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Investigation of Switching Probability in Magnetic Samples Using Pulse Techniques

This study explores the effects of pulse duration and amplitude on the switching probability of magnetic samples in different magnetic field conditions. The methodology involves saturating the sample, applying a measurement field, and measuring the sample resistance using a lock-in amplifier. By iterating through pulse applications, we derive the switching probability for specified pulse conditions. Data analysis includes investigating temperature and field-dependent switching behavior, with results presented in phase diagrams and comparisons to theoretical models.

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Investigation of Switching Probability in Magnetic Samples Using Pulse Techniques

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  1. Long Pulse On 29-19 100nm x 100 nm 2009-08-12

  2. Sample Structure

  3. How We Measure • Saturate sample to AP or P state by applying enough field. • Change the applied field to “measurement field”. • Measure the sample resistance by lock-in at 300 uA. • Switch to pulse circuit that no lock-in current goes into sample. Apply square pulse to sample from AWG at a certain Duration and Amplitude. • Switch to lock-in circuit and measure the sample resistance again. Check if sample has switched: • If yes : repeat from step 1. • If no : repeat from step 4. • After applying the same pulse 100 times, we get the switching probability for a certain pulse duration and amplitude pair.

  4. Types of Experiments • Switching Probability as a function of pulse duration and pulse amplitude: • Ap -> P @ B = 0 T, along easy axis • Ap -> P @ B = -0.01 T, along easy axis • P -> Ap @ B = 0 T, along easy axis • Switching Probability as a function of pulse amplitude at a fixed pulse duration: • Duration = 100 us • Duration = 10 us • Duration = 100 ns • Duration = 5 ns • Duration = 3 ns All Ap -> P @ B = 0 T, along easy axis

  5. How We Analyze • Short Pulse: • I = 1 / Duration + Ic0 * • Long Pulse: • Ic (T) = Ic0 x [1 - (KBT / Uk) x ln (Tpf0)] * * J. Sun, SPIE 2004 (vol. 5339, 445)

  6. Results from Measurement AP -> P 0 T 5ns, 100ns, 1us

  7. Results From Measurement AP -> P 0 T 3ns, 10ns, 20ns, 50ns

  8. Results from Measurement AP -> P 0 T 1us -> 1ms

  9. Phase Diagram From All Zoom in

  10. 1. Compare to High Speed Points crossing 50% for SWITCH: (5 ns, 0.266 V) and (3ns, 0.289 V) Directly read from graph Will be used for next slice Ap -> p @ 0 T For both BST and Switch

  11. 1. Compare to High Speed Zoom In

  12. 1. Compare to High Speed A = 0.2330 + 1.790e-10 / D (Bias Tee) A = 0.2315 + 1.725e-010 / D (Switch)

  13. 2. Compare to DC From extracted boundary value: A(Ap-> P @ 0 T @ 50% @ 1 ms) = 0.167 V Idc / A = 0.0034 + 0.0256 x B / T Ap -> P Idc / A = -0.0069 + 0.0164 x B / T P -> Ap Idc (Ap -> P @ 0 T ) = 3.4 mA I_50% (1ms)= 0.029 * 0.167 A = 4.71 mA I_50% (1s) = 4.01 mA (see next page)

  14. 3. Log D ~ I Fitting Zoom In R2 = 0.98 ~ 6 ns ~ 2 us Intermediate Thermal Dynamic I / A = 0.00401 – 0.000234 x log10(Duration / s)

  15. Uk Calculation Ic0 = 0.2330x 0.028 A= 6.57e -3 A Ic (T) = Ic0 x [1 - (KBT / Uk) x ln (Tpf0)] * - J. Sun, SPIE 2004 (vol. 5339, 445) When Happ + Hdip = 0 KB = 8.617 343e-5eV / K T =300 K KB T = 0.02585eV I / A = 0.00401 – 0.000234 x log10(Duration / s) @ Happ = 0 T @ 50% Uk = 1.53eV f0 = 21.7 GHz I / A = 4.01e-3 – 1.02e-4 x ln (Duration / s)

  16. Uk Calculation z θ m • For H in Z : • E_total = - mu_0 * M_s * H * cos θ + 0.5 * mu_0 * M_s2 * cos2 θ - K_1 * cos2θ • U_k = E_total(θ = 90˚) – E_total(θ = 0˚) = 0.5 * mu_0 * M_s * (2H + 2 * k_1 / (mu_0 * M_s) – M_s) y x

  17. Uk Calculation • If we use:            M_s = 713000 A / m            K_1 = 4.05e5 J / m^3            V = 100nm * 100 nm * 1.6 nm = 1.6e-23 m^3 from in Jean-Jacques report.and:            mu_0 = 1.256637e-6 m * kg * s^-2 * A^-2 ( or T * m / A)            eV = 1.6e-19 JoulesThen we have:            M_eff = -191036 A / m             U_k = 8.56 eV (1.53 eV from experiment) at 0 T applied field.

  18. Results From Measurement Ap -> P @ -0.01 T, 180deg

  19. Phase Diagram From All Zoom in

  20. 1. Compare to High Speed • We have only 3 points below 5e-9 S (all 2ns, the start points), so cannot really compare them.

