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Long Pulse On 29-19

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

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Long Pulse On 29-19

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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 + 0.0164 x B / T Ap -> P Idc / A = -0.0069 + 0.0256 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 -1x 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 -1x 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. Add Sweep Data to Boundary

  48. Results and Fitting Zoom In

  49. 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)

  50. results (2.64 ns, 8.46 mA) @ 50%

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