1. FREQUENCY MODEL FOR AN AMPLIFIER - Pessina …pessina.mib.infn.it/Corsi_del_IV_anno/Dispense 2017/Il... · is done mathematically following different models that better fit the

1Frequency Response of Amplifiers gpessina

1. FREQUENCY MODEL FOR AN AMPLIFIER

The introduction of the frequency behavior into the model of amplifiers is done mathematically following different models that better fit the choice of the device:

Dominant pole amplifier

Two dominant poles amplifier, with τA >> τB

Interpretation of the above settings: the actual number of poles present in the circuit is much more than those considered above (one or two per transistor and the transistors used in an OA are several…) but, if the preamplifier is well designed, only a few are dominant, namely they are located at frequencies much smaller than the others, so that, at first, the effect of the other poles can be neglected.

Circuit modelling is done this way:

+

- Vo

+

-1

Ao R

CτA=RC

+

- Vo +

-

Ao

1RA

CAτA=RACA

1RB

CBτB=RBCB

lunedì 26 giugno 2017

Three dominant poles amplifier. with τA >> τB e τB >> τC

A(s) =Ao

1 + sτA

A(s) =Ao

1 + sτA 1 + sτB

A(s) =Ao

1 + sτA 1 + sτB 1 + sτC

Vo =Ao

1 + sτAV+ − V−

Vo =Ao

1 + sτA 1 + sτBV+ − V−


2. STABILITY CRITERIA FOR A FED BACKED AMPLIFIERLet’s suppose for a while that our amplifier is fed backed with β independent from frequency. The transfer function is, with obvious meaning of the terms:

Af s =)A(s γ s

)1 + βA(s γ s , T = γA, (γ < 0)

)A(s γ s1 + β )A(s γ s =

)A(s γ s1 + | )βA(jω γ s |e )jϑ(jω

We can establish an obviously stability criteria..

The feedback is negative, but now the term βA(s)γ(s) is a complex number, and an unwanted situation can happen if βA(s)γ(s) = -1 for some value of s, indeed this happens at the frequency so such that: |βA(so) γ(s) |=1 and Φ(βA(so) γ(s))=π:

If this condition is met the denominator of Af becomes zero, and we have divergence. Technically, the divergence happens at the frequency so only, but we will see that noise is always present at all frequency and even a very small pulse of noise at so will start the trouble.

We can guarantees to avoid divergence if: “The fed backed network is designed such that at the frequency sm, such that Φ(βA(sm)γ(sm))=π we have |βA(sm) γ(sm)| < 1. Thanks to the fact that A(s) is monotonic and decreasing this assumption guarantees that the denominator cannot be zero for any frequency above sm.”

Rather than considering a value for the loop gain it is generally considered a margin such that it is asked that |βA(s)γ(s)|<1 for frequency where Φ(βA(s)γ(s))=3/2 π (135°) or even Φ(βA(s)γ(s))=2/3 π (120°). In the first case we say that the phase margin is 45° (135°=180°-45°), in the second case we say that the phase margin is 60° (120°=180°-60°).

IMPORTANT NOTE: we must not forget that the stability to consider is for the fed backed network and for that what it is important is the value and the phase of the loop gain , βA(s)γ(s), and not that of the OA gain A(s), only.


3. LOOP GAIN FOR AN AMPLIFIER WITH A SINGLE DOMINANT POLE

Let’s suppose for a while, to simplify, that |γ(s)|=1 and start with a dominant pole amplifier; T is of the form:

where τ1 includes the frequency behavior of the OA and that of f(β, s).

We can see in the above result that when a single pole is present in the loop gain we have not the risk of having nulled the denominator since the phase of T cannot exceeds |90°|.

The slope for frequencies above the pole is 20 dB/dec.

The last interesting characteristic is that the original pole of T is shifted to a larger frequency which depends on the amount of feedback.

The closed loop gain becomes (let’s neglect the direct transmission):

Let’s try to prove what we have concluded about stability in the previous slides.

T = −A0β

1 + sτA, A0 >> 1

Af(s) =1β−T

1 − T =1β

A0β1 + sτA

1

1 + A0β1 + sτA

=1β

A0β1 + sτA + AR0β

=1β

A0β1 + A0β

1

1 + s τA1 + A0β

≈1β

1

1 + s τAA0β

=1β

1

1 + s τA)−T(0


………Loop gain for an amplifier with a single dominant pole 2

ω

dB

1/τ1

Said in other way:

The so called gain-bandwidth product remains constant and equal to that measured when β=1: the larger the factor 1/β, the smaller the bandwidth.

Once again: if there is a single pole in the loop gain the maximum value in the phase is -90°. As a consequence, the real part of the pole cannot change sign and there cannot be divergence in the response.

The frequency behavior of the loop gain is easily studied graphically with the Bode diagrams.

We know that an important point on the curve is that |T|=1 which, in the log-log scale, is easily found:

ω

dB

1/τ1

Gain as it results from the previous page.

dB = 20logA0β

|1 + sτA| = 20logA0

|1 + sτA| + 20log β = 0

⇒ if ωτA ≫ 1 : logA0ωτA

= log1β ⇒ ω3dB =

A0τAβ

1β

Af(s) ≈1β

1

1 + s τAA0β

A0β/τA A0/τAA0β/τA A0/τA

A0

|1 + sτA|A0

|1 + sτA|

1β

1|1 + s ⁄τ𝐴𝐴 AR0β |

ω3dBβ =

A0τA


………Loop gain for an amplifier with a single dominant pole 3

Let’s consider again the result obtained at the beginning of this chapter:

Concerning our feed back representation we can see, from the second line above, that our former calculation 1

β−T1−T

is now replaced by 1β−T(0)1−T(0)

, and that, if |T(0)| is large, as expected, the finale expression is simply the loop gain 1

βmultiplied by the frequency dependence of the system, last line

above.

We, therefore, need only to consider a multiplication term dependent on frequency to obtain an adequate knowledge of our system.

Af s =1β

A0β1 + A0β

1

1 + s τA1 + A0β

=1β

−T(0)1 − T(0)

1

1 + s τA1 + A0β

≈1β

1

1 + s τAA0β

=1β

11 + sτfeed


4. LOOP GAIN FOR AN AMPLIFIER WITH 2 DOMINANT POLESAgain let’s suppose that |γ(s)|=1 and the amplifier has 2 dominant poles, such that the resulting loop gain is:

Now we have:

So we have 2 poles at frequencies:

Starting with β=0, namely no feedback, where the 2 poles are real and negative, the discriminator tries to approach negative values for increasing β, and the poles become CC. The real part of the poles is always non-negative, if they are non-negative at β=0 (indeed ∆=(τ1-τ2)2 when β=0) : when we have 2 poles the real part cannot changes sign, since the phase of π is got only at ω=∞ (remember: poles with positive real part imply divergence in time domain).

Bode diagram is of help also in this situation.

As it was mentioned at the beginning in standard situations one of the 2 poles is at frequency much smaller than the other, and this simplifies the graphical interpretation.

