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∞
Amplifiers and feedback gpessina 2
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
Amplifiers and feedback gpessina 3
………. 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)
Amplifiers and feedback gpessina 4
………. gain of a c-to-c fed backed current OA 3
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
Amplifiers and feedback gpessina 5
………. gain of a c-to-c fed backed current OA 4
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
Amplifiers and feedback gpessina 6
………. gain of a c-to-c fed backed current OA 5
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
Amplifiers and feedback gpessina 7
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
Amplifiers and feedback gpessina 8
………. 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
Amplifiers and feedback gpessina 9
………. input impedance of a c-to-c fed backed current OA 3
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
Amplifiers and feedback gpessina 10
………. input impedance of a c-to-c fed backed current OA 4
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
Amplifiers and feedback gpessina 11
………. input impedance of a c-to-c fed backed current OA 5
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∞
Amplifiers and feedback gpessina 12
………. input impedance of a c-to-c fed backed current OA 6
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
Amplifiers and feedback gpessina 13
………. input impedance of a c-to-c fed backed current OA 7
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∞
Amplifiers and feedback gpessina 14
………. input impedance of a c-to-c fed backed current OA 8
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)
Amplifiers and feedback gpessina 15
………. input impedance of a c-to-c fed backed current OA 9
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.
Amplifiers and feedback gpessina 16
………. input impedance of a c-to-c fed backed current OA 10
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)
Amplifiers and feedback gpessina 17
………. input impedance of a c-to-c fed backed current OA 11
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
Amplifiers and feedback gpessina 18
………. input impedance of a c-to-c fed backed current OA 12
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
Amplifiers and feedback gpessina 19
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
Amplifiers and feedback gpessina 20
………. 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
Amplifiers and feedback gpessina 21
………. output impedance of a c-to-c fed backed current OA 3
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):
Amplifiers and feedback gpessina 22
………. output impedance of a c-to-c fed backed current OA 4
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
Amplifiers and feedback gpessina 23
………. output impedance of a c-to-c fed backed current OA 5
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
Amplifiers and feedback gpessina 24
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
Amplifiers and feedback gpessina 25
………. 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.
Amplifiers and feedback gpessina 26
………. 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
Amplifiers and feedback gpessina 27
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.
Amplifiers and feedback gpessina 28
………. 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:
Amplifiers and feedback gpessina 29
………. output impedance of a c-to-c fed backed OA 3
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
Amplifiers and feedback gpessina 30
………. output impedance of a c-to-c fed backed OA 4
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
Amplifiers and feedback gpessina 31
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
Amplifiers and feedback gpessina 32
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∞
Amplifiers and feedback gpessina 33
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.
Amplifiers and feedback gpessina 34
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
Amplifiers and feedback gpessina 35
………. 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
Amplifiers and feedback gpessina 36
………. the v-to-c fed backed OA 3
…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:
Amplifiers and feedback gpessina 37
………. 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−
Amplifiers and feedback gpessina 38
………. the v-to-c fed backed OA 5
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
Amplifiers and feedback gpessina 39
………. the v-to-c fed backed OA 6
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
Amplifiers and feedback gpessina 40
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
Amplifiers and feedback gpessina 41
………. 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
Amplifiers and feedback gpessina 42
………. the c-to-v fed backed OA 3
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:
Amplifiers and feedback gpessina 43
………. the c-to-v fed backed OA 4
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
Amplifiers and feedback gpessina 44
………. the c-to-v fed backed OA 5
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
Amplifiers and feedback gpessina 45
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.
Amplifiers and feedback gpessina 46
………. 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)
Amplifiers and feedback gpessina 47
………. quadrupole model of an amplifier 3
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∞
Amplifiers and feedback gpessina 48
………. quadrupole model of an amplifier 4
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 →
Amplifiers and feedback gpessina 49
………. quadrupole model of an amplifier 5
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
Amplifiers and feedback gpessina 50
………. quadrupole model of an amplifier 6
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.
Amplifiers and feedback gpessina 51
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!
Amplifiers and feedback gpessina 52
………. 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
Amplifiers and feedback gpessina 53
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.
