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Published byVirginia Dickerson Modified over 9 years ago
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Electronics in High Energy Physics Introduction to electronics in HEP
Electrical Circuits (based on P.Farthoaut lecture at Cern)
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Electrical Circuits Generators Thevenin / Norton representation Power
Components Sinusoidal signal Laplace transform Impedance Transfer function Bode diagram RC-CR networks Quadrupole
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Sources v r R + - I Voltage Generator Current Generator
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Thevenin theorem (1) Rth Vth A A B B
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
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Thevenin theorem (2) V R1 R2 + - A R1//R2 A Voltage divider
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Norton representation
B A Rno Ino Vth Rth Any voltage source followed by an impedance can be represented by a current source with a resistor in parallel
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Power transfer v r R + - I Power in the load R P is maximum for R = r
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Sinusoidal regime
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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
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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
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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 IM2 cost sin t = R IM2 /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 IM2 /2 pq is the reactive power (volt-ampere); pq = Pq sin2t Mean value = 0 Pq = X IM2 /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
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Real capacitance A perfect capacitance does not absorb any active power it exchanges reactive power with the source Pq = - IM2 /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
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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
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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
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Laplace Transform (2) Examples Linearity Derivation, Integration
Translation
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Laplace Transform (3) Change of time scale
Derivation, Integration of the Laplace transform Initial and final value
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Impedances i(t) Z v(t) I(p) Z(p) V(p) Network v(t), I(t)
Generalisation V(p) = Z(p) I(p)
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Transfer Functions I1 I2 Transfer Function V2 V1
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:
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Bode diagram (1) Replacing p with j in F(p), one obtains the imaginary form of the function transfer F(j) = |F| ej() Logarithmic unit: Decibel In decibel the module |F| will be The phase of each separate functions add Functions to be studied
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Bode diagram (2) a F(p) = p + a ; |F1|db= 20 log | j + a|
[rad/s] 20 dB per decade 6 dB per octave a 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
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Bode diagram (3) a |F2|db= - 20 log | j + a|
[rad/s] - 20 dB per decade -6 dB per octave a 3 dB error |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
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Bode diagram (4) Low pass filters As before but:
|F|dB [rad/s] -20 dB per decade Low pass filters -40 dB per decade As before but: Slope 6*n dB per octave (20*n dB per decade) Error at =a is 3*n dB
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Bode diagram (5) Phase of F1(j ) = (j + a) Asymptotic diagram
tg = /a Asymptotic diagram = 0 when < a = /4 when = a = /2 when > a
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Bode diagram (6) Phase of F2(j ) = 1/(j + a) Asymptotic diagram
tg =- /a Asymptotic diagram = 0 when < a = - /4 when = a = - /2 when > a
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Bode diagram (7) |F3|dB = 20 log|b2 - 2 + 2aj| Asymptotic diagram
40 dB per decade b |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
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Bode diagram (8) |F4|dB = - 20 log|b2 - 2 + 2aj| Asymptotic diagram
-40 dB per decade b |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
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Bode diagram (9) Z = 0.1 Z = 1 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
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RC-CR networks (1) V1 V2 R C Integrator; RC = time constant
Dirac response Heaviside response t/RC V C R V1 V2 Integrator; RC = time constant
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RC-CR networks (2) C R V1 V2 Low pass filter c = 1/RC
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RC-CR networks (3) V1 V2 C R Derivator; RC = time constant
Dirac response Heaviside response RC = 1 t/RC V C R V1 V2 Derivator; RC = time constant
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RC-CR networks (4) R2 C1 C2 V1 R1 V2 Dirac response Heaviside response
RiCi = RC t/RC V V1 C1 R1 V2 R2 C2
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RC-CR networks (5) V1 C1 R1 V2 R2 C2 Band pass filter
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Time or frequency analysis (1)
A signal x(t) has a spectral representation |X(f)|; X(f) = Fourier transform of x(t) 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)
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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
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Time or frequency analysis (3)
Time analysis C R X(t) Y(t)
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Time or frequency analysis (4)
Contains all frequencies Most of the signal within 10 MHz To avoid huge distortion the minimum bandwidth is MHz Used to define the optimum filter to increase signal-to-noise ratio
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Quadrupole I1 I2 V1 V2 Passive Active Parameters Network of R, C and L
Internal linked sources Parameters V1, V2, I1, I2 Matrix representation
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Parameters Impedances Admittances Hybrids
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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))
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