T = −A0β

1 + sτA 1 + sτB, A0 >> 1

Af(s) =1β−T

1 − T =1β

A0β1 + sτA 1 + sτB

1

1 + A0β1 + sτA 1 + sτB

=1β

A0β1 + sτA 1 + sτB + A0β

=1β

A0 ⁄β τ1τ2

s2 + τA + τBτAτB

s + 1 + A0βτAτB

p1,2 = −12τA + τBτAτB

1 ∓ 1 − 4τAτB

τA + τB 2 1 + A0β


……… Loop gain for an amplifier with 2 dominant poles 2Let’s start:

From when |T|=1 the closed loop gain behavior reduces to |T/β|=|A(s)f(β,s)|. This comes from the following:

If τ1 >> τ2 the 2 poles reduce to:

If the term under the root is close to 1, namely β was chosen small enough:

and:-

+

ω

dB

|T|=1

Valid if the poles are far away each other.

ω

dB

|T|=11β

Af =1β

−T1 − T ≈

|T|<1 1β | − T| =

Tβ = |A s f β, 𝑠𝑠 )|

1 − x ≈ 1 −x2

-45°

-135°

1/τA 1/τB

A0β/τA

1/τA 1/τB

A0β/τA

0 = 20log T = 20logA0β

1 + sτA 1 + sτB⇒

logA0

|1 + sτA| |1 + sτB|= log

1β

A0

|1 + sτA| |1 + sτB| A0

|1 + sτA| |1 + sτB|1β

1|1 + s ⁄τA A0β | |1 + sτB|

p1,2 ≈ −12

1τB

1 ∓ 1 − 4τBτA

A0β

p1,2 ≈ −12

1τB

1 ∓ 1 − 4τBτA

A0β ≈ −1

2τ21 ∓ 1 − 2

τBτA

A0β ≈

−A0βτA

−1τB


……… Loop gain for an amplifier with 2 dominant poles 3

The phase margin can be found analytically if we ask that:

Therefore:

The sign = gives the maximum value for β to get good margin, larger values of β give rise instability; this is a golden rule: stability constraint translate in an upper limit for the loop gain β, if no other precautions are taken.The 2 poles become, under this assumption:

Concerning the response to a step of the 2 obtained solutions, for the first case we expect an overshoot of about 23%, in the second 9%.

Φ ωs = −π + α, α = 60°, 45°)T(ωs ≤ 1

1 when α = 45°

0.577 when α = 60°

η =

Es. per η=1

p1,2 ≈ −12

1τB

1 ∓ 1 − 4η 1 + η2 =

Φ ωs = −atg ωsτA − atg ωsτB ≈τA>>τB −

π2− atg ωsτB =

teo−π + α,α > 0

atg(ωsτB) =teo π

2 − α ⇒ ωs=tg(π2 − α

τB=

ητB

|T ωs | ≤ 1 ⇒A0βωsτA

1|1 + jωsτB| ≤ 1

A0βτBητ1

1|1 + jη| =

A0βτBητA

11 + η2

≤ 1

β ≤ητAτB

1A0

1 + η2

−12

1τB

1 ∓ 1 − 4 2 = −12

1τB

1 ∓ j2.16 ,α = 45°

−12

1τB

1 ∓ 1 − 4 × 0.667 = −12

1τB

1 ∓ j1.29 ,α = 60°



We can see the effect in the time domain the response to the step function, 1/s, of our fed backed OA with cc poles:

It can be appreciated that, although the frequency of the poles is the same for both solutions, the one with the larger overshoot seems, apparently faster, as the rise time is shorter.

Often, when a short transition time is necessary, it is exploited this characteristics to gain a bit in speed.

F ω =1β

1s

1s − p1

1s − p1

, p1 = −1

2τ21 + jη

Where, according to the result of the last page: η=1,29 and 2,16 (τ2=1 µs in the example) and with β=1 for simplicity:

f t =1β 1 −

e − t2 τ2

η sinη t

2 τ2+ η cos

η t2 τ2

With time domain counterpart:

t 10 -5

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.25Overshoot: 1,23;

M=45°

Overshoot: 1,09; M

=60°



We can now consider what we have to aspect in the direct transmission with frequency.

This is soon understood starting from:

Therefore:

Namely, we know now that:

Af =1β

−T1 − T ≈

|T|<1 1β | − T| =

Tβ = |A s f β, 𝑠𝑠 )|

T →ω→∞

0

Or:

1 − T →ω→∞

1

Af =1β−T(ω)

1 − T(ω)+

ADIR(ω)1 − T(ω)

→ω→∞

ADIR(ω)


5. LOOP GAIN FOR AN AMPLIFIER WITH 3 DOMINANT POLES

As we have seen the better stability condition is obtained if the OA has a dominant pole. This class of OA are widely used, but have the limitation in wide frequency applications. High frequency can be obtained exploiting as much as possible the properties of the technology used. When this is done it is difficult to maintain far away the first pole from the others and a second, and/or a third pole get closer and closer.

We have just studied the case with 2 poles and have verified that an analytical solution is possible. If 3 poles are present only numerical solutions are possible for most of the situations and, if the third pole is far enough, it can be considered to add a perturbation only on the phase margin of the first 2.

Namely, at first we consider the effect of the third pole only on the phase, while the condition |T|≤1 is elaborated with the first 2 poles only:

The above condition will set a phase, hence a frequency, margin smaller than the case θIII were not considered.

Φ(ωs) = −atg(ωsτ1) − atg(ωsτ2) − ϑIII


6. THE DOMINANT POLE COMPENSATION

Let’ consider now the most critical case: a unity buffer amplifier. To simplify let’s assume that Ro, RL, Ri=∞, and the OA has 2 poles.

+

-Vi

Vo

For this oversimplified case the loop gain reduces to:

ω1/τA 1/τB

|T|=1, θM< 10°

Unless the frequency of the second pole is at |T|=1, the phase margin is less than 45° and the stability of the network is critical.

If there is no possibility to work on the OA configuration it must be chosen a value for β smaller than 1, but the design parameters are in this case changed.

An alternative is to choice an OA with a dominant pole only.

Remember: when an OA has more than one dominant pole the larger usable frequency is given or, as an alternative, the minimum close loop gain, 1/β, is given.

Nevertheless some OAs have the additional opportunity of setting the bandwidth or the position of the phase margin, or second pole. Normally this is achieved adding a capacitor at a proper dedicated pin.

dB

Ao1 + sτA 1 + sτB

V+ − V−

T = −Ao

1 + sτA 1 + sτB

A0|1 + sτA| |1 + sτB|

1β

(= 1)

-45°

-135°


.......... The dominant pole compensation 2

We model our OA with the 2-stages network below, with one low pass filter in between to represent the lower frequency pole.

Some commercial OAs allow to add an external capacitance, CC, that allows to modify, by lowering, the value of the lower frequency pole.

Adding CC, the time constant τA becomes now:

By choosing the proper value for CC it is possible to shape the transfer function in such a way that the phase margin takes the value of 45° at β=1:

Obviously, larger values for CC allows to obtain larger phase margins, at expenses of the smaller bandwidth.

Of course, larger phase margins means a lower available frequency bandwidth and a trade-off must be considered.

