16. INPUT/OUTPUT IMP. SUMMARY OF A V-TO-V FED …pessina.mib.infn.it/Corsi_del_IV_anno/Dispense...

Amplifiers and feedback gpessina 1

16. INPUT/OUTPUT IMP. SUMMARY OF A V-TO-V FED BACKED OASummary about the input/output impedances of a v-to-v.Input impedance measured with a voltage source having a series resistance Rs in series:

Rifs = Riol 1 − T

Rifs = Rs + Riol0 1 − To = Rs + Rifs0

Where Riolo and To are Riol and T measured when Rs is set to 0 Ω.

If the input impedance is measured using a current generator with in parallel a source resistor Rs we have:

Rifp = RsRifsooooOOooo

Output impedance measured with a current source having in parallel the load resistor RL:

Rofp =Rol

1 − T

Rofp = RL Rol∞

1 − T∞= RL Rofp∞

Where Rool∞ and T∞ are Rool and T measured when RL is set to ∞ Ω.

If the output impedance is measured using a voltage generator in series to the load RL we have:

Rofs = RL + Rofp∞

RiolT = Riol0T0,

T =Riol0

RS + Riol0T0

TRool

=T∞

Rool∞,

T =RL

RL + Rool∞T∞


17. GAIN OF A C-TO-C FED BACKED CURRENT OA

Just for exercise we can evaluate the loop gain and direct transmission.

For the loop gain we put is=0 A, namely, we open is, since it is a current generator. To simplify let’s consider the OA having R1 in feedback ideal; otherwise, being a feedback in a feedback, it would be recommended to replace it with its model, to avoid to introduce errors (see later for this).

c-to-c= current input and current output amplifier

-

+

iA=Aiivo

Ri RoRL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

R1

vB

+

-

V2+

V2-

-

+

iA=is independentvo

Ri RoRL

Rsis

ii

v+

iF

RF

io


………. gain of a c-to-c fed backed current OA 2

With: RF′ = RF + RSRi

It is:

iA =1

Ro+

1RL

vo

and: RS′ = RSRi

v+ =RS′

RS′ + RFvB =

RS′RS′ + RF

−R1RL

11

Ro+ 1

RL

iA

ii =v+Ri

= −1Ri

RS′RS′ + RF

R1RL

11

Ro+ 1

RL

iA

To obtain:

R1

vB

+

-

V2+

V2-

-

+

iA=is independentvo

Ri Ro RL

Rsis

ii

v+

iF

RF

io

and vB = −R1RL

vo

T = AiiiA

= −ARi

RS′

RS′ + RF

R1RL

11

Ro+ 1

RL

= γA, (γ < 0)



Now for the direct transmission we null the current dependent current generator, iA: note that it is left an open circuit since it is a current generator.

For this particular case the fed backed amplifier A2 shows a very small transmission between its output and its input, being fed backed. Anyway, let’s call α this small term:

io =1β2′

−T21 − T2

iS +ADIR21 − T2

iS

-

+

voRi Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

iA

A2

io

Let’s suppose Ri, Rs, and Ri2 very large to simplify.

We see that if A2=∞, then V2- is 0 and 1/β2’=0. We remain with:

+

Ro RL

is

v+

iF

RF

R1

vB

V2+

V2-

iA io

Ro2

-

+A2

Rs

Ri

vo

io =ADIR21 − T2

iS



If Ro2 were 0 the transmission would be 0. Otherwise:

Now we need to know the loop gain of A2:

Very simple (Rs≈∞):

io = −Ri

Ri + RF′

Ro2Ro2 + R1 + RL + Ro

is ≈ −Ri

RF′

Ro2Ro

is

v2− =RF + Ri ‖ R1 + RL + Ro

RF + Ri ‖ R1 + RL + Ro + Ro2

RL + RoR1 + RL + Ro

VT ≈ VT

Therefore: T2=-A2v2-/VT≈-A2 and:

io = −Ri

RF′

Ro2Ro

11 + A2

= 𝛼𝛼is

+

voRo RL

is

v+

iF

RF

R1

vB

V2+

V2-

iA io

Ro2

-

+A2

Rs

Ri

+

voRo RL

is

v+

iF

RF

R1

vB

V2+

V2-

iA io

Ro2

-

+

VT

Rs

Ri

RF′ = RF + Ro2‖ R1 + RL + Ro



Finally, let’s determine 1/β, considering A=∞:

Remember, if A=∞ it results ii≈0 and v-≈v+, but in this case v-=0, then v+=0, too, and:

is = −vBRF

,

vo =1β−T

1 − T is +ADIR1 − T is

As it can be seen the gain does not depends on both the input source and output load impedances, that suggests that the above is a current output over current input amplifier.

Finally:

io =RFR1

−T1 − T is +

𝛼𝛼1 − T is

-

+

iA=∞iivo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

io = −vBR1

io =RFR1

is

T = AiiiA

= −ARi

RS′RS′ + RF

R1RL

11

Ro+ 1

RL

α ≈ −Ri

RF′

Ro2Ro

11 + A2


18. INPUT IMP. OF A C-TO-C FED BACKED CURRENT OA

The procedure we developed for measuring the input/output impedance for the voltage amplifier remains valid also for this configuration, as it was determined using a general approach.

Being amplified a current we expected that:

Rif =Riol

1 − T

Riol being the impedance measured when the gain is set to 0 and a test current is injected at the input.

Since the output is a current, then:

Again, Rool is the impedance measured at the output, by injecting a test voltage in series to the load, when the gain is set to 0.

Rof = Rool 1 − T

-

+

iA=Aiivo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io


………. input impedance of a c-to-c fed backed current OA 2

Just to exercise let’s verify the statements given in the previous slide.

From the circuit above we expect that:

Rif =v+is

=iiRiis

=iARiAis

Now if we consider iA the branch where to consider the feedback:

iA =1β′

−T1 − T is +

ADIR′1 − T is

iA

We see that if A=0 than iA cannot be different than 0, since the current source is replaced by an open. We have seen that Io=αIs, but IA remains 0 since the impedance in the branch is ∞.

-

+

voRi Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

-

+

voRi Ro

RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

A2

=present evaluated feedback node

iA =1β′

−T1 − T is + ⋯

Previous feedback branch



So we remain with:

Now let’s consider A=∞:

Therefore:

iA =1β′

−T1 − T is

voRL

= Io =RFR1

isiA =voRo

+voRL

and

iA =RLRo

+ 1RFR1

is =1β′ is

Rif =iARiAis

=RLRo

+ 1RFR1

RiA

−T1 − T

But we know that T is proportional to A:

Finally:

CVD!

-

+

iA=∞iivo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

vo =RLRF

R1is

=present evaluated feedback node

T = γA, (γ < 0)

Rif =v+is

=iiRiis

=iARiAis

=RiA

1β′−γA1 − T =

Ri ⁄(−γ β′)1 − T =

Riol1 − T



We can give an evaluation of the open loop input impedance Riol. From above we have that node at voltage VB is almost at 0 V if the main OA gain if nulled:

Therefore:

In a current OA amplifier Ri is small, order of 1 Ω or less, so:

-

+

voRi Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

Rif =RS RFRi

1 − T

Riol = RS RFRi

Rif ≈Ri

1 − T



A consideration (a):

Likewise with the voltage amplifier we would disentangle the input impedance of the fed–backed OA from Rs. Or:

-

+

iA=Aiivo

Ri Ro RL

Rsis

ii

v+

RF

R1

vB

+

-V2+

V2-

io

Norton

-

+

voRi Ro RL

Rsis

ii

v+

RF

R1

vB

+

-

V2+

V2-

io

NortonImpedance, RN

Current nulled

vB ≈0

If iA=0 then vB=0. Therefore the Norton impedance reduces to RF.

