Question on input impedance of measuring instruments?

For example an oscilloscope input channel shows an input impedance of 1 Megohm and 18 pF. I believe these two will appear in parallel to any input signal. Also I believe higher the megohm and lower the capacitance it is the better, to decrease the loading on the input signal. For example lower megohm will cause more power losses and higher capacitance will cause signal distortion at higher frequencies. Could any one kindly confirm?

3 Answers

  • Ecko
    Lv 7
    6 years ago
    Favorite Answer

    Yes you are correct. Consider the reactance of 18pF at 100MHz is only 88 ohms. The X10 probe addresses this issue, especially considering that the input capacitance is usually more than 18pF in parallel with the 1 megohm. It increases the Rin to 10 megohms with a 9 megohm resistor in series. It reduces the signal by 10 times. A small capacitor in parallel with the 9Meg resistor works with the input capacitance to form a capacitive divider so the frequency response is not disturbed. It would be only 2 pF in the example, but still essential. So always use the X10 probe unless there is a reason not to. Note that this probe will need compensating. The capacitor is adjusted for a good response using a square wave signal.

    Thus the X10 probe increases the resistance by 10, reduces the capacitance by 10, and reduces the signal by 10 without affecting the bandwidth.

    Another approach that doesn't lose signal level and has a good frequency response is a coaxial cable terminated by its characteristic impedance at the oscilloscope input. This is usually 50 or 75 ohms. The source being monitored has a special 50 ohm (or 75 ohm) output for monitoring. The input capacitance has little effect.

    This can be taken further with an active probe using a special preamplifier, with as little as 0.8pF input capacitance. The output connects to the oscilloscope using terminated coaxial cable.

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  • 6 years ago

    The oscilloscope provides the bandwidth specified (say 20 MHz) when operated with a source resistance of 50 ohms to 600 ohms. When source impedance is higher one will find that the bandwidth gets restricted. Example. Check the CRO performance with a 10V reduced to 1V using (i) a 900 k and a 100K (ii) a 9k and a 1k. In the first case, the source resistance is 90k while it is 900 ohms. The 18pF of the CRO forms a low pass filter in first case with a cut off frequency of 1/(2.pi.RC) where RC is 90k*18pF= 1.62uS. this gives cut off frequency (bandwidth) of about 100 kHz. (verify by feeding a square wave of 10khz and observing the rise time, time required for output to build from 10% to 90%)

    Using a 10:1 probe with lower input capacitance will help. But the capacitance of a 10:1 probe with a cable might be 3 to 4 times the expected capacitance of 1.8pf

    What is distortion? The filtering effect the arrangement has will reduce the harmonics of a square wave and render at is a sinewave as the frequency fed to attenuator is increased from 10khz to 100khz. When you get a chance try out all these.

    Power loss is not a major concern except in power amplifiers.

    Just imagine what it is like to provide 10V into a 50 ohms, as many standard function generators provide. the current will be a huge 100mA, taking into account the Rout of the generator! Get to know whether this is rms or peak if waveform is sine.

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  • Steven
    Lv 7
    6 years ago

    A passive scope probe is only high impedance at low frequencies. Above the frequency where the probe cable is longer than a quarter wave, because it is impossible to make high impedance cable (>500 Ohms), the probe impedance falls to the cable impedance times the attenuator factor. For a 10x probe that would be about 5K Ohms. Passive scope probe cable is made with a tiny steel center conductor as small as possible to make the cable impedance as high as possible.

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