1+

- RA

CC

Vi Vo

CA

Added capacitance

Ao1 + sτB

Ao1 + sτB 1 + sτA

, τA= RACA, ⁄1 τA << ⁄1 τB

τA′ = RA CA + CC

ω

dB

1/τA 1/τB

|T|=1, θM=45°

Effect of compensationA0

|1 + sτ𝐴𝐴| |1 + sτB|

1β

(= 1) ⁄1 τA′

Over compensation

θM>45°


.......... The dominant pole compensation 3

This phase margin improvement is obtained at the expense of the frequency bandwidth.

If the bandwidth is an issue, and this happens very often, a compromise must be met: the compensation should be set in such a way the phase margin is the minimum needed, 45° or 60°.

ω

dB

1/τB

|T|=1, θM=45°

Effect of compensationA0

|1 + sτ𝐴𝐴| |1 + sτB|

Bandwidth reduction

We can further increase the value of the compensating capacitance CC.What it happens is that the phase margin becomes larger and larger, and in this respect thinks improve.

Compensation gets better and better


We have obtained an important result here; although the OA has only 1 pole, the resulting loop gain has 2 poles, one of which coming from the elements present on the path.

Therefore it is very important to consider what it is connected to our fed backed OA for frequency behavior.

7. Zi DEPENDENCE OF THE FREQUENCY RESPONSE

Let’ consider now a situation with a non ideal OA: the input impedance is not of very large value and depends on frequency, being present a capacitance, Ci. Because of we are interest in the high frequency behavior we suppose to study the frequency range that starts when it is a good approximation:

Let’s suppose negligible the value of Ro:

+

-Vs Ci

R1R2

Vo

R1R2

We need to evaluate the loop gain.

VT independent for loop gain evaluation

Ri‖1

sCi≈

1sCi

Ao1 + sτA

V+ − V−

V− =R1

1 + sCiR1

1

R2 + R11 + sCiR1

VTAo

1 + sτAV+ − V−

T =A(ω) V+ − V−

VT= −

R1R1 + R2

11 + sCiRE

Ao1 + sτA

RE = R1‖R2

R11 + sCiR1

Ci

+

-

vo

+

-

vo


………. Zi dependence of the frequency response 2

Use can be done of the Bode plot to verify the condition |T|=1:

The eq above could be written in several different modes; one is this:

Observation:

Please, take care that the phase we are interested in is:

and not:

ω

dB

1/(CiRE)ωf

|T|=1

ωT

Ex.: 1/βo=2, Ci=10 pF e ωT=Ao/τA=6.28 108 rad/s. Let’ consider 2 situations: RE= 1 kΩ e RE=10 kΩ

Let’s evaluate both the frequency and the phase margin at |T|=1:

The resulting is an IV degree eq to solve:

log −βo

1 + sCiRE

Ao1 + sτA

= 0βo =

R1R1 + R2

logAo

1 + sτA= log

1 + sCiREβo

Ao1 + sτA

|1 + sCiRE|βo

)Φ(ωf) = −atg(ωfCiRE) − atg(ωfτA

)Φ(ωf) = atg(ωfCiRE) − atg(ωfτA

ωf4Ci2RE

2 + ωf2 −

βoAoτA

2

= 0

ωf4Ci2RE

2 + ωf2 − βo2ωT

2 = 0, ωT =AoτA

−βo

1 + sCiRE

Ao1 + sτA

2

= 1βoAo

2

1 + ωf2Ci2RE

21

ωf2τA2

≈ 1

βoAo2

τA2≈ 1 + ωf

2Ci2RE2 ωf

2



from which:

Solutions with ω>0 are those that has the “–” (subtraction symbol):

RE = 10 kΩ → ωf = 55.6 Mrad/s and Φ(ωf) = −170°RE = 1 kΩ →ωf= 163.7 Mrad/s and Φ(ωf) = −148.7°

ω1,22 = −

12Ci2RE

2 1 ∓ 1 + 4Ci2RE2βo2ωT

2

ωf4Ci2RE

2 + ωf2 − βo2ωT

2 = 0

In the above result the only acceptable solution is that with the “-” sign inside the brackets. Let’s call this frequency ωf.

Remembering: 1/βo=2, Ci=10 pF e ωT=Ao/τA=6.28 108 rad/s. Let’s consider 2 situations: RE= 1 kΩ e RE=10 kΩ

In addition:

)Φ(ωf) = −atg(ωfCiRE) − atg(ωfτA ≈ −atg(ωfCiRE) −π2


+

-

vo

+

-

vo

………. Zi dependence of the frequency response 4So far a good phase margin can be obtained by lowering the value of RE, conserving the gain value. We suppose this is not possible and that the increase of the closed loop gain is not possible, too. If the OA is not compensable there is an additional possibility consisting in the connecting a small capacitance in || to R2:

+

-Vs Ci

R1

R2

Vo

CC

Ci

R1

R2

CC

New loop gain:

ω

dB

1/[(Ci+CC) RE]ωf

|T|=1

ωT

We can see now that at the frequency where |T|=1 the phase margin improves:

What we did was a pole/zero cancellation.

VT is independent for T evaluation

Ao

1 + sτAV+ − V−

Ao

1 + sτAV+ − V−

V− =R1

1 + sCiR1

1R2

1 + sCCR2+ R1

1 + sCiR1

VT

T =A(ω) V+ − V−

VT= −

R1

R1 + R2

1 + sCCR2

1 + s Ci + CC RE

Ao

1 + sτARE = R1‖R2

R21 + sCCR2

R11 + sCiR1

Ao

|1 + sτA|

|1 + s Ci + CC RE|βo|1 + sCCR2|

)Φ(ωf) = atg(ωfCCR2) − atg(ωf Ci + CC RE) − atg(ωfτA

T = 1 ⟹

logAo

1 + sτA= log

R1 + R2

R1

1 + s Ci + CC RE

1 + sCCR2


CC + Ci 2RE2 ωf

4 + 1 − β2ωT2CC2R2

2 ωf2 − β2ωT

2 = 0


ω

dB

ωf

|T|=1

ωT

Re-start to calculate ωf:

We obtain a IV degree eq with only even power of the unknown; so we can solve:

a b c

We can see that, adding a small compensating capacitance which works in implementing a pole/zero compensation, we obtain a big improvement. In addition, setting: CCR2=(CC+Ci)RE (in the present case this happens when Ci=CC), a complete suppression of the pole is obtained:

In this situation the maximum phase is 90° and ωf results in:

CC1

CC2 > CC1

CC = Ci

1/[(Ci+CC) RE]

Let’s consider the worst condition with RE=10 KΩ (namely, R2=20 KΩ, if β=0.5):

ωf1,22 = −

b2a

1 ∓ 1 +4acb2

=

ωf = 116 Mra ⁄d s Φ ωf = −113° if CC = 1 pF

ωf = 107Mra ⁄d s Φ(ωf) = −103° if CC = 2 pF

T =VRET

VT= −

R1R1 + R2

Ao1 + sτA

ωf =AoβτA

= 314 Mra ⁄d sAo

|1 + sτA|

|1 + s Ci + CC RE|βo|1 + sCCR2|

|T|2 =AoβτA

2 1 + ωf2CC2R2

2

1 + ωf2 Ci + CC 2RE

21ωf2 ≈ 1


8. LOAD DEPENDENCE OF THE FREQUENCY RESPONSE Another issue that has effect on the frequency response comes from the output load impedance of the network.