Rif = Rs Riol∞

1 − T∞



Now the Norton current, namely, the current in the short set at the node:

-

+

iA=Aiivo

Ri Ro RL

Rsis

ii

v+

RF

R1

vB

+

-

V2+

V2-

io

Norton

That results in:

IN

iN = −Ro

RO + RL

R1RF

iA = −αiA = −αAii

io =Ro

Ro + RLiA vB = −R1io iN =

vBRF

Or:

-

+

iN=-αAii

Ri

Rsis

ii

v+

RF

iR



Now we have that, in case is=0:

-

+

iN=-αAii

Ri

Rsis

ii

v+

RF

iR

ii⁄1 Ri

=iN⁄1 Riol

⇒ii

Riol=

iNRi

|ii Rs=∞⁄1 Ri

=iN

⁄1 Riol∞⇒

|ii Rs=∞Riol∞

=iNRi

But:

iiiN

=ii

−αiA=

T−αA

|ii Rs=∞iN

=|ii Rs=∞−αiA

=T∞−αA

We obtain:

iiiN

=RiolRi

=T

−αA

|ii Rs=∞iN

=Riol∞

Ri=

T∞−αA

From which:T

Riol=

T∞Riol∞

TRiol

=−αA

Ri

T∞Riol∞

=−αA

Ri

T =Rs

Rs + Riol∞T∞



An we have just shown that:

The open loop input impedance results:

From which:

TRiol

=T∞

Riol∞

Riol = Rs RiRF

1Rif

= 1 − TRiol

= 1Riol

− TRiol

= 1Rs

+ 1Riol∞

− T∞Riol∞

= 1Rs

+ 1 − T∞Riol∞

= 1Rs

+ 1Rif∞

(here Riol∞ is Riol evaluated when Rs →∞.)

-

+

voRi Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

iA

A2

io

CVD!

(let’s assume negligible the output impedance of A2 with its feedback)



As with the v-to-v configuration we see here that it is not convenient to adopt a voltage source at the input to measure the input impedance.If we short vs the loop gain is the same found previously:

A consideration (b):

We see that T → 0 if Rs → 0, and this suggest to use a current source.Insisting anyway in using a voltage source we obtain (Norton equivalent):

-

+

iA=Aiivo

Ri Ro RL

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

Is

Rs

vs

T = AiiiA

= −ARi

RS′RS′ + RF

R1RL

11

Ro+ 1

RL

RS′ = RSRi

io =1β−T

1 − T is +ADIR1 − T is =

1β−T

1 − TvsRs

+ADIR1 − T

vsRs

io Rs→0−Tβ

vsRs

+ ADIRvsRs

io Rs→0RFR1

ARi

1RF

R1RL

11

Ro+ 1

RL

vs −Ro2

RFRo

11 + A2

vsNon più indipendente da A.



So that it is not recommended to adopt this configuration if Rs has a negligible value. Nevertheless, we will see that this topology is useful, provided that some precautions are taken. In case Rs has a finite value we can consider to measure the impedance seen by the voltage source Vs:

This input impedance results slightly larger than Rs, suggesting that Rs results in series with a small impedance.This small impedance results Rif∞ as it is shown in the following pages:

Rifs =vsis

= Rs + Rif∞

-

+

iA=Aiivo

Ri Ro RL

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

Is

Rs

vs

or:

is =vs − v+

Rs=

vsRs

−RiiiRs

=vsRs

−RiiARsA

is =vs − v+

Rs=

vsRs

−RiRs

⁄−γ β′′1 − T vs

is = 1 −Ri ⁄(−γ β′′)

1 − TvsRs

Rifs =vsis

=Rs

1 − Ri ⁄(−γ β′′)1 − T

iA =1β′′

−γA1 − T vs =

−A ⁄γ β′′1 − T vs

T = γA, (γ < 0)



To prove that: Rifs =

vsis

= Rs + Rif∞

We start by considering that:

iA =1β′′

−T1 − T =

A ⁄γ β′′

1 − T vs

While, in case the source is a current:

iA =1β′

−T1 − T =

A ⁄γ β′

1 − T is

The 2 configurations are linked by the Norton theorem:

-

+

iA=Aiivo

Ri Ro RL

Rsis

ii

v+

RF

R1

vB

+-

V2+

V2-

io

Norton

Is

Rs

vs

-

+

iA=Aiivo

Ri RoRL

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

Is

Rs

vs

vs = isRs



Results in:

We have shown that:

-

+

iA=Aiivo

Ri Ro RL

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io

Is

Rs

vs

vs = isRs, and

Rs =β′′

β′ or1β′′ =

1Rsβ′

Rif =Ri ⁄γ β′

1 − T

Rifs =Rs

1 − Ri ⁄𝛾𝛾 β′′1 − T

=Rs

1 − 1Rs

Ri ⁄γ β′1 − T

=Rs

1 − RifRs

So that:

Rifs = Rs

1 − RifRs

+ RifRs

1 − RifRs

= Rs 1 +

RifRs

1 − RifRs

= Rs 1 +1

Rs

11

Rif− 1

Rs

By continuing:

Rifs = Rs 1 +1

Rs

11

Rif∞

= Rs + Rif∞CVD!

iA =A ⁄γ β′′

1 − T vs iA =A ⁄γ β′

1 − T is


19. OUTPUT IMPEDANCE OF A C-TO-C FED BACKED CURRENT OA

But, at the input mesh, with A=∞ is ii=0, hence v+=0, iF=0 and:

And we reduces to:

io =1β′′

−T1 − T vT +

ADIR′′1 − T vT

io = 0

io =ADIR′′1 − T vT

where Rool is the impedance measured at the output, by injecting the test voltage in series to the load, see above, when the gain is set to 0.

By applying vT we expect:

Or (take care of the direction chosen for the current) and that V2-=0:

Rofs =vT−io

=1 − T−ADIR′′

= Rool 1 − T

… now let’s determine ADIR’’ that is our -1/Rool…

-

+

iA=Aiivo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io+vT

We want to show now what we introduced in the previous chapter:

Rof = Rool 1 − T


………. output impedance of a c-to-c fed backed current OA 2

But it is evident that:

Assuming that V2- is almost 0 it is:

So that it is proved the last position of the previous page.

-

+

voRi Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io+vT

io = −vT

RL + Ro= ADIR′′vT

Rool =vT−io

= RL + Ro



ii = −αR1

RoolvA

Rof = RL + Rool0 1 − T0

Adopting the usual procedure we can show that it is valid that:

But also:

To arrive at this we must put both vT and iA to see the same impedance Rool or Rool0:

So now if vT is 0 we have that:

-

+

iA independent

voRi Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

iL+vT

Thevenin

-

+

VA=RoiAvo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

iL+vT

Roolii = −αR1vA

Rool0 ii RL=0= −αR1vA

Consideration (c):



iivA

=ii

RoiA=

TRoA

Nevertheless:

We have:

Therefore, from the previous slide:

-

+

VA=RoiAvo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

iL+vT

Roolii = −αR1vA

|ii RL=0vA

=|ii RL=0RoiA

=T0

RoA

Rool0 ii RL=0= −αR1vA

RoolT = Rool0T0

T =Rool0

RL + Rool0T0



Therefore:

Rool = Ro + RLRoolT = Rool0T0

where Rool0 and T0 are Rool and T when RL is 0 Ω.

Rof = Rool 1 − T

= Rool−RoolT

= RL + Rool0−RoolT

= RL + Rool0−Rool0T0

= RL + Rool0 1 − T0

= RL + Rif0

-

+

VA=RoiAvo

Ri Ro RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

iL+vT


20. INPUT IMPEDANCE OF A C-TO-C FED BACKED OA

In the previous configuration studied we have used a current OA for implementing a fed backed current amplifier. This choice is natural, but current OAs are not easy to be found off-the-shelf or, if they are available, are expensive. Normally, the fed backed current amplifier is implemented with a standard voltage OA. We will study now also this configuration and put into evidence the differences.

The evaluation procedure is rather similar to that seen before and, as it can be imagined, the result does not depend on the kind of OA used, so we expect that it will be valid that:

-

+

vA=A(v+-v-)

voRi RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

ioRo

Rif =Riol

1 − TSimilarly as for the previous case, if we put A=0, then VB results 0 V, therefore:

Riol = RS RFRi


………. input impedance of a c-to-c fed backed OA 2

Let’s compare the difference of the results obtained for the 2 types of OAs.In both cases the expression for the input impedance is the same, namely the ratio between the open loop input impedance divided by the 1-T term.What is remarkably different is the value of the open loop input impedance. By considering negligible the output impedance of the second OA, for both circuits we obtained:

Riol = Rs RFRi

For the current OA Ri is of a few Ω and we found:

Riol_current_OA ≈ Ri

For the standard OA Ri can be very large, or a few MΩ at least and the above approximation does not applies and Riol could be limited by RF, a few KΩ.


………. input impedance of a c-to-c fed backed OA 3

Just for giving the order of magnitude let’s suppose that our OAs have a voltage/current gain of the order of 107, a term close to 1 for f(β’) and the closed loop gain is 100, then T results of the order of 105.

As a result:

Rif ≜RiolcurrentOA

1 − T =1

105 ÷ 10−5Ω

Rif ≜RiolstandardOA

1 − T =1 MΩ105

÷ 10 Ω

That is the difference.

