Oscilloscope Attenuator

The $ \times 10 $ attenuator probe employed with oscilloscopes is designed to reduce the magnitude of the input voltage by a factor of 10 . If the input impedance to a scope is $ 1 \mathrm{M} \Omega $, the $ \times 10 $ attenuator probe will have an internal resistance of $ 9 \mathrm{M} \Omega $, as shown in Fig. 1. Applying the voltage divider rule, In addition to the input resistance, oscilloscopes have some internal input capacitance, and the probe will add an additional capacitance in parallel with the oscilloscope capacitance, as shown in Fig. 2. The probe capacitance is typically about $ 10 \mathrm{pF} $ for a 1-m (3.3-ft) cable, reaching about $ 15 \mathrm{pF} $ for a 3-m (9.9-ft) cable. The total input capacitance is therefore the sum of the two capacitive elements, resulting in the equivalent network of Fig. 3. For the analysis to follow, let us determine the Thévenin equivalent circuit for the capacitor $ C_{i} $ : The Thévenin network is shown in Fig. $ 4 $. For $ V_{i}=200 \mathrm{~V} $ (peak), and for $ v_{C}, V_{f}=20 \mathrm{~V} $ and $ V_{i}=0 \mathrm{~V} $, with For an applied frequency of $ 5 \mathrm{kHz} $, with $ 5 \tau=135 \mu \mathrm{s}>100 \mu \mathrm{s} $, as shown in Fig. 5, clearly producing a severe rounding distortion of the square wave and a poor representation of the applied signal.