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))