For the voltage OA we can consider 2 situations. One in which Riol is dominated by Ri:

The second in which Riol is dominated by RF, of a few KΩ:

Rif ≜RiolstandardOA

1 − T =1 KΩ105 ÷ 10−2 Ω


21. OUTPUT IMPEDANCE OF A C-TO-C FED BACKED OA

Also or the output impedance we expect the same behaviour:

Where, being V2- a virtual ground, setting A=0:

Rofs = Rool 1 − T

For the present case Ro is small, and RL normally dominates. If it is at the KΩ value and T≈-105, then:

-

+

voRi RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

io+vT

Ro

vA=A(v+-v-)=present evaluated feedback node

Rool = Ro + RL

Rofs_voltage_OA ≈ 103 × 105 = 108 Ω

In case the current OA were used Ro can be of the order of a few MΩ at least and dominates Rool. Therefore:

Rofs_current_OA ≈ 106 × 105 = 1011 Ω

Again the difference between the 2 types of OAs is in the final value of the fed backed impedance.


………. output impedance of a c-to-c fed backed OA 2

As with the v-to-v amplifier we can try to measure the output impedance using the complementary test source, a current source in this case.We have to take care the way we connect the source. The connection shown below does not allow to measure the output impedance.

Indeed, considering that 1/β’=0:

vo = RLio = RLAR′

1 − T iT

= RLRo

Ro + RL

11 − T iT

Ro_wrong =voiT

=RLRo

Ro + RL

11 − T

A very small value: that is correct as the feedback wish to force a negligible current into RL and, as a consequence, its dropout voltage is negligible: the current iT flows into vA.

-

+

voRi RL

Rsis

ii

v+

iF

RF

R1

vB+

-V2+

V2-

io

Ro

vA=A(v+-v-)

iT

Consideration:



The impedance must be characterized in parallel to the load RL as it is shown below: the terminal of iT previously connected to ground is now connected to the second terminal of RL, see the red connection.

This is an interesting application of the A=∞ case. Considering A=∞ it results that v+=0, then iF=0 and therefore vB=0 and the consequence is that the current in R1 is 0, so it must results that, at node V2-:

-

+

voRi RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

ioRoiT

io = iT

For what concern the direct transmission:

io =Ro

RL + RoiT

-

+

voRi RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

ioRo

vA=A(v+-v-)

iT



Putting together:

io =−T

1 − T iT +Ro

RL + Ro

11 − T iT

Now, the impedance seen by the current generator satisfies:

Rofp =vo − v2−

iT=

RLioiT

=RLiT

−T1 − T +

RoRL + Ro

11 − T iT

= RL−T RL + Ro + Ro

RL + Ro 1 − T

= RL−T RL + Ro + Ro + RL − RL

RL + Ro 1 − T − RL + RL

= RLRofs − RL

Rofs − RL +RL

= RLRofs0

Rofs0 +RL

= Rofs0RL

-

+

voRi RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

ioRo

vA=A(v+-v-)

iT


22. LOOP GAIN OF A C-TO-C FED BACKED OA

Let’s determine T:

In this case Rs’ can be >> RF and RL>>Ro:

T = Av+vA

= −ARS′

RS′ + RF

R1Ro + RL

T = AiiiA

= −ARi

RS′RS′ + RF

R1RL

11

Ro+ 1

RL

Rs′ = RsRi

T≈− AR1RL

We can compare this loop gain with that obtained when the OA is a current amplifier:

For which Ri is << Rs and Ro >> RL:

T = AiiiA

= −AR1RF

So that the loop gains in the 2 cases are similar if RF and RL are comparable.

-

+

voRi RL

Rsis

ii

v+

iF

RF

R1

vB

+

-

V2+

V2-

ioRo

vA=independent


23. INPUT/OUTPUT IMP. SUMMARY OF A C-TO-C FED BACKED OASummary about the input/output impedances of a c-to-c.Input impedance measured with a current source with a parallel resistance Rs :

Rifp =Riol

1 − T

Rifp = Rs Rifp∞

Where Riol∞ and T∞ are Riol and T measured when Rs is set to (open) ∞ Ω.

If the input impedance is measured using a voltage generator with in series a source resistor Rs we have:

Rif𝑠𝑠 = Rs+Rifp∞

oooOOooo

Output impedance measured with a voltage source in series to the load resistor RL:

Rofs = Rol 1 − T

Rofs = RL+Rofs0

Where Rools0 and T0 are Rool and T measured when RL is set to 0 Ω.

If the output impedance is measured using a current generator in parallel to the load RL we have:

Rofp = RLRofs0

RoolT = Rool0T0, T =Rool0

RL + Rool0T0

TRiol

=T∞

Riol∞, T =

RsRs + Riol∞

T∞


24. OTHER CONFIGURATIONS

So far we have studied the v-to-v and c-to-c configurations.

There are other 2 possible and affordable feedback networks: the v-to-c (voltage input, current output) and the c-to-v (current input, voltage output).

There is no reason why these configurations should not behave as the two already studied. Indeed, this is what it happens and we will investigate this shortly.


25. THE V-TO-C FED BACKED OA

Having already shown that the kind of OA does not have impact on the feedback loop we consider only voltage to voltage OAs from now on.

As usual we expect that:

i0 =1βR

−T1 − T vs +

ADIR1 − T vs

Where we must not forget that T is dimensionless, but, this time, βR has the dimension of Ω, as well as ADIR that of Ω-1.

Now, if we assume A=∞ we can evaluate 1/βR:

+

-

vA=A(v+-v-)

voRi Ro RL

Rs

vs

v-

v+

RARB RC

io

vB

v− = vsvB − vs

RB=

vsRA

vB =RA + RB

RAvs

io = iB +VBRC

=vsRA

+RA + RB

RCRAvs

=RC + RA + RB

RCRAvs

(v−≈v+)

(i−≈i+≈0)

iB

iB


………. the v-to-c fed backed OA 2

If we consider the output voltage vo rather than the output current we have:

So we see that the output voltage depends on the load impedance, that suggest that the more suitable output variable is the current.

The output does not depends on the source resistor Rs and this tells us that the input variable is a voltage.

+

-

vA=A(v+-v-)

voRi Ro RL

Rs

vs

v-

v+

RARB RC

io

vB

vo = RLio + vB = RLRC + RA + RB

RCRAvs +

RA + RBRA

vs



…and now ADIR:

Introducing:

ADIR =iovs

= −1

Ro + RL

RC′RC′ + RB

RA′RA′ + Ri′

+

-

vA=0

voRi Ro RL

Rs

vs

v-

v+

RARB RC

io

vB

RA′ = RA‖ RB + RC′ RC′ = RC‖ Ro + RLRi′ = Ri + Rs

v− =RA′

RA′ + Ri′vs vB =

RC′RC′ + RB

v− io = −1

Ro + RLvB

We can write:

Or:

io = −1

Ro + RL

RC′RC′ + RB

RA′RA′ + Ri′

vs

Therefore:


………. the v-to-c fed backed Oa 4

Let’ determine T:

We have:

T = Av+ − v−

vA= −

ARiRi + Rs

RA"RA" + RB

RC"RC" + Ro′

+

-

vA=generic

voRi Ro RL

Rs

vs

v-

v+

RARB RC

io

vB

Again, introducing:Ro′ = Ro + RL RC" = RC‖ RB + RA" RA" = RA‖ Ri + Rs

vB =RC"

RC" + Ro′vA v− =

RA"RA" + RB

vB

v+ − v− = −Ri

Ri + Rsv−

Therefore:

And:

io =RC + RA + RB

RCRA

−T1 − T vs −

1Ro + RL

RC′RC′ + RB

RA′RA′ + Ri′

11 − T vs

REMEMBER: the contribution of ADIR has not necessarily the same sign of the fed backed gain 1/β.

v+ =RS

Ri + Rsv−



Input impedance:

+

-

vA=0

voRi Ro RL

Rs

vs

v-

v+

RARB RC

io

vB

Considering the definitions already given:

RA′ = RA‖ RB + RC′ RC′ = RC‖ Ro + RLRi′ = Ri + Rs

We can write:

Riol = Ri′ + RA′

Of course:

Rif = Riol 1 − T = Ri′ + RA′ 1 − T



Output impedance:

Rool = Ro + RL + RC"

Therefore:

Rofs = Rool 1 − T = Ro + RL + RC" 1 − T

+

-

vA=0vo

Ri Ro RL

Rs

vs

v-

v+

RARB RC

io

vB

+vT

Again, introducing:Ro′ = Ro + RL RC" = RC‖ RB + RA" RA" = RA‖ Ri + Rs


26. THE C-TO-V FED BACKED OA

The expect result:

v0 =1βG

−T1 − T is +

ADIR1 − T is

Again, we must not forget that T is dimensionless, but, this time, βG has dimension: Ω-1, as well as ADIR: Ω.

