EEVblog Electronics Community Forum
Products => Test Equipment => Topic started by: Scr1bbl3 on September 05, 2022, 03:32:20 am
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Hello all,
I just got my first active probe. I have a circuit I want to test it on that has a 9V DC component. The circuit under test is an LC oscillator so it's susceptible to capacitive loading - hence the active probe. The probe passes AC and DC (DC coupling). How do I block DC without adding capacitance to the circuit and pulling it off frequency? Thanks in advance.
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Preferably, you'd block DC at the probe output which is buffered from the DUT. If 9 V is out of the probe's DC range, you could add a series capacitor that is small enough not to detune the circuit but it may cost you some signal attenuation.
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Adding a small capacitor ahead of the probe should not affect directly the shunt capacitance connected to the circuit (which would detune your resonant circuit).
As an example, if your active probe input capacitance is 5 pF (maybe in parallel with 1 megohm), then a 50 pF series capacitance would have an attenuation of approximately 0.9 V/V at high frequencies, and give a high-pass filter that is -3 dB down at 35 kHz (with the assumed 1 megohm input resistance).