  21. 2. Compare to DC Still need longer duration A (Ap -> P @1ms @ 50% @ -0.01 T) = 0.160994059449 V I (I ms) = 0.160994059449 * 0.02821078361584272 A = 0.0045417685745518582 A = 4.5 mA I (I s) = 3.56mA Idc (Ap-> P @ -0.01 T) = 3.26 mA

  22. 3. Log D ~ I Fitting Zoom in I_50% = 6.25 e-3 + 4.55 e-12 / D Dynamic I_50% = 3.5556e-3 - 1.3105e-4 ln(D ) Thermal

  23. Uk Calculation • Ic (T, H) = Ic0 (H) x [1 - (KBT / Uk(H) ) x ln (Tpf0)] • Ic0 (H) = Ic0* (H = 0) * (H + H_k + 0.5 x M_s) / (H_k + 0.5 x M_s) • J. Sun, SPIE 2004 • If we use: • Ic0 (H) = 6.25 mA • Then we get: • f0 = 0.849 GHz • Uk = 1.19 eV • Theoretical Uk at -0.01 T: • Uk = 8.56 eV - 0.71 eV = 7.85 eV • Compare Uk at 0 T and -0.01 T : • Theoretical : ΔUk = B * M_s * V = 0.71 eV • Experimental : ΔUk = 1.53 eV – 1.19 eV = 0.34 eV

  24. Results From Measurement P -> Ap @ 0 T, 180deg

  25. Phase Diagram From All Zoom in

  26. 1. Compare to High Speed • We have only 3 points below 5e-9 S (all 2ns, the start points), so cannot really compare them.

  27. 2. Compare to DC A (P -> Ap @1ms @ 50% @ 0T) = -0.28399085 V I (1 ms)= -0.28399085 * 0.02821078361584272 A = -0.0080116044182292484 A = -8.0 mA I (1 s)= -6.9 mA Idc (P -> Ap @ 0 T) = -6.9 mA

  28. 3. Log D ~ I Fitting r2 = 0.994 Zoom in I_50% = -10.8e-3 – 9.40 e-12 / D Dynamic I_50% = -6.857e-3 + 1.725e-4 ln(D ) Thermal

  29. Uk Calculation • If we use: • Ic0 (H) = -10.8 mA • Then we get: • f0 = 8.50 GHz • Uk = 1.62 eV

  30. Results and Fitting Ap -> P @ 0 T Thermal Exponential fitting from all raw data points which have P != 1

  31. Uk Calculation • Psw = 1 – exp{(-t / τ0)exp[(-Uk/kBT)(1 – I /Ic0)]} * • J. Sun, SPIE 2004 (vol. 5339, 445) • t / τ0= 5.24e5 , Uk/kBT = 1 / 1.88e-2 = 53 • If we use: • t = 1e-4 s , T = 300 K • We have: • Uk = 1.38 eV • f0 = 1/ τ0 = 5.24 GHz

  32. Results and Fitting Ap -> P @ 0 T Thermal Exponential fitting from all raw data points which have P != 1

  33. Uk Calculation • t / τ0 = 2.11e5 , Uk/kBT = 1 / 1.62e-2 = 62 • If we use: • t = 1e-5 s , T = 300 K • We have: • Uk = 1.60 eV • f0 = 1/ τ0 = 21.1 GHz

  34. Results and Fitting Ap -> P @ 0 T Intermediate Doesn't fit for small I Exponential fitting from all raw data points which have P != 1

  35. Uk Calculation • t / τ0 = 4.33 , Uk/kBT = 1 / 0.0523 = 19 • If we use: • t = 1e-7 s , T = 300 K • We have: • Uk = 0.49 eV • f0 = 1/ τ0 = 0.0433 GHz

  36. Results and Fitting Ap -> P @ 0 T Dynamic Doesn't fit for small I Exponential fitting from all raw data points which have P != 1

  37. Uk Calculation • t / τ0 = 0.184 , Uk/kBT = 1 / 0.0870 = 11 • If we use: • t = 5e-9 s , T = 300 K • We have: • Uk = 0.30 eV • f0 = 1/ τ0 = 0.0368 GHz

  38. New Fitting for Short Time Pulse • Equation: • We assume the initial spin angle θ0 would be distributed as: • f(θ) = A * exp( - E_totle(θ) * V / kBT) * sin(θ) if: 0 < θ < 90° • f(θ) = 0 else • where: • E_totle(θ) is the total energy defined previously, • V is the total volume of the spin valve junction. • A is just a constant and will satisfy: • Integrate(f(θ) dθ) = 1 • Probability of switching equals to probability of initial spin angle larger than θ0 • Therefore, we can get all the θ0 from Probability data and then do I ~ θ0 fitting. Where : ----Jonathan Z. Sun June 22nd, 2009

  39. Fitting for 5 ns Reminder: I_c0 = 6.57 mA from BST pulse

  40. Results and Fitting Ap -> P @ 0 T Doesn't fit for small I Exponential fitting from all raw data points which have P != 1

  41. Uk Calculation • t / τ0 = 0.0514 , Uk/kBT = 1 / 0.0907 = 11 • If we use: • t = 3e-9 s , T = 300 K • We have: • Uk = 0.28 eV • f0 = 1/ τ0 = 0.017 GHz

  42. Fitting for 3 ns Reminder: I_c0 = 6.57 mA from BST pulse

  43. I ~ Initial Angle

  44. Data Table

  45. All in One

  46. All in One

  47. Results and Fitting Zoom In

  48. Calculation • Psw = 1 – exp{(-t / τ0)exp[(-Uk/kBT)(1 – I /Ic0)]} * • J. Sun, SPIE 2004 (vol. 5339, 445) • exp[(-Uk/kBT)(1 – I /Ic0)] / / τ0 = 3.44e4 • I = 5.1 mA, Ic0 = 6.57 mA • Uk / kBT = ln f0 / 0.227 – 46 • Bring I ~ P fitting data: • Uk / kBT = ln f0 / 0.227 – 45 (100 us) • Uk / kBT = ln f0 / 0.227 – 43 (10 us)

  49. Add Sweep Data to Boundary

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