Setting A=∞ we can get 1/βG:

v-=0, then the current is flows all into RF and:

is =v− − vo

RF

Or:

vo = −RFis =1βG

is

Note that vo does not depends on RL and Rs, proving that the input variable is a current while the output a voltage.

+

-

vA=A(v+-v-)

voRi Ro RLv-

v+

RF

io

Rsis


………. the c-to-v fed backed OA 2

Loop gain T calculation:

Be:

Ri′ = RiRs

To give:

And:

T = Av+ − v−

vA= −A

Ri′RF′

RLRF′

RLRF′ + Ro

+

-

vA=Generic

voRi Ro RLv-

v+

RF

io

Rsis

and: RF′ = RF + Ri′

We have:

vo =RLRF′

RLRF′ + RovA v− =

Ri′Ri′ + RF

vo =Ri′RF′

vo

v− =Ri′RF′

RLRF′

RLRF′ + RovA



Direct transmission:

Starting from:

v− =1Ri

+1

RS+

1RF + RLRo

−1

is vo =RLRo

RF + RLRov−

vo =RLRo

RF + RLRo

is⁄1 Ri + ⁄1 Rs + ⁄1 RF + RLRo

= ADIRis

We obtain:

+

-

vA=0

voRi Ro RLv-

v+

RF

io

Rsis

vo = −RF−T

1 − T is +

+RLRo

RF + RLRo

1⁄1 Ri + ⁄1 Rs + ⁄1 RF + RLRo

11 − T is

In the end:



Input impedance:

And:

Riol =1Ri

+1

Rs+

1RF + RLRo

−1

Riof =1Ri

+1

Rs+

1RF + RLRo

−1 11 − T

+

-

vA=0

voRi RoRL

v-

v+

RF

io

Rsis



Output impedance:

Rool =1

Ro+

1RL

+1

RF + RsRi

−1

Rofp =1

Ro+

1RL + RF + RsRi

−1 11 − T

Therefore

+

-

vA=0

voRi Ro RLv-

v+

RF

io

Rsis

iT


27. QUADRUPOLE MODEL OF AN AMPLIFIER

We are now in a condition of giving a simpler representation of our fed backed amplifier, useful in several applications and particularly important when we design a system for a customer.

+

-Vi RL

+ vo

voq=aqvi

vs

Rs

The model we would like to obtain is a quadrupole:

Riq

Roq

Such that:

vo =Riq

Riq + Rsaq

RLRL + Roq

vs

Where Riq, Roq and aq are parameters extracted from our previous characterization of the fed backed network.

At minimum we need the 3 parameters above. If we have to consider that the feedback network is not ideal, but bi-directional, some additional parameter must be added.


………. quadrupole model of an amplifier 2

The simplest approximation is like this:

+

-Vi RL

+ vo

voq=Afvi

vs

Rs

Rif0

Rof∞

Such that:

vo =Rif0

Rif0 + RsAf

RLRL + Rof∞

vs

Where, if we represent T in the form: T=T(Rs, RL):

vo =Rif0

Rif0 + Rs

1β−T(0,∞)

1 − T(0,∞)RL

RL + Rof∞vs

(let’s omit for a while the term given by the direct transmission)



We have to remember that:

Unfortunately, because the feedback is not unidirectional, we expect that:

Rif0= Rif0(RL), Rof∞= Rof∞(Rs), ADIR=ADIR(Rs, RL).

Rif = Rs + Riol0 1 − T0 =Rs + Rif0

Rof = RL Rool∞

1 − T∞= RL‖Rof∞

vβ =1β−T

1 − Tvs +

ADIR1 − T vs

T(Rs, RL) = T 0, RLRiol0

Rs + Riol0

T(Rs, RL) = T Rs,∞RL

RL + Rool∞



A first step is to disentangle, for instance, the source from the input:

We know from the theory developed:

RiolT = Riol0T 0, RL

vo =1β−T

1 − T vs

=1β−T 0, RL

1 − T 0, RL

Rif0Rif0 + Rs

vs

Proof:−T 0, RL

1 − T 0, RL

Rif0Rif0 + Rs

=−T 0, RL

1 − T 0, RL

Rif0Rif

=−T 0, RL

1 − T 0, RL

Riol0 1 − T 0, RL

Riol 1 − T 0, RL

But we have already shown that it is valid:

Therefore:

−T 0, RL Riol0Riol 1 − T

=−RiolT

Riol 1 − T=

−T1 − T

CVD!

+

-Vi RL

+ vo

voq=Afvi

vs

Rs

Rif0

Rof∞

Remember: Rif0=Rif0(RL).

Proved below →



Now to the output:

From the model above:

+

-Vi RL

+ vo

voq=Afvi

vs

Rs

Rif0

Rof∞

With:

voq = vo RL=∞=

1β−T 0,∞

1 − T 0,∞|Rif0 RL=∞

|Rif0 RL=∞ + Rsvs

Rof∞ = Rof∞(Rs)=|Rof∞ Rs=0

1 − T 0,∞ Ril0Ril0 + Rs

Rif0 RL=∞= Ril0 RL=∞

1 − T 0,∞

The 2 critical parameters are: |Rof∞ Rs=0 and |Rif0 RL=∞.Things simplifies if Ril0 is >> Rs and Rof∞ and Rif0 are tiny dependent on Rs and RL respectively. Under this assumption we can give the parameters Rif0, Rol∞ and Af=1/β(-T(0,∞)) /(1- T(0,∞) ), to give:

voq =1β−T 0,∞

1 − T 0,∞vi = Afvi

vo = AfRif0

Rif0 + Rs

RLRof∞ + RL

vs



A similar evaluation can be done for ADIR.

A similar approach can be used when the fed backed OA is a current amplifier, or transimpedance amplifier or transconducatnce amplifier.


28. LOOP GAIN EVALUATION OF CASCADED STRUCTURESSometimes the fed backed amplifier is composed by the cascade of several amplifiers. In this case there are 2 possible configurations. The first is when every stage is an open loop stage itself. In this case there is no need for a particular precaution in choosing the source where to open the loop and find the loop gain. Here an example:

Some precautions are instead to be taken when inside the loop there are one or more closed loop amplifiers. In this case it is not recommended to choose the controlled source to be made independent at a nested fed backed amplifier: that amplifier is subjected to a double feedback and the procedure we developed so far does not apply to it, unless some other rules are taken.

vA=is generic now

Ri Ro+

-

+

-

β

vA=is generic now

Ri Ro+

-

+

-

ββ’

Not recommended here!


………. loop gain evaluation of cascaded structures 2

To avoid any possible misleading it is a good choice to make generic a source at a stage not fed backed, first:

Then, a second useful step is to replace the inner local closed loop amplifier(s) with its model:

+

-

+

-

β

vA=is generic now

Ri Ro

β’

+

-

β

vA=is generic now

Ri Ro


29. BIBLIOGRAPHY

La storia del Silicio,F S Norman e G Einspruch, Boringhieri88-339-1117-9

Sergio FrancoAmplificatori 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;

Sergio Franco Electric Circuits FundamentalsOxford University Press, USA,1994, C. Biblio 621.3815 FRAS.ELE/1995

Sergio Franco Design with Operational Aplifiers and Analog Integrated CircuitsMcGraw-Hill, 2002, C. Biblio 621.3815 FRAS.DES/2002.


APPENDIX A: voltage amplifier

Let’s study T.

The aim here is to measure the input impedance Rif and the output impedance ROf of the below network, for which we already studied the gain:

A=104, Ri=106, Rs=RO=100 Ω, R1=1000 Ω, R2=9 KΩ, RL=200 Ω.

Let’s write the node equations:

With:

R1’=R1||(Rs+Ri)=999 Ω

+

-

R1

R2

vs=0 voRivi Ro

+

RL

vA=is generic nowRs

vA − voRo

=voRL

+vo − v−

R2

vo − v−R2

=v−R1

+v−

Rs + Ri

vi = v+ − v−

vARo

=1

Ro+

1RL

+1

R2vo −

v−R2

voR2

=1

R2+

1R1

+1

Rs + Riv−

vi = v+ − v−

vARo

=1

Ro+

1RL

+1

R2vo −

v−R2

v− =R1′

R2 + R1′vo

vi = v+ − v−


………. Appendix A (2)

In addition:

R1’=R1||(Rs+Ri)=999 Ω e RE=RL||(R2+R1’).