Let’s study the case the load is resistive.

R1R2

RL has effect only if the output impedance of the OA is not zero, therefore a small value for the OA output impedance is a good figure of merit.

If we impose that |T|=1 when ω=1/τB, the phase margin that we obtain is 45°:

+

-Vs

R1R2

Vo

RL

+

-

RoVT

RL

Vo

Ao1 + sτA 1 + sτB

V+ − V−

VT independent for a while

RL′ = RL‖ R1 + R2

Vo =RL′

RL′ + Ro

VT

V− =R1

R1 + R2Vo =

R1R1 + R2

RL′

RL′ + Ro

VT

T =A(ω) V+ − V−

VT= −

R1R1 + R2

RL′

RL′ + Ro

Ao1 + sτA 1 + sτB

|T| =R1

R1 + R2

RL′

RL′ + Ro

AoτBτA 2

≤ 1

R1 + R2R1

≥RL′

RL′ + Ro

AoτBτA 2


………. Load dependence of the frequency response 2

Graphical interpretation, starting from:

A low value for RL allows to chose lower value for 1/β as there is a translation to lower value of the gain, maintaining the frequency performance unaffected.

Low values for RL could increase power dissipation, unless AC connected to the output.

ω

dB

1/τA 1/τB

Ro= 0

RL< ∞

R1 + R2R1

≥RL′

RL′ + Ro

AoτBτA 2

A0

|1 + sτA| |1 + sτB|

Lower limits forR1 + R2

R1

RL′

RL′ + Ro

A0|1 + sτA| |1 + sτB|

+

-Vs

R1R2

Vo

RLCAC

We have to take into account or, better, take care, when RL is not alone and has also a capacitance in parallel…

T = 1 ⇒ logRL′

RL′ + Ro

Ao1 + sτA 1 + sτB

= logR1 + R2

R1

T = −R1

R1 + R2

RL′

RL′ + Ro

Ao1 + sτA 1 + sτB

,


We can see that if RL=0 Ω there is an additional pole in T at CLRo, which will lead to instability. Therefore it is a standard to consider the presence of a resistance in series to a cable. Typical values are in the range 50 – 100 Ω.

………. Load dependence of the frequency response 3A capacitive load is often present, in particular if a cable, or transmission line is used. In the latter case we could have transmission delays and reflections, so care must be considered anyway.

+

-Vs

R1R2

Vo

R1

R2

A series resistance is recommended in series to the load capacitance; the reason for that will be evident below.

We can appreciate the importance for the presence of resistor RL: it determine a pole/zero compensation. If the time constant CLRL is properly set it would remain valid the previous result obtained with CL=0 pF:

RL

+

-

RoVT

RL

Vo

CL

CL

Ao1 + sτA 1 + sτB

V+ − V−

VT independent for a while

RL′ =

1 + sCLRLsCL

‖ R1 + R2 ≈1 + sCLRL

sCL

Vo ≈1 + sCLRL

1 + sCL RL + RoVT

T =−A(ω)V−

VT= −

R1

R1 + R2

1 + sCLRL

1 + sCL RL + Ro

Ao

1 + sτA 1 + sτB

R1 + R2R1

≥RL′

RL′ + Ro

AoτBτA 2


A first important question is when the line can be considered a delay line, namely a medium through which the electro-magnetic field travels at the speed of light (in the medium), and when a capacitive load?

9. TRANSMISSION LINE DIGRESSION

If the bandwidth of our signal is << than the inverse of the time delay, or travelling time, (strictly speaking 2 time the travelling times as the signal runs back and forth) of the line we can assume valid the so called «condensed parameters» approximation in which the geometry of the system does not come into play and all the elements are condensed in one point. Under this assumption a delay line, which is modeled by a capacitance, a resistance and an inductance per unit length, C/m, R/m, L/m is modelled by its total capacitance, C×Length, total resistance R×Length, and total inductance L×Length.

If the speed of the signal becomes comparable with the travelling time of light in the chosen line, the geometrical factors cannot be contracted anymore and the differential eqs governing the system must account of both time and position.

From the point of view of our concern whenever the line is outside the feedback loop and is properly terminated it behaves only as a delay line and does not influence the response of the system. Vice-versa, reflections add effects similar to a capacitive load and must be considered.

Managing of fast signals in presence of transmission lines requires to terminate the line with the proper resistor (usually 50 Ω) at both ends.

Vo

+

-Vs

R1R2

RL

Zo ZoRLF


………. Transmission line digression 2

Again: with fast signals the length of the transmission line cannot be neglected.

This happens because the electromagnetic waves travel at the speed of light in the medium, smaller than the speed of light in vacuum.

To put some numbers: the travelling time through 10 cm in a medium having relative dielectric constant of 5, typical value of the substrate of the printed circuit boards, is 1.7 ns.

Travelling lines could be the tracks on a PCB, connecting cable, etc.

The parameters/concepts we are interested in a line are:

Electromagnetic waves can be transmitted and reflected, and the generic wave is composed of a mix of such a 2 components which move back and forth:

The speed of light in the medium is limited:

therefore the propagation time through the line, one direction, is given by:

A very simple example of transmission line is a pair of wires. In this case the medium is the insulation which cover each wire

Vs

Rs

)f(x, t) = f1(x − vt) + f2(x + vt

υL =1εµ

T = εµ ∗ L



The easiest way to model a line is by considering it composed of a regular repetition of a circuit pattern defined per unit length: inductance, resistance, capacitance and conductance.

When we cannot neglect the resistance value per unit length because of the DC voltage drop per unit length due to a large current, we say that the line is dissipative.

Vs

Rs L R

C G

L R

C G

L R

C GRL

x

L

C

L

C

L

C

L

CVs

Rs

x

In our study we deal with small signals and currents and the model can be simplified most of the times omitting the series resistance and conductance (this last omission is valid if the line is terminated or the speed of the signal is fast enough):

If the line is non-dissipative the energy delivered to it is not lost during its travelling and its behavior can be described by a simple balance of energy.

First step:

Supposing that a short pulse is given by the source, it is applied across the first inductance and then shared with the first capacitance, than the second inductance and the second capacitance and so on.



Since with are in the hypothesis of non-dissipative medium, taken any point x in the line it must be satisfied that the energy that the inductance has received is delivered to the following capacitance. Written in a mathematical form:

namely:

From which we can write:

At every point of the line the ratio between the voltage and the current is constant and dependent on the line manufacturing. This constant ratio has the dimension of an impedance and is called the «characteristic impedance of the line» (NOTE: this is not a dissipative impedance).

Examples (the wires are intended to exit the paper): a coaxial cable

)UL(x) = UC(x

12 LdxiT x 2 =

12 CdxvT x 2

Zo =)vT(x)iT(x =

LC

The section of a coaxial cable



Our signal travels at the speed of light within the medium and the propagation time through a line whose length is x, one direction, is:

Geometrical dependence of the capacitance and inductance of the line are similar, but one the inverse of the other such that, it results:

Examples (the wires are intended to exit the paper): track or strip line on a PCB:

υL =1εµ =

1LC

T =xυL

= x εµ = x ZoC

Examples (the wires are intended to exit the paper): a standard pair of wires:



Let’s suppose to apply a signal to an infinite length line. We expect to have only the transmitted signal as there is no way for the signal to be reflected.

