise Sources and Noise Gain Analysis

The thermal noise sources of the circuit are modeled in .Fig.7

The amplifier has two non-coherent noise sources, a differential voltage noisesource and the input current noise source . The values of those noise sources are@ 3n nprovided by the amplifier’s manufacturer in the data sheets. The resistor has its ownthermal noise voltage source . All the values are given in terms of noise density@nV


- 15 -

functions (per 1 Hz bandwidth), and to know the actual rms noise we have to accountfor the system’s bandwidth.

Cf

Rf

A

i i

o

i

Ci

n

nR

in

mCf

Rf

Ao

i

Ci

nem

Fig.7: Thermal noise sources and the equivalent total noise source

As shown in , all the three noise sources are inside the amplifier’sFig.7feedback loop, therefore we can approximate all the noise sources by a singleequivalent source in place of , with a value of:@ @ne n

@ œ @ 3 V @ne n fn nÉ a b# #V# (71)

Because the noise sources are caused by independent random processes, they are non-coherent, non-correlated, so it is appropriate to sum their powers (voltage or currentsquared); otherwise the components could be summed directly.

Amplifiers with MOSFET or jFET input transistors usually have their inputcurrent noise very low, so in cases when:

3 V Ÿ@

$n f

n

the input current noise can be neglected.

The feedback resistor’s thermal noise depends on the resistor’s value, itstemperature, and the circuit bandwidth . Often , therefore the?0 œ 0 0 0 ¥ 0H L L Hbandwidth can usually be approximated by the upper cutoff frequency of the circuit,?0 ¸ 0 œ 0H h.

@ œ %5 XV 0n B fVÈ ? (72)

where:5 5 œ "Þ$) ÎB B is the Boltzmann thermodynamic constant, ×10 J K;#$

X X ¸ $!! is the absolute temperature in K; in low power circuits K.

In addition to these noise sources the signal source itself can have its ownnoise componets, i.e., the dark current white noise of a photodiode, which increaseswith reverse bias and temprature, and also a 1 noise that becomes important inÎ0cases where both the amplifier’s bandwidth and the signal source bandwidth extenddown to DC. However, the signal source noise cannot be distinguished from the signaland is processed by the system in the same way. In contrast, the system’s equivalentnoise source is inside the feedback loop, and is being processed by the circuit’s@nenoise gain.

It is very important to note that the ,noise gain is not equal to the signal gainin fact it can often be much higher!


- 16 -

The system’s noise gain is found by analysing the system’s response to thenoise source . We start again from the amplifier’s inverting input:@ne

@ œ œ @ @

E = E=

mo oa b

!!

!

=

=

(73)

The voltage at the node is then:@i

@ œ @ @i m ne (74)

The current summing at the node is:@i

@ @ @" "

=G

œ

"

V =G

i o i

i

ff

(75)

We regroup the coefficients of the voltage variables:

@ =G V " =G V œ @ " =G Vi i f f f o f fc d a ba b (76)

By inserting (74) into (76) we obtain:

a bc d a ba b@ @ " = G G V œ @ " =G Vm ne i f f o f f (77)

and by replacing with (73):@m

Œ c d a ba b @ @ " = G G V œ @ " =G V=

Eo ne i f f o f f

=

=!

! !(78)

We again regroup the voltage variables:

@ " = G G V œ @ " =G V " = G G V=

Ene i f f o f f i f fc d c da b a bœ =

=!

! !(79)

From (79) we obtain the noise gain expression:

@ " = G G V

@œ

" =G V " = G G V=

E

o i f f

ne f f i f f

a bc da b=

=!

! !

(80)

By some rearranging we get the final expression:

œ œK@

@

E " E

" E G G V= " E

= = " E " G " E

G G V G G G G V

no

ne

i f f

i f f i f i f f

f

! ! !

!! !

#! !

! !

a b a b” •a b a b” •a b a bŒ a b

==

==

(81)

The first thing we note is that the noise gain is a non-inverting function,because the sign is positive, which means that the phase of the source at lowfrequencies is the same as the phase of the output voltage. Further, the termE Î " E! !a b is the frequency indepencent gain error owed to the finite amplifier’sgain; since , this term can be neglected.E ¦ "!


- 17 -

In a similar way as the input impedance, the noise gain is a sum of twocomponents, one is a band-pass component:

K œ= " E

= = " E " G " E

G G V G G G G V

nBP

i f f i f i f f

f

=

==

! !

#! !

! !

a b” •a b a bŒ a b (82)

and the other is a low-pass component:

K œ

" E

G G V

= = " E " G " E

G G V G G G G V

nLPi f f

i f f i f i f f

f

=

==

! !

#! !