+

-

R1

R2

vs=0 voRivi Ro

+

RL

vA=is generic nowRs

vARo

=1

Ro+

1RL

+1

R2vo −

v−R2

v− =R1′

R2 + R1′vo

vi = v+ − v−

vARo

=1

Ro+

1RL

+1

R21 −

R1′R2 + R1′

vo

v− =R1′

R2 + R1′vo

vi = v+ − v−vARo

=1

Ro+

1RL

+1

R2 + R1′vo

v− =R1′

R2 + R1′vo

vi = v+ − v−

vARo

=1

Ro+

1RE

vo

v− =R1′

R2 + R1′vo

vi = v+ − v−

vo =RE

RE + RovA

v− =R1′

R2 + R1′vo

vi = v+ − v−



Therefore:

And then:

with an error of the order of 0.2%

+

-

R1

R2

vs=0 voRivi Ro

+

RL

vA=is generic nowRs

vo =RE

RE + RovA

v− =R1′

R2 + R1′vo

vi = v+ − v−

vo =RE

RE + RovA

v− =R1′

R2 + R1′vo

vi =Rs

Rs + Riv− − v−

vo =RE

RE + RovA

v− =R1′

R2 + R1′vo

vi = −Ri

Rs + Riv−

v+ − v− = −Ri

Rs + Ri

R1′R1′ + R2

RERE + Ro

vA = βf(β)vA

T =A v+ − v−

vA= −A

RiRs + Ri

R1′R1′ + R2

RERE + Ro

= −661.6

Af =1β

−T1 − T =

1β 0.99849



It is interesting to verify which values takes T0∞:

+

-

R1

R2

vs=0 voRivi Ro

+

RL

vA=is generic nowRs

The effect of Rs over T is marginal when becomes negligible, as

R1’=R1||(Rs+Ri)=999.001 Ω and R1’0=R1||(Ri)=999.0009 Ω, namely 999 Ω.

Different is the effect of RL when it takes very large values:

RE=RL||(R2+R1’)= 196 Ω, while RE∞= R2+R1’=9999 Ω.

The loop gain T, from the value of -661.6, becomes, after that Rs is set to 0 and RL to ∞:

The factor –T/(1-T) from the value 0.99849 now takes:

T = −AR1′0

R1′0 + R2

RE∞RE∞ + Ro

= −989.2

T0∞1 − T0∞

= 0.99899



The input impedance is determined starting by nulling the gain and measuring the input current consequent to a source excitation:

Input current Is comes from:

+

-

R1

R2

vsvo

Rivi Ro

+

RL

Is Rs

vs − v−Rs + Ri

=v−R1

+v− − vo

R2

v− − voR2

=voRL

+voRo

Is =vs − v−Rs + Ri

vs − v−Rs + Ri

=v−R1

+v− − vo

R2

v−R2

=voR2

+vo

RL‖Ro


vsRs + Ri

=v−

Rs + Ri+

v−R1

+v−R2

1 −RL‖Ro

R2 + RL‖Ro

vo =RL‖Ro

R2 + RL‖Rov−


vs − v−Rs + Ri

=v−R1

+v− − vo

R2

vo =RL‖Ro

R2 + RL‖Rov−




+

-

R1

R2

vsvo

Rivi Ro

+

RL

Is Rs

vsRs + Ri

=v−

Rs + Ri+

v−R1

+v−R2

1 −RL‖Ro

R2 + RL‖Ro

vo =RL‖Ro

R2 + RL‖Rov−


vsRs + Ri

=v−

Rs + Ri+

v−R1‖ R2 + RL‖Ro

vo =RL‖Ro

R2 + RL‖Rov−


v− =R1‖ R2 + RL‖Ro

Rs + Ri + R1‖ R2 + RL‖Rovs

vo =RL‖Ro

R2 + RL‖Rov−




+

-

R1

R2

vsvo

Rivi Ro

+

RL

Is Rs

Therefore:

And::

v− =R1‖ R2 + RL‖Ro


vo =RL‖Ro

R2 + RL‖Rov−


Is =1

Rs + Ri1 −

R1‖ R2 + RL‖RoRs + Ri + R1‖ R2 + RL‖Ro

vs

Is =1


Rir =vsIs

= Rs + Ri + R1‖ R2 + RL‖Ro ≈ 1.0012 × 106 Ω

Rif = Rs + Ri + R1‖ R2 + RL‖Ro 1 − T≈ 1.0012 × 106 Ω 1− T = 662.7 × 106 Ω



The output impedance can be evaluate starting by nulling the input and the gain:

R1’=R1||(Ri+Rs)

and:

+

-

R1

R2

vTvoRivi Ro

RL IT

Rs

iT =vTRL

+vTRo

+vT − v−

R2

vT − v−R2

=v−R1

+v−

Ri + Rs

iT =vTRL

+vTRo

+vT − v−

R2

vTR2

=v−R2

+v−R1′

iT =vTRL

+vTRo

+vT − v−

R2

v− =R1′

R1′ + R2vT

iT =vTRL

+vTRo

+vT − v−

R2

v− − vT = −R2

R1′ + R2vT

iT =vTRL

+vTRo

+vT

R1′ + R2

v− − vT = −R2

R1′ + R2vT

Rool =vTiT

= RL Ro R2 + R1′ ≈ 66.25 Ω

Rof =Rool

1 − T≈

66.25661.6

= 0.1 Ω



+

-

R1

R2

vsvo

Rivi Ro

+

RL

Rs

Therefore:

And::

Finally the direct transmission:

Let’s try to determine it looking at the topology of the network. Let’s define:

RL′ = RLRo R2′ = R2 + RL′ R1′ = R1R2′

Then:

v− =R1′

R1′ + Ri + Rsvs vo =

RL′RL′ + 𝑅𝑅2

v−

vo =RL′

RL′ + 𝑅𝑅2R1′

R1′ + Ri + Rsvs

ADIR =vovs

=RL′

RL′ + R2

R1′R1′ + Ri + Rs

= 6.6 × 10−6



So that:

The overall gain is then:

ADIR =RL′

RL′ + R2

R1′R1′ + Ri + Rs

= 6.6 × 10−6

vo =R1 + R2

R1

−T1 − T vs +

ADIR1 − T vs

A=104, Ri=106, Rs=RO=100 Ω, R1=1000 Ω, R2=9 KΩ, RL=200 Ω.

T = −ARi

Rs + Ri

R1′R1′ + R2

RERE + Ro

= −661.6

With:

vo = 10 × 0.9985 × vs +9.99 × 10−9 × vs


APPENDIX B: unity gain voltage amplifierLet’s determine now the input and output impedance of the unity gain buffer amplifier:

A=104, Ri=106, RO=100 Ω, RL=200 Ω, Rs=50 Ω.

By assuming for a while that A=∞ we know that v+≈v- and:

vo=vs, namely β=1.

Let’s start with T:

Finally:

+

-

voRivi Ro

+

RL

Avi

Vs

Rs

+

Vo =RL′

RL′ + Ro

VA

RL′ = RL‖ Ri + Rs

= 199,96 Ω

V+ − V− = −Ri

Rs + RiVo

T =A V+ − V−

VT= −

RL′

RL′ + Ro

RiRs + Ri

A ≈ −6666

+

-

voRivi Ro

RL

Rs

+

vA=is generic now

+-vs

vo

RL

V+ =R𝑠𝑠

Rs + RiV− therefore:


………. Appendix B (2)

From which:

Finally the output impedance:

+

-

voRi

Ro

RL

Rs

IT

From which:

Rif = Rir 1 − T = Rs + Ri + RoRL (1 − T) = 6,67 GΩ

Ror = RL Ro Rs + Ri = 66.66 Ω

Rof =RL‖Ro‖ Rs + Ri

1 − T≈ 10 mΩ

Now the input impedance in open loop condition:

+

-

voRi Ro

+

RL

VT

Rs


………. Appendix B (3)

+

-

voRi

Ro

+

RL

Vs

Rs

It results:

vo =Ro′

Ro′ + Ri + Rs

vs = ADIRVs

Ro′ = Ro‖RL = 66.66 Ω

vo ≈Ro′

Rivs = 70 × 10−6 Vs

And, considering the loop closed:

vo =−T

1 − T vs +ADIR1 − T vs

vo =6666

1 + 6666vs +

70 × 10−6

1 + 6666 vs

vo = 0,99985 vs + 10,5 × 10−9vs

Finally the direct transmission:


APPENDIX C: current amplifier

In the circuit on the left we have: A=1000, Ro=100 Ω, R1=1000 Ω, R2=4000 Ω, RL=200 Ω, Ri=105 Ω, Rs=107 Ω. Determine β, T, AOL, Rif and Rof.

