Principle analysis of oscilloscope passive probes
Engineers commonly used 10 times the passive voltage probe Rp (9MΩ) and Cp is located in the probe tip, Rp for the probe input impedance, Cp for the probe input capacitance, R1 (1MΩ) represents the input impedance of the oscilloscope, C1 represents the input capacitance of the oscilloscope and the equivalent capacitance of the coaxial cable as well as the probe compensation box capacitance of the combination of values. In order to measure accurately, the two RC time constants (RpCp and R1C1) must be equal; any imbalance will bring about a distortion of the measured waveform, and never cause inaccurate measurements of some parameters such as rise time and amplitude. Therefore, it is necessary to calibrate the work of the oscilloscope probe before measurement to ensure the accuracy of the measurement results. From the signal model of the probe we can analyse that for DC measurement of the signal, the input capacitance Cp and C1 are equivalent to an open circuit. The signal is divided by Rp and R1, and the final oscilloscope input is: Vout=[R1/Rp+R1]*Vin=1/10*Vin
Oscilloscope The input signal to the oscilloscope is attenuated to 1/10 of the input signal to be measured. for higher frequency input signals, the effect of capacitive reactance on the signal will be greater than the impedance. For example, a standard 1MΩ ~ 10pF passive voltage probe, the input signal frequency of 100MHz, at this time, the probe input capacitive impedance Xc (Cp) = 1 / (2 × π × f × C) = 159Ω, capacitive impedance is much smaller than the probe impedance of 9MΩ, the signal current will be more through the input capacitance provided by the low-resistance circuit, the high-resistance circuit of the 9MΩ impedance is equivalent to bypass. Can also be understood as 159Ω and 9MΩ parallel connection after the equivalent impedance of 159Ω. At this time, the actual input to the oscilloscope signal amplitude (AC / high frequency) is determined by the input capacitance of the probe as well as the ratio of the total capacitance of the loop, the equivalent of: Vout = [Cp/Cp+C1]*Vin
Typically, a passive probe cable has a capacitive load of 8-10pF/foot (1 foot foot = 12inches inch = 0.3048metre metre) and a rise time of 1.5nS/foot. For a 6feet cable there is 60pF of capacitance, plus the 20pF input capacitance of a typical oscilloscope and some stray, roughly 90pF or so. According to the 1:10 voltage division, the input capacitance of the probe should be about 10pF to satisfy the characteristic of Vout/Vin=[10/10+90]=1/10 input attenuation by a factor of 10. Taking into account the probe and cable capacitance of some errors, the need to use the probe compensation capacitance box to carry out a circuit compensation, due to the error, passive voltage probe input capacitance is generally between 8 ~ 12pF. Currently the mainstream 10 times passive voltage probe input load model is generally input capacitance 8~12pF, input resistance 9M ohm.