! !

a ba b” •a b a bŒ a b (83)

In we have plotted the noise gain (81) and its two components (82), (83).Fig.8

10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9-100

-50

0

50

f [Hz]

G [

dB] GnBP

GnLP

Gn

Cf

Rf

Ao

i

Ci

nem

A0 = 10 5

ω π2 ⋅10 40 = rad s/

Ci = 70 pF

Cf = 0.58 pFRf = 100 kΩ

Gnp ≈ CiCf

+ 1+20dB/10f

-20dB/10f

-40dB/10f

Fig.8: Noise gain and its componentsKn

Having determined the noise gain, we want to see its effect on the system’snoise spectrum. It is often assumed that the amplifier’s shot noise and the resistor’sthermal noise are essentially ‘white’, which (analogous to the white light) means thatall the noise frequencies are of equal power. In reality, certain resistor types, likecarbon film resistors, have an additional low frequency component noise (‘red’ or‘excess’ noise), inversely proportional to frequency, ~ . Likewise, all amplifiers"Î0have the ~ noise (below about 300 Hz), but certain amplifier types also have a ~"Î0 0(‘blue’) noise component, above 100 kHz. If the system’s low frequency cutoff (usingan additional high-pass filter after the amplifier) is above 1 kHz, we do not needVGto worry about the low frequency noise. However, the high frequency noise spectrumwill be within the system’s bandwidth in most cases, so its part below the upper cutofffrequency cannot be neglected.


- 18 -

From the system’s noise optimization view it is important to distinguishbetween the amplifier’s shot noise (short for Schottky, or quantum noise) and theresistor’s thermal noise (Johnson noise). The shot noise is the consequence of currentflow and structural imperfections of the conductor, but in metal conductors it is verylow because of the large number of free charge carriers; in semiconductors the numberof free charge carriers is much lower and because of the dopants there are relativelymany structural imperfections. The shot noise voltage is inversely proportional to thecurrent flow; in FETs it is also independent of temperature, but in bipolar junctiontransistors it is proportional to temperature because of the base-emitter equivalentresistance . In contrast, the thermal noise is present even if there is no< œ 5 XÎ; Me B e ecurrent flow in the resistor, and as the name implies it is a function of temperature,actually , as will be seen soon. The excess noise in carbon film resistors isÈX "Î0proportional to current flow. The noise in amplifiers is mostly proportional to"Î0junction leakage because of the reverse bias voltage (say, the collector-base voltage orthe drain-gate voltage in jFETs).

If the noise spectrum is not constant with frequency, we must take into accountthat we will be plotting it in the logarithmic frequency scale, which may influence theshape of the plot. It must be undestood that the white noise power spectrum (equalpower per 1 Hz bandwidth) can be drawn by a flat line only in a plot with a linearfrequency scale. In the usual scale every frequency decade is of equal size, so aloga b0decade between 1–10 kHz has 10× more 1 Hz bands than the decade between 100 Hzand 1 kHz. Likewise, in the octave between 1–2 kHz there are twice as many 1 Hzbands as between 500 Hz and 1 kHz. This means that a white noise power ploted inthe scale will be apparently rising in proportion with .loga b È0 0

However, in amplifiers we are mostly interested in the signal to noise voltageratio. Since voltage is proportional to the square root of power, the scale plotloga b0of the white noise voltage will be again constant with frequency. But other types ofnoise have different spectral distribution, so when plotting those in the scaleloga b0those differneces must be taken into account.

Amplifier manufacturers specify the amplifier’s current and voltage noisealready as a noise density function per 1 Hz bandwidth.

Obviously, in order to have the noise density for the resistor’s thermal noise,we must eliminate the from (72). Our of 100 k will thus have the noise? H0 Vfdensity:

/ œ %5 XV œ % † "Þ$) † "! † $!! † "! œ %!Þ( În B fV#$ &È È ÈnV Hz (84)

Assume that our amplifier has a voltage noise density of nV Hz , and/ œ ( În Èa current noise density of fA Hz , thus nV Hz . Our total3 œ "& Î 3 V œ "Þ& În n fÈ Èequivalen voltage noise density will be:

/ œ / 3 V / œ ( "Þ& %!Þ( ¸ %" Îne n f nnÉ a b È È# # # ##V nV Hz (85)

If the amplifier’s noise voltage contains a blue component above some frequency ,0b/ " 0Î0n b should be multiplied by . The noise voltage will start to decreaseÈbeyond amplifier’s transition frequency owed to the presence of secondary poles (notaccounted for in our simplified model).


- 19 -

So our dominant noise source is the resistor’s thermal noise. This will beamplified by the system’s noise gain:

/ œ K † /ns n ne (86)

Because this function is not constant with frequency, it is not appropriate tosimply multply it by . To obtain the actual noise voltage, the expression (86)È?0must be integrated in frequency. Analytical integration will be in most cases verycomplicated, so it is often done numerically. Alternatively, because the noise sourcesare not correlated, we may intgerate the individual components and then take the rootof the sum of squared values. The spectrum of the equivalent noise voltage and itscomponents (some suitably multiplied by the noise gain) is shown in .Fig.9

10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 910 -9

10 -8

10 -7

10 -6

10 -5

f [Hz]

Eqiv

alen

t Noi

se V

olta

ge D

ensi

ty

inRfRfCf

12π

en 1 + ffb

fb

enRene

ens = Gn ene

Fig.9: The spectrum of the equivalent noise voltage and its components.

The noise spectral density shows that the dominant noise will be in thefrequency range between 5×10 and 5×10 Hz, where the noise gain has its peak, with& (

values between 2 and 4 µV . Because of this pronounced peak the noise willÎÈHzappear to be not completely random, but rather having an oscillating component at ornear the frequency of the noise gain maximum. This is characteristic for all amplifiershaving a pronounced noise gain.

transimpedance amplifier analysis - ijsmargan/articles/trans_z_amplifier.pdf · transimpedance...

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