Amplifiers and feedback gpessina 54
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−
Amplifiers and feedback gpessina 55
………. 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−
Amplifiers and feedback gpessina 56
………. Appendix A (3)
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
Amplifiers and feedback gpessina 57
………. Appendix A (4)
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
Amplifiers and feedback gpessina 58
………. Appendix A (5)
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
Is =vs − v−Rs + Ri
vsRs + Ri
=v−
Rs + Ri+
v−R1
+v−R2
1 −RL‖Ro
R2 + RL‖Ro
vo =RL‖Ro
R2 + RL‖Rov−
Is =vs − v−Rs + Ri
vs − v−Rs + Ri
=v−R1
+v− − vo
R2
vo =RL‖Ro
R2 + RL‖Rov−
Is =vs − v−Rs + Ri
Amplifiers and feedback gpessina 59
………. Appendix A (6)
+
-
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−
Is =vs − v−Rs + Ri
vsRs + Ri
=v−
Rs + Ri+
v−R1‖ R2 + RL‖Ro
vo =RL‖Ro
R2 + RL‖Rov−
Is =vs − v−Rs + Ri
v− =R1‖ R2 + RL‖Ro
Rs + Ri + R1‖ R2 + RL‖Rovs
vo =RL‖Ro
R2 + RL‖Rov−
Is =vs − v−Rs + Ri
Amplifiers and feedback gpessina 60
………. Appendix A (7)
+
-
R1
R2
vsvo
Rivi Ro
+
RL
Is Rs
Therefore:
And::
v− =R1‖ R2 + RL‖Ro
Rs + Ri + R1‖ R2 + RL‖Rovs
vo =RL‖Ro
R2 + RL‖Rov−
Is =vs − v−Rs + Ri
Is =1
Rs + Ri1 −
R1‖ R2 + RL‖RoRs + Ri + R1‖ R2 + RL‖Ro
vs
Is =1
Rs + Ri + R1‖ R2 + RL‖Rovs
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
Amplifiers and feedback gpessina 61
………. Appendix A (8)
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
Amplifiers and feedback gpessina 62
………. Appendix A (9)
+
-
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
Amplifiers and feedback gpessina 63
………. Appendix A (10)
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
Amplifiers and feedback gpessina 64
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:
Amplifiers and feedback gpessina 65
………. 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
Amplifiers and feedback gpessina 66
………. 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:
Amplifiers and feedback gpessina 67
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:
Amplifiers and feedback gpessina 68
………. 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
Amplifiers and feedback gpessina 69
………. Appendix C (3)
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
Amplifiers and feedback gpessina 70
………. Appendix C (4)
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
Amplifiers and feedback gpessina 71
………. Appendix C (5)
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
Amplifiers and feedback gpessina 72
………. Appendix C (6)
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
Amplifiers and feedback gpessina 73
………. Appendix C (7)
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:
Amplifiers and feedback gpessina 74
………. 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
Amplifiers and feedback gpessina 75
………. Appendix C (9)
Direct transmission:
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
Amplifiers and feedback gpessina 76
………. Appendix C (10)
Direct transmission:
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
Amplifiers and feedback gpessina 77
………. Appendix C (11)
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
Amplifiers and feedback gpessina 78
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′
Amplifiers and feedback gpessina 79
………. 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
Amplifiers and feedback gpessina 80
………. 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
Amplifiers and feedback gpessina 81
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
Amplifiers and feedback gpessina 82
………. 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
Amplifiers and feedback gpessina 83
………. 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
Amplifiers and feedback gpessina 84
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
Amplifiers and feedback gpessina 85
………. 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
Amplifiers and feedback gpessina 86
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:
Amplifiers and feedback gpessina 87
………. 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
Amplifiers and feedback gpessina 88
………. 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
Amplifiers and feedback gpessina 89
………. Appendix G (4)THE DIFFERENTIAL AMPLIFIER
+-
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
Amplifiers and feedback gpessina 90
………. Appendix G (5)THE DIFFERENTIAL AMPLIFIER
+-
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+
Amplifiers and feedback gpessina 91
………. Appendix G (6)THE DIFFERENTIAL AMPLIFIER
+-
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
Amplifiers and feedback gpessina 92
………. Appendix G (7)
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
Amplifiers and feedback gpessina 93
………. Appendix G (8)
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
Amplifiers and feedback gpessina 94
………. Appendix G (9)
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
Amplifiers and feedback gpessina 95
………. 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
Amplifiers and feedback gpessina 96
………. Appendix G (11)
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
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-
+
-
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-+
-
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-
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
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-
+
-
+
-
+
-+
-
+
-
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
Amplifiers and feedback gpessina 97
………. Appendix G (12)
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.
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+
-
+
-
+
-+
-
+
-
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.
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-
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RA
RB
RB
R
R
R
R
VA
VB
Vovs
vM
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-
+
-
+
-
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−
Amplifiers and feedback gpessina 98
………. 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.
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RA
RB
RB
R
R
R
R
VA
VB
Vovs
vM
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+
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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
Amplifiers and feedback gpessina 99
………. Appendix G (14)
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.
+
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+
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+
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+
-+
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+
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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
Amplifiers and feedback gpessina 100
………. Appendix G (15)
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.
Amplifiers and feedback gpessina 101
………. Appendix G (16)
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:
Amplifiers and feedback gpessina 102
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:
Amplifiers and feedback gpessina 103
………. 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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