Electronics in High Energy Physics Introduction to electronics in HEP Electrical Circuits (based on...
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Transcript of Electronics in High Energy Physics Introduction to electronics in HEP Electrical Circuits (based on...
Electronics in High Energy PhysicsIntroduction to electronics in HEP
Electrical Circuits(based on P.Farthoaut lecture at Cern)
2
Electrical Circuits
Generators Thevenin / Norton representation Power Components Sinusoidal signal Laplace transform Impedance Transfer function Bode diagram RC-CR networks Quadrupole
3
Sources
Voltage Generator
vr
R+
-
I
rR
VI
R
VIthenRr
cter
VIthenRr Current Generator
4
Thevenin theorem (1)
Any two-terminal network of resistors and sources is equivalent to a single resistor with a single voltage source
Vth = open-circuit voltage Rth = Vth / Ishort
A
B
VthRth
A
B
5
Thevenin theorem (2)
Voltage divider
VR1
R2+
-
AR1//R2
A
2R1R
2RV
2R//1R2R1R
2R1R
Ishort
VthRth
1R
VinIshort
2R1R
2RVinVthVopen
6
Norton representation
Any voltage source followed by an impedance can be represented by a current source with a resistor in parallel
B
A
RnoInoVthRth
A
B
RthRno.e.iVthRnoInoRth
VthIshortIno
7
Power transfer
Power in the load R
Power in the load
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
R/r
Po
wer
vr
R+
-
I
P is maximum for R = r
222
rR
RVRIP
8
Sinusoidal regime
-1.5
-1
-0.5
0
0.5
1
1.5
0 2 4 6 8 10
Time
Am
pli
tud
e
f2;AmplitudeV)tcos(V)t(V
shiftphase')'tcos(V)t('V
delayais
tVtVtV
'
'cos)'cos()('
9
Complex notation
Signal : v1(t) = V cos( t + )
v2(t) = V sin( t + )
v(t) = v1 + j v2 = V e j( t + ) = V ej ej t = S ej t
Interest:– S = V ej contains only phase and amplitude– ej t contains time and frequency
Real signal = R [ S ej t ]
In case of several signals of same only complex amplitude are significant and one can forget ej t – One can separate phase and time
10
Complex impedance
In a linear network with v(t) and i(t), the instantaneous ratio v/i is often meaningless as it changes during a period
To i(t) and v(t) one can associate J ej t and S ej t
S / J is now independent of the time and characterizes the linear network– Z = S / J is the complex impedance of the network
Z = R + j X = z ej – R is the resistance, X the reactance– z is the module, is the phase– z, R and X are in Ohms
Examples of impedances:– Resistor Z = R– Capacitance (perfect) Z = -j / C; Phase = - /2
» 100 pF at 1MHz 1600 Ohms » 100 pF at 100 MHz 16 Ohms
– Inductance (perfect) Z = jL; Phase = + /2» 100 nH at 1 MHz 0.63 Ohms» 100 nH at 100 MHz 63 Ohms
11
Power in sinusoidal regime
i = IM cos t in an impedance Z = R + j X = z ej
v = z IM cos( t + ) = R IM cos t - X IM sin t
p = v i = R IM2 cos2t – X IM
2 cost sin t = R IM
2 /2 (1+cos2t ) - X IM2 /2 sin2 t
p = P (1+cos2 t ) - Pq sin2 t = pa + pq
