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))