The non-inverting terminal is at constant potential and we know that the inverting terminal is at the same potential due to the feedback action: therefore we have a current input sensitivity.

Assuming A=∞: v-∼0, results:

Start:

vo = −R2is β = −1

R2

R1′ = R1RiRs = 990 Ω

vA − voRo

=voRL

+vo − v−

R2

vo − v−R2

=v−R1′

Now set:

vA − voRo

=voRL

+vo − v−

R2

vo − v−R2

=v−R1

+v−Ri

+v−Rs

+

-

RL

R1

R2

vo

is

v-

A

Rs

+

-

R1

R2

voRivi Ro RL

v-

vA=is generic now

Rs

Now the loop gain T:


………. Appendix C (2)

vA − voRo

=voRL

+vo − v−

R2

vo − v−R2

=v−R1′

vARo

=voRo

+voRL

+v−R1′

vo = v− +R2R1′ v−

vARo

=RL + Ro

RLRo

R1′ + R2

R1′ v− +

v−R1′

vo =R1′ + R2

R1′ v−

vA =RL + Ro R1

′ + R2 + RLRoRLR1

′ v−

vo =R1′ + R2

R1′ v−

+

-

R1

R2

voRivi Ro RL

v-

vA=is generic now

Rs



vA =RL + Ro R1

′ + R2 + RLRoRLR1

′ v−

vo =R1′ + R2

R1′ v−

vA =RL R1

′ + R2 + R1′ + R2 + RL Ro

RLR1′ v−

vo =R1′ + R2

R1′ v−

vA = R1′ + R2

1 + ⁄Ro RER1′ v−

vo =R1′ + R2

R1′ v−

RE = RL‖ R2 + R1′ = 192.3 Ω

With:

+

-

R1

R2

voRivi Ro RL

v-

vA=is generic now

Rs



vA = R1′ + R2

1 + ⁄Ro RER1′ v−

vo =R1′ + R2

R1′ v−

Or:

vA =R1′ + R2

R1′

RE + RoRE

v−

vo =R1′ + R2

R1′ v−

v− =R1′

R1′ + R2

RERE + Ro

vA

+

-

R1

R2

voRivi Ro RL

v-

vA=is generic now

Rs



The closed loop gain, or 1/β, does not depends on R1, but R1 takes part in

the loop gain T, lowering its absolute value.

In this configuration R1 is not necessary and worse the loop gain and should be avoided. Putting R1=∞ we obtain:

T = Av+ − v−

vA= −A

R1′

R1′ + R2

RERE + Ro = −130.53

T = −ARiRs

RiRs + R2

RERE + Ro = −640.6

v− =R1′

R1′ + R2

RERE + Ro

vA

Then:

+

-

R1

R2

voRivi Ro RL

v-

vA=is generic now

Rs

RE = RL‖ R2 + R1′

R1′ = R1RiRs



Solving:

and:

Rif =Riol

1 − T = 6.05 Ω

+

-

R1

R2

voRivi Ro

RL

v-iTRs

iT =v−Rs

+v−Ri

+v−R1

+v− − vo

R2

v− − voR2

=voRL

+voRo

Riol =v−iT

iT =v−Rs

+v−Ri

+v−R1

+v− − vo

R2

v−R2

=voR2

+vo

RL‖Ro

Riol =v−iT

Input impedance evaluation:

Riol = RS R1 Ri R2 + RLRo = 796.18



Solving:

and:

iT =vTRL

+vTRo

+vT − v−

R2

vT − v−R2

=v−R1

+v−Ri

+v−Rs

Rool =vTiT

iT =vTRL

+vTRo

+vT − v−

R2

vTR2

=v−R2

+v−

R1‖RiRs

Rool =vTiT

Rool = RL Ro R2 + R1‖RiRs = 65.8 Ω

Rof =Rool

1 − T = 0.5 Ω

+

-

R1

R2

voRivi Ro

RL

v- vT

ITRs

Output impedance evaluation:


………. Appendix C (8)Direct transmission:

iT =v−Ri

+v−R1

+v−Rs

+v− − vo

R2

v− − voR2

=voRL

+voRo

iT =v−Ri

+v−R1

+v−Rs

+v− − vo

R2

v−R2

=voR2

+vo

RL‖Ro

iT =v−R1′ +

v− − voR2

v− =RL‖Ro + R2

RL‖Rovo

iT =R1′ + R2R2R1

′RL‖Ro + R2

RL‖Rovo −

voR2

v− =RL‖Ro + R2

RL‖Rovo

iT =R1′ + R2 RL‖Ro + R2 − R1

′ RL‖RoR1′ R2RL‖Ro

vo

v− =RL‖Ro + R2

RL‖Rovo

+

-

R1

R2

voRivi Ro

RL

v-iTRs

R1′ = R1RiRs




iT =R1′ + R2 RL‖Ro + R2 − R1

′ RL‖RoR1′ R2RL‖Ro

vo

v− =RL‖Ro + R2

RL‖Rovo

iT =R1′ R2 + R2 RL‖Ro + R2

R1′ R2RL‖Ro

vo

v− =RL‖Ro + R2

RL‖Rovo

iT =R1′ + RL‖Ro + R2

R1′ RL‖Ro

vo

v− =RL‖Ro + R2

RL‖Rovo

+

-

R1

R2

voRivi Ro

RL

v-iTRs




Finally:

vo =R1′ RL‖Ro

R1′ + RL‖Ro + R2

iT

iT =R1′ + RL‖Ro + R2

R1′ RL‖Ro

vo

v− =RL‖Ro + R2

RL‖Rovo

vo =R1′ RL‖Ro + R2

R1′ + RL‖Ro + R2

RL‖RoRL‖Ro + R2

iT

vo = R1′ ‖ RL‖Ro + R2

RL‖RoRL‖Ro + R2

iT = ADIRiT = 13 iT

+

-

R1

R2

voRivi Ro

RL

v-iTRs



Summary:

Or:

+

- RL

R1

R2

vo

is

v-

A

Rs

T = −AR1′

R1′ + R2

RERE + Ro = −130.53

vo = −R2−T

1 − T is + R1′ ‖ RL‖Ro + R2

RL‖RoRL‖Ro + R2

11 − T is

vo = −R2130.5

1 + 130.5 is +13

1 + 130.5 is

vo = −R20.992is + 0.099is

The ratio between the gain and the direct transmission is:

R20.992is0.099is

= 40 × 103


APPENDIX D: inverting amplifier

+

- RL

R1

R2

vov-

A

Vs

Let’s assume A=∞. Then, v-∼0, since v+=0 and:

This is called inverting configuration and, if R1=R2, it results: vO=-vs. Note that the gain depends on the impedance in series to the source, so the feedback is sensitive to the input current.

The loop gain is the same as that of the previous example of Appendix B, but R1 and Rs are now one resistor:

Output impedance is the same as that of Appendix B, as the output networks are similar.

As we saw at the end of chapter 18, the impedance seen from the voltage source satisfies:

Rifs = R1 + Rifp∞ ≈ R1

Being:

Rif∞ = limR1→∞

Rifp

vsR1

= −voR2

⇒ vo= −R2R1

vs

T = −AR1′

R1′ + R2

RERE + Ro

R1′ = R1‖Ri

RE = RL‖ R2 + R1′


………. Appendix D (2)

+

- RL

R1

R2

vov-

A

vs

Rif

Once-again a remark: remember that the impedance seen from the voltage source is not merely R1 divided by (1-T).

Remember the evaluation done at the note seen at the end of chapter 18, whose result is reported in the pervious page


………. Appendix D (3)

It has to be remarked that with this kind of feedback the connection at the input of a voltage source is unnatural and must be considered with care.

As it can be seen above if the source has a non negligible series impedance Rs its value enters in the gain since it is in series with R1:

This means that the source impedance must be known with precision or must be negligible.

Normally this setting is used when the stage is driven by a voltage output fed backed stage, with known negligible impedance, as in this example:

vo= −R2

Rs + R1vs

+

- RL

R1

R2

vov-

A

Vs

Rs

+

- RL

R1

R2

vov-

A+-

R1R2

vs

Rs


APPENDIX E: consideration about large feedback gain

+

- RL

Rs

R2

vov-

A

vs

Side effect:

Let’s suppose that R1 is the source impedance Rs, namely a resistor that assumes small values, having a zero value in the ideal case.

In this situation the gain is subjected to 2 effects for both β and T:

Therefore, if β→0 the closed loop gain can not increase indefinitely since, at the same time, T →0: the resulting gain is limited by the OA gain and it results as if the network is not closed loop.

