Electronics in High Energy Physics Introduction to electronics
- Slides: 48
Electronics in High Energy Physics Introduction to electronics in HEP Operational Amplifiers (based on the lecture of P. Farthoaut at Cern) 1
Operational Amplifiers u u u Feedback Ideal op-amp Applications – – u Non-ideal amplifier – – – u u Voltage amplifier (inverting and non-inverting) Summation and differentiation Current amplifier Charge amplifier Offset Bias current Bandwidth Slew rate Stability Drive of capacitive load Data sheets Current feedback amplifiers 2
Feedback u Y is a source linked to X – Y=mx u Open loop – x=de – y=mx – s=sy=sdmx u e Closed loop x m d y s s b u u m is the open loop gain bm is the loop gain 3
Interest of the feedback e x m d s s b u In electronics – m is an amplifier – b is the feedback loop – d and s are input and output impedances u If m is large enough the gain is independent of the amplifier 4
Operational amplifier e -A e + u u Gain A very large Input impedance very high – I. e input current = 0 u A(p) as shown 5
How does it work? u R 2 Direct gain calculation R 1 u I - e Feed-back equation + -A e Vout Vin u Ideal Op-Amp 6
Non-inverting amplifier R 2 u Gain R 1 u u I - Called a follower if R 2 = 0 + Input impedance Vout Vin 7
Inverting amplifier R 2 u Gain R 1 u u Input impedance Gain error I - Vin + Vout 8
Summation R u Transfer function R 1 V 1 I - Rn Vn In u If Ri = R + Vout 9
Differentiation R 2 R 1 V 1 I 1 R 1 V 2 I 1 + Vout R 2 10
Current-to-Voltage converter (1) C R Iin + u u Vout = - R Iin For high gain and high bandwidth, one has to take into account the parasitic capacitance 11
Current-to-Voltage converter (2) R 1 R 2 r Iin + Vout u High resistor value with small ones u Equivalent feedback resistor = R 1 + R 2 * (R 1/r) – ex. R 1 = R 2 = 100 k ; r = 1 k ; Req = 10. 2 M u Allows the use of smaller resistor values with less problems of parasitic capacitance 12
Charge amplifier (1) R C u Requires a device to discharge the capacitor – Resistor in // – Switch I + Vout 13
Charge amplifier (2) R C I + Input Charge In a few ns V 1 C 1 R 2 R 1 Output of the charge amplifier Very long time constant C 2 V 2 Shaping a few 10’s of ns 14
Miller effect u Charge amplifier – – u Vin = e Vout = -A e The capacitor sees a voltage (A+1) e It behaves as if a capacitor (A+1)C was seen by the input C Vin –Two circuits are equivalent X Z e -A e Vout + Miller’s theorem –Av = Vy / Vx - Y Y X Z 1 Z 2 » Z 1 = Z / (1 - Av) » Z 2 = Z / (1 -Av-1) 15
Common mode u The amplifier looks at the difference of the two inputs – Vout = G * (V 2 - V 1) u The common value is in theory ignored – V 1 = V 0 + v 1 – V 2 = V 0 + v 2 u In practice there are limitations – linked to the power supplies – changes in behaviour u Common mode rejection ratio CMRR – Differential Gain / Common Gain (in d. B) 16
Non-ideal amplifier u Input Offset voltage Vd u Input bias currents Ib+ and Ib. Ib- u Limited gain u Input impedance u u e Zd Output impedance Common mode rejection Noise Bandwidth limitation & Stability Zc -A e Zout + Vd Ib+ Zc 17
Input Offset Voltage u u “Zero” at the input does not give “Zero” at the output In the inverting amplifier it acts as if an input Vd was applied R 2 I R 1 - – (Vout) = G Vd u Notes: – Sign unknown – Vd changes with temperature and time (aging) – Low offset = a few m. V and Vd = 0. 1 m. V / month – Otherwise a few m. V Vd + Vout 18
Input bias current (1) u u u (Vout) = R 2 Ib (Vout) = - R 3 (1 -G) Ib+ Error null for R 3 = (R 1//R 2) if Ib+ = Ib- R 2 Ib- R 1 + R 3 Ib+ Vout 19
Input bias current (2) u u u In the case of the charge amplifier it has to be compensated Switch closed before the measurement and to discharge the capacitor Values – less than 1. 0 p. A for JFET inputs – 10’s of n. A to m. A bipolar Ib- C + R 3 Ib+ Vout 20
Common mode rejection u u u Non-inverting amplifier Input voltage Vc/Fr (Vc common mode voltage) Same effect as the offset voltage R 2 R 1 Vc/Fr I + Vout 21
Gain limitation R 2 R 1 Vin I - e + u -A e Vout A is of the order of 105 – Error is very small 22
Input Impedance R 2 R 1 Zc- + Vin Zd Vout Zc+ u Non-inverting amplifier u Zin = Zc+ // (Zd A / G) ~ Zc+ G= (R 1+R 2)/R 1 23
Output impedance u R 2 Non-inverting amplifier R 1 I 0 + Iout - e -A e + I 0 Iout Z 0 Vout 24
Current drive limitation Maximum Output Swing R 2 R 1 I + Vout RL Vin u u u RL*Imax RL Vout = R I = RL IL The op-amp must deliver I + IL = Vout (1/R + 1/RL) Limitation in current drive limits output swing 25
Bandwidth f 3 db= f. T/G f. T u Gain amplifier of non-inverting G(p) = G A(p) / (G + A(p)) – A(p) with one pole at low frequency and -6 d. B/octave » A(p) = A 0 / (p+w 0) – G = (R 1+R 2)/R 1 40 d. B – Asymptotic plot » G < A G(p) = G » G > A G(p) = A(p) 26