Nevertheless at its input the ratio of the inductance and capacitance give the characteristic impedance Zo, so we expect a voltage partition between the series impedance of the source, RSRC, and Zo.

Now suppose the line has a limited length. The condition of not having a reflected signal is that the incoming energy, the energy stored in line, is totally dissipated at the end of the line and this is possible if the line is terminated with a resistor exactly equal to the characteristic impedance (the line is terminated this way).

That energy is dissipated when the energy approaches the load, after the propagation delay.

Now we can then imagine what does happen when the line is terminated with an impedance different from the characteristic impedance Zo…



We look at the signals of as a pair of pulses of energy travelling in opposite directions.

RL

L

C

L

C

L

C

L

CVs

Rs

x

VT, IT

VR, IR

The incoming wave bring with it an energy ET. If this energy is not completely dissipated at the load, ZL, part of it will come back; call it ER:

The better is to consider power (taking the derivative of both sides with respect to time):

and:

Therefore:

From which:

(the same consideration can be done for currents)

The reflected signal is therefore a measure of the amount of difference between the load impedance and Zo.

The same consideration applies also to the input terminal of the line, as the signal can be reflected back and forth.

In case the load is a short or an open connection the power dissipated is zero and the energy is completely reflected.

ER = ET − ELOAD

PR = PT − PLOAD

PR =dERdt =

d 12 CVR2

dt = CVRdVRdt = IRVR =

VR2

Zo

VR2

Zo=

VT2

Zo−

VR + VT2

RL

VR =RL − ZoRL + Zo

VT


………. Transmission line digression 8Let’s see what happens when the line is loaded with a short, while being properly terminated at its input.We know that in this case the energy is completely reflected since not dissipated at the load. The reflected signal changes sign since the sum of the incoming and reflected signals must be zero at the load (the voltage across a short is zero).

If T is the propagation delay, then after an interval equal to 2T the signal is came back to the input and cancels out. This technique is often exploited for creating short pulses:

The second situation is when the load is an open, while the input is terminated with Zo.

The reflected and transmitted signals are equal and after 2T the signal double in amplitude at the input:


The differentiator 1

-

+

C

R

Vi Ci

Let’s suppose Ro of negligible value.

Assuming A=∞ we find 1/β:

C

R

Ci

Although the OA has only one dominant pole the loop gain has 2 poles. The differentiator is critical for what concern stability, unless care is taken.

We can adopt 2 approaches for its compensation, both of them lead the network to behave as a quasi-differentiator.

-

+

C R

Vi Ci

CC

A capacitor CC is put in parallel to R so that, at high frequency, the gain approach the capacitance ratio. Indeed T becomes:

The value of CC must not result too close to that of C, otherwise the properties of the network are lost. Indeed, when A is assumed ∞ we get now:

Let’s suppose that : ωT=Ao/τA=6.28 Mrad/s, C=500 pF, Ci=10 pF and R=10 kΩ.

-

+

VT independent

Vo = −sCRVi

Ao

1 + sτA

Ao

1 + sτA

V− =1

1 + s C + Ci R VT

T =−A(ω)V−

VT= −

Ao1 + sτA

11 + s C + Ci R

T = −Ao

1 + sτA1 + sCCR

1 + s C + Ci + CC R

Now the loop gain T:

Vo = −sCR

1 + sCCR Vi



-

+

C R

Vi Ci

CCWe have to take the module of T:

Let’s consider the case CC=0 first:

T = −Ao

1 + sτA1 + sCCR

1 + s C + Ci + CC R

|T|2 =A02

1 + ω2τA21 + ω2CC2R2

1 + ω2 C + Ci + CC 2R2 =1

ωf2 =

12 C + Ci 2R2 −1 ± 1 + 4ωT

2 C + Ci 2R2

|T|2 =A02

1 + ω2τA21

1 + ω2 C + Ci 2R2 =1

|T|2 ≈⁄A0

2 τA2

ω21

1 + ω2 C + Ci 2R2 =1

ω4 C + Ci 2R2 + ω2 − ⁄A02 τA2 = 0

At the frequency where we will have |T|=1 this approximation is valid:

The poles are obtained solving:

Calling ωf the solution:

ω4 C + Ci 2R2 + ω2 − ωT2 = 0



Substituting the numbers (the sign “+” within brackets is the one to be considered for obtaining a positive value of frequency):

The phase, with the values found for ωf becomes:

Let’s give to CC the value such that the phase margin is 45° at ωf0=1/ RCC.

ω

dB

1/(CCR)

|T|=1

By setting in |T| ω=ωf0=1/CCR and supposing (CC+Ci+C)R>>1/ωf0, we get that:

When (neglecting Ci):

From the latter second order eq:

ω

dB

1/(CCR)

135°

Φ ωf ≈ −atg ωf Ci + C R −π2 = −170°

Ao

|1 + sτA|

|1 + s C + Ci + CC R||1 + sCCR|

Ao

|1 + sτA|

)Φ(ωf0) = atg(1) − atg(ωf0 Ci + CC + C R

−π2 ≈ −135°

ωT 2CC

CC + Ci + C CCR ≈ 1

CC2RωT 2 − CC − C = 0

CC = 81 pF, about 20% of C.

ωf2 =

12 C + Ci 2R2 −1 ± 1 + 4ωT

2 C + Ci 2R2

⇒ ωf = 1.1 Mra ⁄d s

CC =1 ± 1 + 4CωT 2

2ωT 2

-45°

-90°

-45°

-45°0°

-45°

-90°

-45°

-45°


The differentiator 4Second approach.

A resistor, RC, is put in series to C and the high frequency gain is the ratio between R and RC. Capacitance Ci takes part in the stability, but has a small value. Now, when A=∞:

This new eq must be solved with respect to RC now:

-

+

C R

ViCi

RC

The parallel combination of Ci and the series of RC and C is:

1sCi

1 + sCRC

sC=

1 + sCRC

s C + Ci 1 + s CCiC + Ci

RC

Ci small 1 + sCRC

sC

With the last approximation:

T ≈ −A0

1 + sτA1 + sCRC

1 + sC R + RC

Again, if τA and C(R+RC) are >> CRC the 45°-phase margin is close to ωf=1/CRC and:

|T|2 ≈ ⁄A02 τA2 C2RC

2 2

1 + R + RC2

RC2

= 1

|T|2 ≈ ωT2C2RC

2 2

1 + R2RC2

= 1

2ωT2C2RC

4 − 𝑅𝑅𝐶𝐶2 − 𝑅𝑅2 = 0

RC2 =

14ωT

2C21 ± 1 + 8ωT

2C2R2 ⇒ RC = 1500 Ω

Well, it does not result RC<<R, that means that the result is a fair approximation. Both this approach and the previous take advantage by large value of ωT, namely large bandwidth OAs, as both RC and CC reduce.