So:

But we know that T=-Af( β)β:

β = −R𝑠𝑠R2

R𝑠𝑠→00, hence1β→ ∞

vo = −R2Rs

−T1 − T vs

Rs→0 −R2[0]

[−0]1 vs

vo =1β−Af(β)β

1 + Af(β)β vs =)−Af(β

1 + Af(β)β vsβ→0

− Af(β)vs

vo = −Af β vs = −ATβ

vs =

= −AR𝑠𝑠′

R𝑠𝑠′ + R2

RERE + Ro

R2R𝑠𝑠

vsRs→0 − A

RERE + Ro vs

T = −ARs′

Rs′ + R2

RERE + Ro

Rs→00 Rs′ = RsRi


………. Appendix E (2)

Another side effect:

Let’s suppose to increase the closed loop current gain by increasing and increasing R2.

+

- RLR2

vo

is

v-

A

Again the network behaves as it were operating open loop.

To conclude: every time T→0 the circuit behaves as it is in open loop, as it is expected since T is the measure of the loop operation.

We know that if T=∞ we get:

Coming back to the case T is not very very large:

But:

Therefore:

β = −1

R2

R2→∞0

T = −AR1′

R1′ + R2

RERE + Ro

R2→∞0

1β−T

1 − T ≈−Tβ = Af β = A

R1′

R1′ + R2

RERE + Ro R2

R2→∞AR1′ RE

RE + Ro

vo =1β−T

1 − T is =⁄−T β

1 − Tis =

)Af(β1 − T

isR2→∞Af(β)is = −A

R1′ RE

RE + Rois

vo = −R2is


………. Appendix E (3)

A last side effect:

Once again the network behaves as it is open loop.

To conclude: every time T→0 the circuit behaves as it is in open loop, as it is expected since T is the measure of the loop operation.

Again, if T=∞ we have:

If now R1→0 the close loop gain →∞, but:

But:

+

-

R1R2

vs

vo

RL

And so:

vo =R1 + R2

R1Vs

β =R1

R1 + R2

R1→00

T = −AR1′

R1′ + R2

RERE + Ro

R1→00

Tβ = −Af(β) = −A

R1′

R1′ + R2

RERE + Ro

R1 + R2R1

R1→0 − ARE

RE + Ro

vo =1β−T

1 − T Vs =⁄−T β

1 − T Vs =)Af(β

1 − T VsR1→0Af(β)Vs = A

RERE + Ro Vs


APPENDIX F: Miller theoremThe Miller effect

Zv1 v2

IWe want to show that the simple scheme on the left is equivalent to this:

I I

v1 v2

Z1 Z2

The current I must be equal in both branches:

From which:

Let’s apply the Miller effect to resistor R2 of the example of Appendix B:

+- RL

R2

vo

is

v-

A By assuming for a while negligible the output impedance of the OA:

+- RL

Z1

vo

is

v-

A

Z2

Hence:

And we get the order of magnitude of the expected results

I=v1Z1

=−v2Z2

=v1 −v2

Z

vOv−

≈ −A

1Z1

=1Z 1 −

v2v1

, Z1 =Z

1 − v2v1

1𝑍𝑍2

=1Z 1 −

v1v2

, Z2 =Z

1 − v1v2

Z1 =R2

1 − vov−

≈R2

1 + A = 4 Ω


………. Appendix F 2

is

+

-

R2

voRivi Ro

RL

Avi

v-

+Let’s assume that v- is about zero::

To evaluate the Miller effect we must know the ratio between VO and V-.

The result of the previous page has been obtained with some approximation and we can try now to improve the precision:

We have that:

therefore:

With a greater precision.

vO = −iSR2

−Av− − vORO

≈vOR2

+vORL

−Av−RO

= vO1

R2+

1RL

+1

RO= vO

1RO

+1

R2‖RL

−Av−RO

= vOR2‖RL + RORO R2‖RL

⇒vOv−

= −AR2‖RL

R2‖RL + RO

Z1 =R2

1 − vov−

≈R2

1 + ARLR2

Ro + RLR2

=4000

1 + 1000 × 0.656 = 6.09 Ω


APPENDIX G: the differential amplifierTHE DIFFERENTIAL AMPLIFIER

+

-

RARB

RC RD

vov1

v2

Let’s consider A=∞, from which we know it is valid that v+ ≈ v-, and i+≈i-≈0, we get the following eq:

Solving:

We must arrange the coefficients that multiply v1 and v2 to be equal for obtaining a true differential amplification. We have some degrees of freedom, namely:

From which now it is valid that:

v+ =RD

RD + RCv1

v2 − v−RA

=v− − vo

RB

vo =RA + RB

RA

RDRC + RD

v1 −RBRA

v2

RD = RBRA = RC

vo =RB

RAv1 − v2

v2RA

−1

RA+

1RB

v− = −voRB

−v2RA

+1

RA+

1RB

RDRD + RC

v1 =voRB

From which:


………. Appendix G (2)THE DIFFERENTIAL AMPLIFIER

The impedance that the source v1 sees is always RA+RB, whatever is v2, in the approximation that Ri>>RA, RB.

Source v2 has a different perspective and:

The impedances seen from the 2 sources.

+

-

RARB

RA RB

vov1

v2

i2

v2 = v1 i2 =1

RAv2 −

RBRA + RB

v2 =v2

RA + RB

R12 =v2i2

= RA + RB

v2 = −v1 i2 =1

RAv2 +

RBRA + RB

v2 =1

RA

RA + 2RBRA + RB

v2

R12 =v2i2

= RARA + RB

RA + 2RB

v1 = 0 i2 =1

RAv2

R12 =v2i2

= RA


………. Appendix G (3)

The fact that the impedance seen from the 2 sources are not equal, and, more important, not very large, can lead to a not negligible effect that comes from the source resistors Rs1 and Rs2 of the 2 input sources.

Our original differential gain is modified this way, now:

We can se that if the 2 source resistors are equal the gain look different, but, very important, if the 2 resistors are different the gain is no more differential and a large common mode signal results.

The above result can be put in the following form, that highlights the effect:

+

-

RARB

RA RB

vov1

v2

i2Rs2

Rs1

vo =RB

RA + Rs2

RA + Rs2 + RBRA + Rs1 + RB

v1 − v2

vo =RB

RA + Rs2v1 − v2 +

Rs2 − Rs1RA + Rs1 + RB

v1

=RB

RA + Rs2v1 − v2 +

RBRA + Rs2

Rs2 − Rs1RA + Rs1 + RB

v1



+-

RARB

RC RDvov1

v2

Let’s suppose now that the 4 resistors have not a very precise value. To simplify let’s suppose that only one is not precise, RB: RB=RB(1-ε).

From the general result of the previous page:

If now v1=v2=vC:

In the ideal case of perfect matching:

Therefore we can say that the Common Mode Rejection Ratio, CMRR is:

vo =RA + RB 1 − ε

RA

RDRC + RD

v1 −RB 1 − ε

RAv2

=RB=RD RB

RA

RA + RBRC + RB

v1 −εRB

RC + RBv1 − 1 − ε v2

=RA=RC RB

RAv1 −

εRBRA + RB

v1 − v2 + εv2

=RBRA

v1 − v2 + ε v2 −RB

RA + RBv1

vo = εRB

RA + RBvc

= Acfvc

vo =RBRA

v1 − v2 = Adf v1 − v2

CMRR = 20log10AdfAcf

= 20log10RBRA

1 + ⁄RA RBε ≈

Adf=10ε=0.1% 81 dB



+-

RARB

RA RB

vovc

vc

Another source of uncertainty is given by the OA itself, since, in practice, it is not a perfect differential OA. In general it can be written:

Let’s not neglect the common mode gain now:

vo = Ad v+ − v− + Acv+ + v−

2

v+ =RB

RB + RAvc

vc − v−RA

=v− − vo

RB⇒ v− =

RBRB + RA

vc +RA

RB + RAvo

vo = Ad +Ac2

v+ − Ad −Ac2

v−

Now:

vo = Ad v+ − v− +Ac2 v+ + v−

voAd

= v+ − v− +Ac

2Adv+ + v−

But Vo/Ad ≈0, then:

v−≈ v+ +Ac

2Adv+ + v−

v−≈1

1 − Ac2Ad

v+ +Ac

2Adv+

v−≈ v+ +Ac

2Adv+



+-

RARB

RA RB

vovc

vc

v+ =RB

RB + RAvc

v− =RB

RB + RAvc +

RARB + RA

vo

v−≈ v+ +Ac

2Adv+

Elaborating a bit:

So that:

RBRB + RA

vc +Ac

2Ad

RBRB + RA

vc =RB

RB + RAvc +

RARB + RA

vo

Ac2Ad

RBRB + RA

vc =RA

RB + RAvo

vo =RBRA

Ac2Ad

vc

+

-

RA

RB

RA RBvo

v1

-v1

vcomCommon mode signal is reflected at the input as it is a differential signal in open loop condition..