Slew Rate u u Limit of the rate at which the output can change Typical values : a few V/ms A sine wave of amplitude A and frequency f requires a slew rate of 2 p. Af S (V/ms) = 0. 3 f. T (MHz); f. T = frequency at which gain = 1 27
Settling Time u Time necessary to have the output signal within accuracy – ±x% u Depends on the bandwidth of the closed loop amplifier – f 3 d. B = f. T / G u Rough estimate – 5 t to 10 t with t = G / 2 p f. T 28
Stability u G(p) = A(p) G / (G + A(p)) – A(p) has several poles u u If G = A(p) when the phase shift is 180 o then the denominator is null and the circuit is unstable Simple criteria – On the Bode diagram G should cut A(p) with a slope difference smaller than -12 d. B / octave – The loop gain A(p)/G should cut the 0 d. B axe with a slope smaller than -12 d. B / octave u -12 d. B/octave Phase margin – (1800 - Phase at the two previous points) u Unstable amplifier The lower G the more problems - Open loop gain A(p) - Ideal gain G - Loop gain A(p)/G 29
Stability improvement -6 d. B/octave Compensation u -6 d. B/octave Pole in the loop Move the first pole of the amplifier – Compensation u Add a pole in the feed-back u These actions reduce the bandwidth 30
Capacitive load u Buffering to drive lines R 2 R 1 - C = 20 p. F 10 + u C Load = 0. 5 m. F The output impedance of the amplifier and the capacitive contribute to the formation of a second pole at low frequency – A’(p) = k A(p) 1/(1+r C p) with r = R 0//R 2//R – A(p) = A 0 / (p+w 0) u Capacitance in the feedback to compensate – Feedback at high frequency from the op-amp – Feedback at low frequency from the load – Typical values a few p. F and a few Ohms series resistor 31
Examples of data sheets (1) 32
Examples of data sheets (2) 33
Current feedback amplifiers e - -A e Zt ie + u Voltage feedback + u u ie Current feedback Zt = Vout/Ie is called the transimpedance gain of the amplifier 34
Applying Feedback R 2 R 1 I - Zt ie + ie Vout Vin u u Non-inverting amplifier Same equations as the voltage feedback 35
Frequency response R 2 R 1 I - Zt ie + ie Vout Vin u The bandwidth is not affected by the gain but only by R 2 – Gain and bandwidth can be defined independently u Different from the voltage feedback – f 3 d. B = f. T / G 36
Data sheet of a current feedback amplifier 37
Data sheet of a current feedback amplifier (cont’) u Very small change of bandwidth with gain 38
Transmission Lines u u u Lossless Transmission Lines Adaptation Reflection Transmission lines on PCB Lossy Transmission Lines 39
Lossless transmission lines (1) u L, C per unit length x Impedance of the line Z u Pure resistance u Lx Cx Z 40
Lossless transmission lines (2) u Propagation delay u Pure delay Lx V 1 Cx I V 2 Z 41
Lossless transmission lines (3) u Characteristic impedance pure resistance u Pure delay u Capacitance and inductance per unit of length u Example 1: coaxial cable – Z = 50 – t = 5 ns/m – L = 250 n. H/m; C = 100 p. F/m u Example 2: twisted pair – Z = 100 – t = 6 ns/m – L = 600 n. H/m ; C = 60 p. F/m 42
Reflection (1) Source generator u Zs Zo – V, Output impedance Zs u Line appears as Z 0 u All along the line Vs = Z 0 Is If the termination resistance is ZL a reflection wave is generated to compensate the excess or lack of current in ZL u u V Vs Is ZL The reflected wave has an amplitude 43
Reflection (2) u The reflected wave travels back to source and will also generate a reflected wave if the source impedance is different from Z 0 – During each travel some amplitude is lost u ZS = 1/3 Z 0 ZL = 3 Z 0 The reflection process stops when equilibrium is reached – VS = V L u u Zs < Z 0 & Z L > Z 0 Dumped oscillation Zs > Z 0 & Z L > Z 0 Integration like ZS = 3 Z 0 ZL = 3 Z 0 44
Reflection (3) u Adaptation is always better – At the destination: no reflection at all – At the source: 1 reflection dumped 1 transit time » Ex. ZL = 3 Z 0 u Can be used to form signal – Clamping Zs V 2 transit time Zo Vs 45
Transmission lines on PCB u Microstrip u Stripline 46
Lossy transmission lines u Idem with Rs. L instead of L, Rp//C instead of C Rs L C u Rp Characteristic impedance depends on w – Even Rs is a function of w because of the skin effect u u Signal is distorted Termination more complex to compensate cable characteristic 47
Bibliography u The Art of Electronics, Horowitz and Hill, Cambridge – Very large covering u An Analog Electronics Companion, S. Hamilton, Cambridge – Includes a lot of Spice simulation exercises u Electronics manufacturers application notes – Available on the web » (e. g. http: //www. national. com/apnotes_all_1. html) u For feedback systems and their stability – FEED-2002 from CERN Technical Training 48
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