hi all,

i am using probe 85024A to measure DUT and notice that there is 10dB drop for the result measurement using my N9000A spectrum analyzer.

as this is a resistive probe 10:1 i am expecting a 20dB drop instead of 10dB drop.

i also realize that it's a 1Mohm impedance versus my instrument 50ohm.

how would i calculate and would expect a 10dB drop rather 20dB drop?

thanks you.

i am using probe 85024A to measure DUT and notice that there is 10dB drop for the result measurement using my N9000A spectrum analyzer.

as this is a resistive probe 10:1 i am expecting a 20dB drop instead of 10dB drop.

i also realize that it's a 1Mohm impedance versus my instrument 50ohm.

how would i calculate and would expect a 10dB drop rather 20dB drop?

thanks you.

I’ve attached the 85024A Product Overview. There are loading effects of the shunt capacitor that is used inside the 85024A at various frequencies. The graph on page 2 of the attachment is difficult to understand since the writing is small and the scale is log, but the numbers in the example can be computed. Here are the effects at two different frequencies.

At 500 MHz, Xc=1/(2pi)(F)(C), where F is 500 MHz and C is the value of the shunt capacitor used in the 85024A. So, 1/(6.28)(500 MHz)(0.7pF), so Xc or the capacitive reactance at 500 MHz of the probe itself is 455Ω.

The voltage drops can be calculated using the capacitive reactance value at 500 MHz when used in a 50 ohm system as (Xc/Xc+50 ohms), 455/455+50, 455/505 = 90%, where 90% of the voltage is going to the active probe. Another way of putting this is 10% of the voltage is dropped by the capacitor at 500 MHz.

Then the graph tries to illustrate what happens if the shunt capacitor value was 3 pF, the capacitive reactance at 500 MHz would be 106Ω . The graph illustrates minimal loading effects when the 85024A probe is used due to its low value shunt capacitance of 0.7 pF.

106/106+50 ohms = 68%, where 68% of the voltage is going to the active probe. Another way of putting this is 32% of the voltage is dropped by the capacitor at 500 MHz if a 3 pF capacitor was used in the 85024A.

At 2.4 GHz, Xc=1/(6.28)(2.4 GHz)(0.7pF)=94.7Ω. 94.7/94.7+50= 65.4%. 65.4% of the voltage is going to the active probe and 34.6% is dropped across the 0.7pF capacitor at 2.4 GHz.

These examples can be leveraged for different frequency measurement points.

Regards -