Closed loop gain

vcom =Ac

2Advc



It is possible to improve the performance of the differential amplifier, at expense of a greater complexity.

Now the 2 sources see each one a very large impedance, due to the feedback.

Supposing that for the 3 OA is A=∞:

The output stage is simply a unity gain differential amplifier:

+

-

+

-+

-

RA

RB

RB

R

R

R

R

v1

v2

vA

vB

vo

Rs1

Rs2

vA − v1RB

=v1 − v2

RAv1 − v2

RA=

v2 − vBRB

vA =RA + RB

RAv1 −

RBRA

v2

vB =RA + RB

RAv2 −

RBRA

v1

vo = vB − vA =RA + 2RB

RAv2 − v1

vo = vB − vA =RA + RB

RAv2 −

RBRA

v1 −RA + RB

RAv1 +

RBRA

v2



Now it is possible to appreciate what happens to the common mode signal: v1=v2=vc.

If the 2 inputs are equal the current through RA is zero; as a consequence the current will be zero also in the 2 resistors RB’s. Therefore vA=vBwhatever is the tolerance of RA and RB’s and the value of the Rs’s.

If ε is the tolerance of the resistors R of the output OA, their CMRR is:

If the inputs OAs are very similar, that is the case when they share the same package, then the common mode gain Ac is similar, giving similar contribution to the nodes vA and vB. This way the CMRR is limited in accuracy by the output OA and:

That is to say that the CMRR is larger and larger if the gain of the input stage is larger and larger.

CMRRR = 20log102RB + RA

RA

1ε

CMRRAMPLI = 20log102RB + RA

RA

AdAc

+

-

+

-+

-

RA

RB

RB

R

R

R

R

vc

vc

vA

vB

vo

Rs1

Rs2



Here there is another differential configuration having a large input impedance:

Let’s apply the superimposition principle and set v1=0 first, that implies that vA=0 V and:

Now suppose that v2=0 V, which sets to 0 V the non-inverting terminal of the output OA:

Finally, by setting:

Therefore:

we get:

vO2 =RA + RB

RAv2 = 1 +

RBRA

v2

vO1 = −RBRA

RC + RDRC

v1 = −RBRA

1 +RDRC

v1

vO = vO1 + vO2 = 1 +RBRA

v2 −RBRA

1 +RDRC

v1

= 1 +RBRA

v2 −RBRA

RDRC

+RBRA

v1

RBRA

=RCRD

vO = 1 +RBRA

v2 − v1

+

-

+

-

Rs2

RDRB

v2

v1

vAvo

RC

Rs1RA


………. Appendix G (10)Application of the differential configuration.

Errors rise when the source and the amplifier are connected to a ground references far away each other: since currents may flow around in the environment the 2 references could have slightly different potential.

This can be modelled this way:

This way the disturbance is rejected since the amplifier behaves as it is virtually floating: vo=A( (Vs+vM) – vM)=Avs. But…

The amplified input has superimposed the disturbance vM to the signal vs.If now we introduce our differential amplifier we have 2 reading terminals that can be exploited to read the signal across the source only:

+

-

RA

RB

vs

vo

vM

+

-

RA

RB

vs

vo

vM

vs

vM

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vo

vs+vM

AvM

A vs+vM

vo=A[(vs+vM) – vM]=Avs

vo =RA + RB

RAvs + vM



As it can be seen above, the Faraday’s law is proportional to the coil area and one way to suppress its effect is by minimizing the sensitive area and or organizing the connections in a series of coils such that in each coil the current is opposed to that of the previous coil, as is the case with the twisted pair connections:

The way the source is connected to the reading amplifier is important. If in the environment are present sources of EMI disturbances they can generate spurious signals in the connecting loop:

vs

vM

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vo

In this way the signal current induced in one coil cancels in the following coil as the current reverses.

vs

vM

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vo

EMIvEMI ∝

𝜕𝜕Φ𝜕𝜕t Area

According to the Faraday/Lenz’s law when a time varying electro-magnetic field interacts with a coil it generates an electric field, or voltage, in the coil. In our example above this EMI voltage, vEMI, adds to the signal voltage.The common cause of magnetic field disturbances are from the main line, 50 (Europe)/ 60 (USA)Hz.Disturbances from the main line are always present and one of the task when we instrument a low noise system is to suppress them with shields and other stuffs that we will try to address in the following.

Zero result



Even better the source of the signal is made differential, to drive the twisted line:

In this way the signal current induced in one coil cancels in the following coil as the current reverses.

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vovs

vM

+

-

+

-

+

-

+

-

RA

RB

RB

VA

VB

Now let’s consider a practical consideration that very often induces errors. In principle the scheme above is working since there is a true differential signal readout from a true differential input pair.

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vovs

vM

+

-

+

-

+

-

+

-

RA

RB

RB

VA1

VB1

System on the right and on the left can be floating each other and the 2 grounds could be different in DC voltage: this is not a disturbance, but it depends only on the environment settings.This means that:

Vin1

Vin2

VA1 = VGND_OFFSET + Vs+VB1 = VGND_OFFSET + Vs−


………. Appendix G (13)+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vovs

vM

+

-

+

-

+

-

+

-

RA

RB

RB

VA1

VB1

Vin1

Vin2

VA1 = VGND_OFFSET + Vs+VB1 = VGND_OFFSET + Vs−

VGND_OFFSET is the voltage difference between the 2 grounds.

Vin1-Vin2 is equal to Vs+-Vs-, and this is the signal of interest.

But the absolute value of Vin1 and Vin2 can be large and even close to one of the rails. This determines a saturation of the input stage and a signal distortion.

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vovs

vM

+

-

+

-

+

-

+

-

RA

RB

RB

VA1

VB1

Vin1

Vin2

The workaround to the problem is to set a common reference for both grounds with a feeble connection through which the current cannot flow, to avoid disturbances. A small resistor, RG, helps in doing this.

RG



Twisted cables are useful to reject EMI disturbances. But they are not enough and our apparatus, and cables, need to be shielded.

We need to shield electro-magnetic filed that may consists of both electric and magnetic fields.

We know that the electric field is shielded by enclosing our circuits in metallic box, exploiting the Faraday-cage principle.

+

-

+

-

+

-

+

-+

-

+

-

RA

RB

RB

R

R

R

R

VA

VB

Vo

Electric field is well rejected by shielding both the amplifier and the connection.

For better results, the Shield must be connected to a constant potential, ground, to avoid any residual field variation coming, from example, from large currents around.

Metallic shield

Coaxial, shielded, cable

Incident Electric field

Reflected Electric field



Electric field is easily shielded. But magnetic field has the ability to penetrate, especially at small frequency: 50 Hz is very difficult to shield, indeed.

The attenuation of the magnetic field at low frequency, the coefficient that multiplies the field, is given by:

A ∼ exp −tδ δ =

0.0137πfµrσr

Therefore, at low frequency 𝑓𝑓 we need

• High magnetic permeability 𝝁𝝁𝒓𝒓

• High electrical conductivity 𝝈𝝈𝒓𝒓

with:

Conclusion: to obtain an efficient attenuation, namely A small, we need a small value of δ. If a small value of δ is not available in the metal shield we are using, then, we need a large thickness of shield, otherwise the disturbances is not shielded, but only slightly attenuated.



Very large relative permeability, µr, is found in Iron and some steel, around 1000. There are a few special alloys, studied for this purpose: Mumetal, Skudotech, … which allow to obtain a value close to and larger than 20000.This is important as in this way a shield can be obtain with a few mm of material, against the cms needed for iron. This has impact on the weight of the final apparatus.

A ∼ exp −tδ δ =

0.0137πfµrσr

with:


APPENDIX H: OA specifications

An actual OA is not ideal and shows several characteristics that can have affect in the application. Its datasheet list them and one can select the best suited off-the-shelf OA for the own application.

The characteristics are listed in the datasheet and, for a good OA, several are listed with a good detail.

Generally one focalizes the parameters important for the own application and do not consider the others.

Here some examples:


………. Appendix H (2)

Other important parameters are the noise;

The frequency response and the ability / inability to drive low value impedances and capacitive impedances;

The value of the output impedance;

The upper and lower limits of the output signals with respect to the supply voltages;

Power dissipation;

The maximum output current and short circuit current;

Etc.

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