– pa is the active power (Watts); pa = P (1+ cos2t)» Mean value > 0; R IM
2 /2
– pq is the reactive power (volt-ampere); pq = Pq sin2t » Mean value = 0» Pq = X IM
2 /2 » In an inductance X = L ; Pq > 0 : the inductance absorbs some reactive energy» In a capacitance X = -1/C; Pq < 0 : the capacitance gives some reactive energy
12
Real capacitance
A perfect capacitance does not absorb any active power– it exchanges reactive power with the source Pq = - IM
2 /2C In reality it does absorb an active power P Loss coefficient
– tg = |P/Pq|
Equivalent circuit– Resistor in series or in parallel
– tg = RsCs
– tg = 1/RpCp
Cs
Rp
Rs
Cp
13
Real inductance
Similarly a quality coefficient is defined– Q = Pq/P
Equivalent circuit– Resistor in series or in parallel
– Q = Ls/Rs
– Q = Rp/Lp
Ls
Rp
Rs
Lp
14
Laplace Transform (1)
v = f(i) integro-differential relations In sinusoidal regime, one can use the complex notation and the complex
impedance– V = Z I
Laplace transform allows to extend it to any kind of signals
Two important functions– Heaviside (t)
» = 0 for t < 0 » = 1 for t 0
– Dirac impulsion (t) = ’(t) » = 0 for t 0
»
0
1dt)t(
15
Laplace Transform (2)
0
pt dte)t(h)p(F)t(f)t()t(h
)p(Fe)at(h);p(F)t(h
)ap(F)t(he);p(F)t(h
p
)p(Fdt)t(h);p(F)t(h
)p(pF)t('h);p(F)t(h
)p(bF)p(aF)t(bh)t(ah
ptsin)t(
ap
1e)t(
p
1)t(
1)t(
ap
at
2121
22
at
Examples»
Linearity
Derivation, Integration
Translation
16
Laplace Transform (3)
Change of time scale
Derivation, Integration of the Laplace transform
Initial and final value
)t(flim)p(pFlim
)t(flim)p(pFlim
t
)t(hdp)p(F
)t(th)p('F
)ap(aF)a
t(h);p(F)t(h
0tp
t0p
17
Impedances
Network v(t), I(t)
V(p)Z(p)
I(p)
Generalisation– V(p) = Z(p) I(p)
I(p) Lp V(p):Inductance–
I(p) Cp1
V(p) i.e I(p) C1
V(p)p :Capacitor–
I(p) R V(p):Resistor–
transform Laplace the Applyingdtdi
L v(t) :Inductance–
i(t) C1
dtdv
:Capacitor–
i(t) R v(t) :Resistor–
i(t) and v(t) between Relation
v(t)Z
i(t)
18
Transfer Functions
Input V1, I1; Output V2, I2
Voltage gain V2(p) / V1(p)
Current gain I1(p) / I2(p)
Transadmittance I2(p) / V1(p)
Transimpedance V2(p) / I1(p)
Transfer function Out(p) = F(p) In(p)– Convolution in time domain:
V1 V2
I2I1
TransferFunction
d)t(F)(In)t(F*)t(In)t(Out
19
Bode diagram (1)
Replacing p with j in F(p), one obtains the imaginary form of the function transfer– F(j) = |F| ej()
(Pole) Zero also is (Pi*) *Zi complex, (Pi) Zi if
)Pp)...(Pp)(Pp(
)Zp)...(Zp)(Zp(K)p(F
n21
n21
0db
2
0db
V
Vlog20V;
R
VP
P
Plog10P:Power
|PjPj|log20|ZjZj|log20|Pj|log20|Zj|log20Klog20 *ii
*iiii
ja2b
1)j(F;ja2b)j(F
aj
1)j(F;aj)j(F
22422
3
21
Logarithmic unit: Decibel
In decibel the module |F| will be
The phase of each separate functions add Functions to be studied
20
Bode diagram (2)
0
20
40
60
80
100
120
1 10 100 1000 10000 100000
|F|dB
[rad/s] 0
5
10
15
20
25
30
35
40
45
1 10 100
20 dB per decade 6 dB per octavea
3 dB error