VO −sCR

1 + sCRCVi


The integrator or charge sensitive preamplifier 1

-

+

CD

Rf

IiCi

Cf

Detector

The time constant RfCf is set as large as possible if we wish to obtain a theoretical integrator and for noise optimization (see later for this).

Therefore, being the time constant large, but not ∞, the network behaves as a quasi-integrator.

Very often the detector signal is a very short pulse. If it is very short with respect to the minimum width that the following stages are designed to manage, it can be well approximated with a δ-like distribution:

here Q is the charge released into the detector by the incident particle.

Let’ start to study the expected output signal when the OA gain is assumed ∞:

Therefore:

Ao

1 + sτA

)ID(t) = Qδ(t

Vo = −Rf

1 + sCfRfQ ≈ −

QsCf

β = −1 + sCFRF

RF≈ −sCF

⇒ Vo t = −QCf

e− ⁄t τf ≈ −QCf

1 t , τf= CfRf



-

+

CD

Rf

IiCi

Cf

Detector

We understand the reason for a capacitive feedback: the output signal is inversely proportional to the feedback capacitance and the charge gain can be very large if this capacitance is small.

As an instance. Let’s consider a Si detector on which it is release an energy of 100 keV. The conversion charge within a Si detector is about 2.7 eV per generated electron, so that we have about Q=27.8 kel.If the feedback capacitance Cf is 1 pF the output signal results in:

About Rf : Rf has to satisfy different constraints that we will study: its upper limit is tied to the DC input bias current of the detector /preamplifier (ex. Rf=1 GΩ, input bias current 1 nA, then the dropout across Rf is 1 V), discharge of the signal, thermal noise.

Ao

1 + sτA

Vo t = −QCf

e− ⁄t τf ≈ −QCf

1 t , τf= CfRf

A few considerations:

Vo =QCf

=27800 × 1.6 × 10−19

10−12 = 4.5 mV



Let’s start to study the dynamic behaviour of the our Charge Sensitive Preamplifier, CSP from now on.

We can consider 2 classes of applications: the first class covers the range of the particles energy limited to a few tens of MeV at maximum (passive physics such as neutrino physics, dark matter study, medical radioscopy, et.); the second class is the domain of the accelerator physics, whose range of energy is well above the GeV energy.

The range of energy sets the value of the feedback capacitance and the philosophy to adopt for stability of the CSP.

Step 1: the study of the CSP at low energy.

In this case we can simplify by considering negligible the value of the OA output impedance, as, if combined with Cf of the order of 1 - 10 pF gives a time constant of 0.1 – 1 ns in case Ro is of the order of 100 Ω. This time constant is normally outside the frequency range of the standard applications.

The loop gain T:

CD

Rf

Ci

Cf

-

+

VT independent

V− =1

s CD + Ci1

1s CD + Ci

+ Rf1 + sCfRf

VT

T =−A(ω)V−

VT= −

Ao1 + sτA

1 + sCfRf1 + s CD + Ci + Cf Rf

Ao

1 + sτA



ω

dB

1/((Ci+CD+ Cf)Rf) ωf

|T|=1

ωT

Bode analysis:

Since there is a pole/zero compensation we have not problems with stability, if the OA has a single dominant pole and the feedback capacitance is in the range of a few tens of pF at its maximum.

Under the condition that Rf is very large we can approximate:

So far the fed backed amplifier remains with a single pole, if the starting OA has a single pole.

CD

Rf

Ci

Cf

-

+

VT independent

T =−A(ω)V−

VT= −

Ao1 + sτA


20log T = 0 ⇒

20logAo

|1 + sτA| = 20log|1 + s CD + Ci + Cf Rf|

|1 + sCfRf|

Ao

1 + sτA

Ao

1 + sτA

|1 + s Ci + CD + Cf Rf||1 + sCfRf|

T ≈ −Cf

CD + Ci + CfAo

1 + sτA= −γ

Ao1 + sτA

-45° di τA

-45°-90° di τA

-45° 0°



Therefore, if RfCf is large enough:

ω

dB

ωf

|T|=1

ωT

From:

We get:

As already seen, this is a quite general result: if –T(0) has a large value, the quantity –T(0)/(1-T(0)) is close to 1 and:

Namely, since at large frequency |T| reduces, its effect is put into evidence only with frequency.

Ao

1 + sτA

Ci + CD + CfCf

T ≈ −Cf

CD + Ci + CfAo

1 + sτA= −γ

Ao1 + sτA

Vo ≈ −1

sCf−T

1 − T = −1

sCfγAo

1 + sτA1

1 + γAo1 + sτA

Q

= −1

sCfγAo

1 + sτA + γAoQ = −

1sCf

γAo1 + γAo

1

1 + s τA1 + γAo

Q

≈ −1

sCfγAo

1 + γAo

1

1 + s τAAo

CD + Ci + CfCf

Q

≈ −1

sCf1

1 + s τAAo

CD + Ci + CfCf

Q

Af =1β

1anωn + an−1ωn−1+. . . +1



Coming back to the time domain:

Let’s evaluate the input impedance:

-

+

CD

Rf

ITCi

0Ao =Cf

Vo=0

Let’s null the loop gainVT (let’s assume that Ro≅0)

Vo ≈ −Q

sCf1

1 + s τAAo

CD + Ci + CfCf

= −QCf

1s

11 + sτfe

, τfe =τAAo

CD + Ci + CfCf

⇒ Vo(t) =QCf

1 − e− ⁄t τfe 1(t)

IT = VTsCD + VTsCi + VT1 + sCfRf

Rf

=1 + s Cf + CD + Ci Rf

RfVT ⇒ Zi =

Rf1 + s Cf + CD + Ci Rf

The output signal rises with time constant τfe, then remain at the value Q/Cf in first approximation. Actually the signal decay with a slow time constant CfRf.



We see that the input impedance depends on frequency and shows an inductive (namely impedance proportional to frequency) and capacitive (namely, impedance inversely proportional to frequency) behaviour.

-

+

CD

Rf

ITCi

0Ao =Cf

Vo=0

Let’s null the loop gainVT (let’s assume that Ro≅0)

Zi =Rf

1 + s Cf + CD + Ci Rf

Zif =Rf


11 − T

=Rf


1

1 + Ao1 + sτA


= Rf1 + sτA

1 + sτA 1 + s CD + Ci + Cf Rf + Ao 1 + sCfRf

=Rf

1 + Ao

1 + sτA

s2 τA CD + Ci + Cf Rf1 + Ao

+ sτA + CD + Ci + Cf 1 + Ao Rf

1 + Ao+ 1

And:



Let’s elaborate a little finding first the 2 poles at the denominator:

Zif =

Rf1 + Ao

1 + sτA

s2 τA CD + Ci + Cf Rf1 + Ao

+ sτA + CD + Ci + Cf 1 + Ao Rf

1 + Ao+ 1

Zif =1

τA CD + Ci + Cf1 + sτA

s2 + sτA + CD + Ci + Cf 1 + Ao Rf

τA CD + Ci + Cf Rf+ 1 + AoτA CD + Ci + Cf Rf

Zif ≈1


s2 + s 1 + ωTCfRfCD + Ci + Cf Rf

+ ωTCD + Ci + Cf Rf

Zif ≈1


s2 + s ωTCfCD + Ci + Cf


Collect the term that multiplies s2:

