EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: gogoman on August 12, 2018, 05:01:55 pm
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Hi I soldered a 50ohm resistor to the end of a coax and monitored the voltage and current across the resistor.
Next I stepped the freq and compared the measured current to the calculated, the current probe was off by 20-30%
Is this a validate measurement technique?
thanks
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How did you measure the voltage and current?
And what are the measured values?
What current probe did you use?
What frequencies are we talking about?
What does the signal look like?
Without the basic information, it is hard to give any feedback or advise.
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the probe is Tektronix P6021
signal Source is a sine wave
The voltage probe is across cross the resistor, the resistor is encased in the current probe jaw.
The signal from the current probe is a sine wave. At 40mhz there voltage peaking.
the I R Drop does not is equal the peak-peak scope voltage
thanks
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That can work and it is how current probes are calibrated but the signal source needs to be leveled and the resistor should be one with low inductance and capacitance. I think you will have to extend the ground side of the resistor and place the probe there instead of over the resistor to reduce electrostatic coupling at high frequencies.
The Tektronix 067-0559-00 current probe test fixture uses a transmission line environment. I might try to construct something similar out of double sided copper board and a chassis mount BNC connector.
http://w140.com/tekwiki/wiki/067-0559-00 (http://w140.com/tekwiki/wiki/067-0559-00)
Read the P6201 calibration instructions to see how it is used.
http://www.produktinfo.conrad.com/datenblaetter/1100000-1199999/001171294-an-01-en-TEKTRONIX_P6021A_AC_STROMTASTKOPF.pdf (http://www.produktinfo.conrad.com/datenblaetter/1100000-1199999/001171294-an-01-en-TEKTRONIX_P6021A_AC_STROMTASTKOPF.pdf)
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the probe is Tektronix P6021
signal Source is a sine wave
The voltage probe is across cross the resistor, the resistor is encased in the current probe jaw.
The signal from the current probe is a sine wave. At 40mhz there voltage peaking.
the I R Drop does not is equal the peak-peak scope voltage
thanks
Do you have the 6021 termination module #011-0105-00 or type 134 amplifier ?
One or the other must be used in conjunction with a P6021.
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Do you have the 6021 termination module #011-0105-00 or type 134 amplifier ?
One or the other must be used in conjunction with a P6021.
Oh, that is a good point. The probe will not have the proper response without the passive termination box or one of the amplifiers designed for it.
If you are desperate, you could build a passive termination for it.
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thanks you for the input :)
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In addition to the great advice by David, you should also consider that connecting any test instrument will alter the signal you are testing. This observer effect (https://en.wikipedia.org/wiki/Observer_effect_(physics)) is unavoidable.
So consider carefully how your voltage probe is affecting the current through the resistor. A basic "10 MOhm" passive probe has about 15 pF of tip capacitance. At 40 MHz (which is what I assume you meant), the reactance of that capacitance is about 250 Ohm, and this is in parallel with your 50 Ohm resistor, so some current will be shunted through it. Also consider that the current probe is actually a transformer, and by clipping this onto the resistor, you have effectively added a series inductance to the resistor, which increases its impedance. So you should expect the current to be lower than expected, if that makes sense.
Also keep in mind the frequency response of the probe. It isn't a perfect flat line to its rated 60 MHz. At 60 MHz, it should be about 3 dB down (about 30% low), and at 40 MHz it will definitely be down a bit already.