F(p) = p + a ; |F1|db= 20 log | j + a|
Bode diagram = asymptotic diagram
< a, |F1| approximated with A = 20 log(a)
> a, |F1| approximated with A = 20 log()
»6 dB per octave (20 log2) or 20 dB per decade (20 log10)
Maximum error when = a
–20 log| j a + a| - 20 log(a) = 20 log (21/2) = 3 dB
ap)p(F;aj)j(F 11
21
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
1 10 100
-120
-100
-80
-60
-40
-20
0
1 10 100 1000 10000 100000
|F|dB
[rad/s]
Bode diagram (3)
|F2|db= - 20 log | j + a| Bode diagram = asymptotic diagram
< a, |F2| approximated with A = - 20 log(a)
> a, |F2| approximated with A = - 20 log()
» - 6 dB per octave (20 log2) or - 20 dB per decade (20 log10)
Maximum error when = a– 20 log| j a + a| - 20 log(a) = 20 log (21/2) = 3 dB
- 20 dB per decade
-6 dB per octave
a
3 dB error
ap
1)p(F;
aj
1)j(F 22
22
Bode diagram (4)
As before but:– Slope 6*n dB per octave (20*n dB per decade)
– Error at =a is 3*n dB
-250
-200
-150
-100
-50
0
1 10 100 1000 10000 100000
|F|dB
[rad/s]
-20 dB per decade
-40 dB per decade
Low pass filters
n2n )ap(
1)p(F;
aj
1)j(F
23
Bode diagram (5)
Phase of F1(j ) = (j + a)
– tg = /a Asymptotic diagram
= 0 when < a = /4 when = a = /2 when > a
0102030405060708090100
0.1 1 10
/a
Ph
ase
[d
eg
re]
24
Bode diagram (6)
Phase of F2(j ) = 1/(j + a)
– tg =- /a Asymptotic diagram
= 0 when < a = - /4 when = a = - /2 when > a
-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
0.1 1 10
/a
Ph
ase
[d
eg
re]
25
0
50
100
150
200
250
1 10 100 1000 10000 100000
w [rad/s]
|F| d
B
-20-15-10-5051015202530
0.1 1 10
/b
|F| d
B
z=0.1
z=2
Bode diagram (7)
|F3|dB = 20 log|b2 - 2 + 2aj| Asymptotic diagram
--> 0 A = 40 log b --> ∞ A’ = 20 log 2 = 40 log – A = A’ for = b
Error depends on a and b– p2 + 2a p + b2 = b2[(p/b)2 + 2(a/b)(p/b) + 1]– Z = a/b U = /b
40 dB per decade
b
ap2b)p(F;ja2b)j(F 223
223
26
-250
-200
-150
-100
-50
0
1 10 100 1000 10000 100000
w [rad/s]
|F| d
B
-30-25-20-15-10-505101520
0.1 1 10
/b
|F| d
B
z=0.1
z=2
Bode diagram (8)
|F4|dB = - 20 log|b2 - 2 + 2aj| Asymptotic diagram
--> 0 A = - 40 log b --> ∞ A’ = - 20 log 2 = - 40 log – A = A’ for = b
Error depends on a and b– Z = a/b U = /b
-40 dB per decade
b
ap2b
1)p(F;
ja2b
1)j(F
224224
27
Bode diagram (9)
Phase of F3(j) = (b2 - 2 + 2aj) and F4(j) = 1/(b2 - 2 + 2aj)
– tg = 2a/ (b2 - 2) Asymptotic diagram
= 0 when < b = ± /2 when = b = ± when > b
-200
-180
-160
-140
-120
-100
-80
-60
-40
-20
0
0.1 1 10
/b
Ph
ase
[d
eg
re]
Z = 0.1
Z = 1
28
RC-CR networks (1)
Integrator; RC = time constant
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6
Dirac response
Heaviside response
t/RC
V
CR
V1 V2
RC
1p
1
RC
)p(V
Cp
1R
Cp
1
)p(V)p(V 112
RC
t
2
2
1
eRC
1)t()t(V
RC
1p
1
RC
1)p(V
1)p(VDirac
RC
t
2
2
1
e1)t()t(V
RC
1p
1
p
1
RC
1p
1
p
1
RC
1)p(V
p
1)p(VHeaviside
29
RC-CR networks (2)
Low pass filter c = 1/RC
-70
-60
-50