Now collect 1+Ao and put into evidence ωT:

Consider that ωTCfRf is >> 1:



Zif ≈1


s2 + s ωTCfCD + Ci + Cf


Now find the poles:

ω1,2 = −ωTCf

2 CD + Ci + Cf±

12

ωTCfCD + Ci + Cf

2

−4ωT

CD + Ci + Cf Rf

ω1,2

= −ωTCf

2 CD + Ci + Cf1 ∓ 1 −

4ωTCD + Ci + Cf Rf

CD + Ci + Cf 2

ωTCf 2

ω1,2 = −ωTCf

2 CD + Ci + Cf1 ∓ 1 −

4 CD + Ci + CfCf

1ωTCfRf

ω1,2 ≈ −ωTCf

2 CD + Ci + Cf1 ∓ 1 −

2 CD + Ci + CfCf

1ωTCfRf

Expand at the first order the term under the square root:

Finally the 2 poles:

ω1 ≈ −1

CfRfω2 ≈ −

ωTCfCD + Ci + Cf



An interesting situation is got when CfRf=τA:

Under this condition the input impedance is resistive and flat till large frequencies with no approximations.

Therefore, the input impedance is constant till the pole in closed loop is approached; its value is small if the OA bandwidth is large.

Zif =1

τA CD + Ci + Cf1

s + ωTCfCD + Ci + Cf

=1

ωTCf1

1 + s CD + Ci + CfωTCf

Substituting the poles just found:

Zif ≈1


s + 1CfRf

s + ωTCfCD + Ci + Cf

Zif ≈1

τA CD + Ci + CfCfRf CD + Ci + Cf

ωTCf1 + sτA

1 + sCfRf 1 + s CD + Ci + CfωTCf

Zif ≈Rf

τAωT

1 + sτA

1 + sCfRf 1 + s CD + Ci + CfωTCf

ZifRfCf, τA large 1

ωTCf1




Let’s study now when we have to take care about stability.

The first case is when the OA has 2 poles, at least.

Since at high frequencies Rf has no influence, if large enough:

ω

dB

|T|=1

Under the hypothesis that τA>>τB the phase margin at 45° is close to ωs= 1/τB, where we have to force |T|≤1 to find out our limits:

In case the OA has 2 poles it results that we have to consider an upper limit on the feedback capacitance Cf.

Of course, we cannot change the value of CD as it depends on the detector and on Ci, that will be constrained to the noise performances (we will see why).

A1

τ B1

τ

T = −Cf

CD + Ci + CfAo

1 + sτA 1 + sτB

Ao

|1 + sτA||1 + sτB|

Ci + CD + CfCf

CfCD + Ci + Cf

ωTτB2

≤ 1

⇒ Cf ≤CD + CiωTτB

2− 1

Step 2: the study of the CSP at large energy, or with a 2 poles OA.

-45°

-135°



In the experiments with accelerators the particle energy is very large, GeV and/or TeV.

As an instance. A 10 GeV particle develops about 2.8 Gel in a Si detector. If we suppose that the feedback capacitance is 100 pF the expected signal at the preamplifier output is 4.44 V: very very big!

Preamplifiers which manage large particle energy are therefore forced to work with large feedback capacitances for which the output impedance of the OA cannot be neglected anymore.

CD

Ci

Cf

-

+

RoVT

We have 2 poles in the fed backed system, starting from a single pole OA; the second pole being dependent on the OA open loop output impedance, if not negligible.

Ao

1 + sτA

V− =1

s CD + Ci1

1 + sCfRosCf

+ 1s CD + Ci

VT

T = −Cf

CD + Ci + CfAo

1 + sτA1

1 + s CD + Ci CfCD + Ci + Cf

Ro


Let’s suppose that: Ao/τa=5.28 108 rad/s, Ro=500 Ω and Ci=CD=50 pF, Cf=100 pF and, finally, γ=1/2.


ω

dB

|T|=1

ωs

Ao

|1 + sτA|

CD + Ci + CfCf

1 + sCD + Ci Cf

CD + Ci + CfRo

CD + Ci + CfCD + Ci CfRo

τe =CD + Ci Cf

CD + Ci + CfRo γ =

CfCD + Ci + Cf

T = −Cf

CD + Ci + CfAo

1 + sτA1


Ro

T = −γAo

1 + sτA1

1 + sτe

To obtain:

Let’s make the following positions:

log T = logAo

1 + sτA+ log

CfCD + Ci + Cf

1


Ro

= 0

logAo

1 + sτA= log

CD + Ci + CfCf

1 + sCD + Ci Cf

CD + Ci + CfRo



ω

dB

|T|=1

Forcing |T|=1 implies that:

The above is a IV degree equation whose solutions are:

So far the phase margin is not adequate. One way to improve the situation is by decreasing Ro. Lowering Ro from 500 Ω to 100 Ω we obtain:

This solution works, but has a drawback as it increases the power dissipation since RO is the output impedance of the output transistor of the OA and this impedance is inversely proportional to its bias current (RO=1/gm ÷ 1/IBIAS), we will see this.

ωs

Ao

|1 + sτA|

CD + Ci + CfCf

1 + sCD + Ci Cf

CD + Ci + CfRo

CD + Ci + CfCD + Ci CfRo

γAoτA

2 1ωs2

11 + ωs

2τe2= 1

ωs = 108 Mrad/s and Φ(ωs) = −160°

ωs = 214Mrad/s and Φ(ωs) = −137°

T = −γAo

1 + sτA1

1 + sτe


0 0.5 1 1.5 2x 10-5

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Tempo

AU

tauF/tau= 100tauF/tau= 50tauF/tau= 10tauF/tau= 5tauF/tau= 1.1


Effects of the feedback resistor RF.The presence of RF makes the amplifier a quasi-integrator.

If we suppose that the output of the amplifier is shaped by an CR-RCn, with n=1,we will have:

With respect to the ideal case now we have a zero at zero frequency which determines an under-elongation which is more and more evident as soon ass τF approaches τ.

This derives from the fact that the zero tries to cancel the DC component of the signal..

VO s =RF

1 + sτFsτ

1 + sτ 2 Q with τF = RFCF

Rf

IT

-

+

CD

Ci

CfVT

CR-RCnVo



RF must be present for 2 reasons. The first is practical:

The second reason regards the recovery of the output voltage consequent to the impinging of a particle.

Supposing IDC- completely negligible, setting Rf=∞ the imping of any particle would be stored on Cf and the output will saturate also this way after that a certain number of particles arrive.

+

-RF

IDC-

IDC+

CF The 2 biasing currents IDC+ ≈IDC- are static, DC currents needed to bias the input transistors of the OA.

We see that IDC+ cannot develops a dropout voltage.

The DC current IDC- flows through RF, rising an offset voltage at the output:

If RF=∞ at the preamplifier output we would have:

Which would saturate the preamplifier toward a rail after a while.

In some applications Rf is set ∞ and a switch is put in parallel to Cf that is shorted periodically, or when necessary, to cancel the charge stored on it.

NOTE: RF must have the larger possible admitted value for noise reasons (we will see why).

Effects of the feedback resistor RF.

