53230A Agilent Technologies Test Equipment, 53230A Datasheet - Page 19

• Manufacturer Part Number: 53230A
• Description: Universal Frequency Counter/Timer, 350 MHz, 12 digit/sec, 20 ps
• Manufacturer: Agilent Technologies Test Equipment
• Type: Counter/Timerr

Specifications of 53230A

• Frequency Range: DC coupled 1 mHz to 350 MHz; AC coupled 10 Hz to 350 MHz
• Input Impedance: Selectable 1 MΩ ±1.5% or 50Ω ±1.5%
• Input Type: Front panel BNC
• Input, Range: ±5 V (± 50 V) full scale ranges
• Noise: 500 μVrms (max), 350 μVrms (typ)
• Number Of Channels: 3 Channels
• Sensitivity: DC - 100 MHz: 20 mVpk > 100 MHz: 40 mVpk

Appendix A - Worked Example
Basic Accuracy Calculation for Frequency Measurement
Parameter assumptions:
• 53220A
• 95% Confidence
• 100MHz signal, 1sec gate
• AUTO
• Level: 5V
• OCXO standard timebase for unit plugged in for 30 days
• Assumes 100 samples per trigger taken
Process:
Basic Accuracy = k * (Random Uncertainty ± Systematic Uncertainty ± Timebase Uncertainty)
1. Use k=2 for 95% confidence and k=2.5 for 99% confidence calculations)………………..k = 2
2. Random Uncertainty for Frequency Measurement =
3. Systematic uncertainty for frequency measurement = If RE>=2: 10
Note: Typical is achieved with an average of 100 readings with 100 samples per trigger. Worst case is trigger and sample
count set to 1.
4. Timebase uncertainty = (aging + temperature + calibration uncertainty) = (0.2 ppm + 1 ppm + 0.5 ppm) = 1.7 ppm
Basic accuracy = k * (random uncertainty ± systematic uncertainty ± timebase uncertainty) =
2(23.3 ps ± 2 ps ± 58.8 ms) = .1176 s
Note: Using a higher accuracy timebase or locking to an external timebase standard will have the biggest impact on
improvement to accuracy calculations.
T
T
E
Vx = N/A (remove signal from other channel)
SR
R
Gate time = 1 sec
Aging: 0.2 ppm
Temperature: 1 ppm
Calibration uncertainty: 0.5 ppm
(10 MHz)(1.7*10-6) = 17 Hz = 58.8 ms
SS
E
N
E
(for 5 V)
-TRIG POINT
= 6
= Assume input signal RMS noise voltage is 0.
= 100 ps
= maximum slew rate (sine)SR= 2πF*V
=
(500 μV
SR
2
-TRIG POINT
+ E
N
2
+Vx
2
)
½
=
0 to PK
1.4* (T
R
= 2π(100 MHz)*5 V = 3.14*10
E
(500 μV
3.14 * 10
x Gate Time
19
SS
2
+ T
2
)
-11
½
9
E
/gate max, 2*10
2
)
½
=
=
.159 ps
1.4* (100ps
-12
9
/gate (typ) = 2 ps
6 x 1
2
+ .159ps
2
)
½
=
23.3 ps