-40
-30
-20
-10
0
0.1 1 10 100 1000
* RC
V d
BCR
V1 V2
RC
1p
1
RC
1)p(F
30
RC-CR networks (3)
Derivator; RC = time constant
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6
Dirac response
Heaviside response
RC = 1
t/RC
V
C RV1 V2
RC
1p
p)p(V
Cp
1R
R)p(V)p(V 112
RC
t
2
2
1
eRC
11)t()t(V
RC
1p
1
RC
11
RC
1p
p)p(V
1)p(VDirac
RC
t
2
2
1
e)t()t(V
RC
1p
1)p(V
p
1)p(VHeaviside
31
RC-CR networks (4)
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
Dirac response
Heaviside response
RiCi = RC
t/RC
V
V1C1
R1 V2
R2C2
;CR;CR;CR 123222111
21321
21
12
1111pp
p)p(V)p(V
)tsinh(e1
)t()t(V
ap
1)p(V
1;p
1)p(VHeaviside
at2
222
i1
)tsinh(2
3)tcosh(e
1)t()t(V
ap
a
ap
ap
ap
p
1p3p
p)p(V
1;1)p(VDirac
at2
22222222
i1
32
RC-CR networks (5)
Band pass filter
V1C1
R1 V2
R2C2
0
10
20
30
40
50
60
70
80
90
1 10 100 1000 10000 100000 1000000
1E+07 1E+08
rad/s
dB
22ap
p)p(F
22a )alog(20 22
33
Time or frequency analysis (1)
A signal x(t) has a spectral representation |X(f)|; X(f) = Fourier transform of x(t)
dte)t(x)f(X ft2j
The transfer function of a circuit has also a Fourier transform F(f)
The transformation of a signal when applied to this circuit can be looked at in time or frequency domain
x(t)
X(f)
y(t) = x(t) * f(t)
Y(f) = X(f) F(f)
f(t)
F(f)
34
Time or frequency analysis (2)
The 2 types of analysis are useful Simple example: Pulse signal (100 ns width)
– (1) What happens when going through a R-C network?» Time analysis
– (2) How can we avoid to distort it?» Frequency analysis
0
2
4
6
8
10
12
-30 -20 -10 0 10 20 30
time (*10 ns)
x(t)
35
Time or frequency analysis (3)
Time analysis
CR
X(t) Y(t)
)e1)(100t()e1)(t()t(y
RC1
p
1
RC
1)p(x)p(y
ep
1
p
1)p(x
)100t()t()t(x
RC
)100t(
RC
t
p100
RC = 20 ns
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250 300
Amplitude
Tim
e (n
s)
0
2
4
6
8
10
12
-30 -20 -10 0 10 20 30
time (*10 ns)
x(t)
36
Time or frequency analysis (4)
Contains all frequencies Most of the signal within 10 MHz To avoid huge distortion the minimum bandwidth is 10-20 MHz Used to define the optimum filter to increase signal-to-noise ratio
0
2
4
6
8
10
12
-30 -20 -10 0 10 20 30
time (*10 ns)
x(t)
-40
-20
0
20
40
60
80
100
120
-50 -40 -30 -20 -10 0 10 20 30 40 50
Frequency (MHz)
X(f
)
37
Quadrupole
Passive– Network of R, C and L
Active– Internal linked sources
Parameters– V1, V2, I1, I2
– Matrix representation
B'
A'
x4x3
x2x1
B
A
V1 V2
I1 I2
38
Parameters
Impedances
Admittances
Hybrids
I2
I1
Z22Z21
Z12Z11
V2
V1
V2
V1
Y22Y21
Y12Y11
I2
I1
V2
I1
h22h21
h12h11
I2
V1
39
Input and output impedances
Input impedance: as seen when output loaded – Zin = Z11 - (Z12 Z21 / (Z22 + Zu))
– Zin = h11 - (h12 h21 / (h22 + 1/Zu)) Output impedance: as seen from output when input loaded with the
output impedance of the previous stage– Zout = Z22 - (Z12 Z21 / (Z11 + Zg))
– 1/Zout = h22 - (h12 h21 / (h11+ Zg))