VO = −RFIDC−

VO = −IDC−CF

t



Effect of the preamplifier bandwidth on the signal shape

Let’ suppose that RF=∞, and tr=2.2τr is the preamplifier rise time. Then, the signal at the CR-RC shaper output is:

0 1 2 3 4 5 6x 10-6

0

0.2

0.4

0.6

0.8

1

Tempo

AU

taur/tau= 0.01taur/tau= 0.05taur/tau= 0.1taur/tau= 0.5taur/tau= 0.9

A pole is added that will shift the peak and lower the amplitude of the signal.

Obviously, the larger the rise time, the larger is the effect on the peaking time/amplitude of the output signal.

The preamplifier bandwidth could change with temperature (the bandwidth is at the limit of the feedback effect), affecting the accuracy.

0 1 2 3 4 5 6x 10-6

0

0.2

0.4

0.6

0.8

1

Tempo

AU

taur/tau= 0.01taur/tau= 0.05taur/tau= 0.1taur/tau= 0.5taur/tau= 0.9

In the first plot above the amplitude is normalized to emphasize the peaking time shift.

In the second plot below the amplitude is not normalized and it can be appreciated that the preamplifier bandwidth has effects on both the peaking time and amplitude, if not larger of at least ten times that of the shaper.

VO(s) =1

1 + sτrsτ

1 + sτ 2Q

sCf


Where the resistance 1gm

CoCf

is called «cold resistance» because it could be larger than ⁄1 gm , although the series input noise is proportional to ⁄1 gm (see later the reason for this).


About the input impedance.

We have evaluated the input impedance as:

In case we can consider both Rf and τA large enough:

As we will see, the frequency where the OA gain drops to unity is:

Where gm is the transconductance of the input transistor and Co is the compensating capacitance, then:

Zif ≈Rf

1 + Ao

1 + sτA

1 + sCfRf 1 + s τA CD + Ci + Cf1 + Ao Cf

Zif ≈1

sCfsτAAo

1

1 + s τA CD + Ci + Cf1 + Ao Cf

=1

ωTCf1


ωT =gmCo

Zif ≈1

gmCoCf

1

1 + s Co CD + Ci + CfgmCf



About the bandwidth with respect to the layout.

A very very important concern is the detector – preamplifier connection, especially when the detector capacitance is small.

From the electrical point of view the link between the detector and the preamplifier is seen as a capacitance, a parasitic capacitance, Cpar in the figure above. This Cpar is normally proportional to the length of the link and must be as small as possible: the golden rule is to locate the preamplifier as close as possible to the detector.

This evaluation is soon verified if we remember the fed backed amplifier bandwidth:

-

+

CD

Rf

IiCi

Cf

Detector

Ao

1 + sτA

Cpar

ωfe=Cf

CD + Ci + Cf + CparωT

(ωT being the OA unity gain bandwidth). As it can be seen now the parasitic capacitance Cpar has been taken into consideration showing that ωfe reduces as soon as it increases.

We will see that the parasitic capacitance Cpar must be small also for noise reason, namely, it must be minimized in any case.


Appendix A: classroom study of an amplifier 1

+

-

CC

VCC

CF2

RA

RB

VS

VCC

CF1 RF1

RF2

The amplifier under study is biased with a single rail, the other rail being grounded.

When working with a single rail it is recommended to try to maintain the DC biasing at half of the supply voltage, by AC coupling the signal, if necessary.

To block the DC part some capacitances can be used. Of course, the low frequency information of the signal is lost and we are now in this situation as our study is about stability, at high frequency.

What it is implemented is a double loop: first the amplifier is biased ad DC in a proper way, then the signal is AC coupled.

In the above network there are 2 capacitors used to block the DC component of the signal: CF1 and CF2.



Assuming that the AO has gain ∞, we know that v+≈v- and i+≈i-≈0, while a small DC (static or biasing) current flow very similar from/to both inputs, negligible most of the times.

With a good approximation we can write::

To study the loop at DC every capacitors and Cf1 and Cf2, in particular, are omitted, or considered to have a value of zero..

In our circuit RF1 and RF2 set the DC voltage of node V+, therefore V-, since the loop is closed by RB at the inverting node.

Note the at DC the gain is 1 since RA is floating and there is no current flowing through it.

+

-

VCC

RA

RB

VCC

RF1

RF2

I=0

VO

Being RA floating at DC it results negligible the current through RB. As a consequence, VO will be close to V-, hence VO will satisfies:

V+ =RF2

RF1 + RF2VCC =

RF1=RF2 VCC2

VO = V− = V+ =RF2

RF1 + RF2VCC =

RF1=RF2 VCC2



Now we can go back to the AC signal. It will be fully amplified for all the frequencies above those established by the 2 high frequency filters.

+

-

CC

VCC

CF2

RA

RB

VS

VCC

CF1 RF1

RF2

The 2 AC coupling time constants are:

Now the simpler approach is to consider DC nodes at ground potential, as AC signals are considered small perturbations of the equilibrium conditions.

τ1 = CF1 RF1‖RF2τ2 = CF2RA



In this last view the amplifier reduces to configuration we have already well studied.

In the circuit above there is the possibility to modify the bandwidth changing the value of capacitance CC, we have to remember that this option is not available in all the commercial OAs.

In our example we chose the AD829 that has this option that allows to set the below working conditions:

At frequencies well above those of the AC coupling the 2 filtering capacitances CF1 and CF2 can be considered short circuit:

+

-

CC

RA

RB

VS

RF1

RF2



This is the schematic of our test circuit:

Sheetof

INFN - Uni - Fis- MIBP.zza della Scienza, 3. 20126 (MI) Italy 1 1

schematic name = SCHEMATIC1

Tuesday , December 14, 2010

schematic path = /

AD829

-

+

U1AD829

3

26

74

15

8

R151

R247

R31k

R451

C468p

C327p

C26.8p

C13.3p

J1BNC_IN

12

J3POWER

123

J2BNC_OUT

12

R90

VCC

R121k

VCC

R131k

SW2SW DIP-4/SM

C5100n

VCC

R5ELIPOT-1k

12

C71u

C8100n

R62k

R72k

VCC

C61u

SW3SW DIP-4/SM

R11150

R1047

Guadagno Variabile

Compensazione Variabile

Accoppiam

ento A

C


Bibliography

Franco Sergio,

Amplificatori operazionali e circuiti integrati analogici : tecniche di progetto, applicazioni,

U. Hoepli, c1992.

P.R.Gray, R.G.Meyer, Analysis and Design of Analog Integrated Circuits, John Wiley & Sons;

Jacob Millman, Arvin Grabel

Microelectronics 2. ed

Mcgraw-Hill , 621.3815MILJ.MIC /1987

E.Gatti e P.F.Manfredi,

Processing the signals from solid-state detectors in elementary-particle physics,

La Rivista del Nuovo Cimento, Volume 9, Serie 3, 1986.

S.Franco

Electric Circuits Fundamentals

Oxford University Press, USA,1994, C. Biblio 621.3815 FRAS.ELE/1995

S.Franco

Design with Operational Aplifiers and Analog Integrated Circuits

McGraw-Hill, 2002, C. Biblio 621.3815 FRAS.DES/2002.

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