Passive probes, compensation & loading

Let the probe contain a series resistance R_p and let the oscilloscope input resistance be R_i. At sufficiently low frequency, ignore capacitances. The channel sees

For a nominal 10× divider,

The oscilloscope or user interface then rescales the displayed voltage by the declared probe factor. If the scope is configured for 1× while a 10× probe is connected, the display will be wrong by a factor of ten even if the probe itself is electrically perfect.

The oscilloscope input is not purely resistive. Model it as

If the probe were only a series resistor, the divider ratio would change with frequency because C_i progressively shunts the input resistance. High-frequency components would be attenuated differently from low-frequency components, distorting pulse shape.

A compensated passive divider adds a capacitance C_p in parallel with R_p:

The divider transfer is

If the two RC time constants are equal,

then both impedances contain the same factor 1/(1+sτ). It cancels from numerator and denominator:

Thus the first-order divider has the same attenuation at low and high frequency. For a 10× probe with R_p=9R_i, the condition becomes

This is the mathematical meaning of probe compensation.

Under exact first-order compensation, the series combination has

This is equivalent to

For a 10× divider with R_p=9R_i and C_p=C_i/9,

The probe therefore increases the resistive input impedance by ten and, in this simplified model, reduces the effective capacitance by ten. That is why a compensated 10× probe can load a high-impedance/high-frequency node much less than a direct 1× connection.

Use deliberately generic teaching parameters: the 1× channel model is 1 MΩ ∥ 20 pF. A first-order idealized 10× probe therefore presents approximately 10 MΩ ∥ 2 pF at its tip. Let the source resistance be 100 kΩ and f=100 kHz.

For the direct 1× connection:

For the compensated 10× tip loading:

The 10× probe strongly improves the disturbance at the DUT node in this example, but its internal divider also sends only one tenth of that tip voltage to the oscilloscope front end. Loading and measurement noise/dynamic-range consequences must therefore be considered together.

Define

If τ_p<τ_i, the probe capacitance is too small relative to the input capacitance: high-frequency components are attenuated too strongly and a square-wave edge appears rounded/slow. This is under-compensation.

If τ_p>τ_i, high-frequency components are emphasized relative to the desired ratio and the step can show excessive peaking/overshoot. This is over-compensation.

The square-wave compensation signal is useful because a square wave contains a DC/low-frequency level plus many harmonics. A flat top and faithful edge test whether the divider treats those frequency components consistently.

Real probes also contain cable effects, distributed capacitance, connector inductance and a ground-return geometry. A long ground lead contributes inductance. In the simplest local model,

Fast edge current through L_g can create an additional voltage error. Together with local capacitance it can form a resonant structure. A rough LC frequency scale is

This is an intuition scale, not a complete probe model. It explains why shortening the return connection and minimizing loop area often reduces ringing and pickup on fast edges.

The correct choice depends on source impedance, signal amplitude, edge speed/frequency content, required accuracy and actual probe/scope specifications.

A useful empirical check is to observe a node with one probe, store/record the result, and then connect a second similar probe to the same node. If the waveform changes materially, the first probe was already loading the circuit and the doubled load made the effect visible.

This test does not quantify every parasitic by itself, but it is a direct falsification attempt: an ideally non-invasive measurement would not change when another identical high-impedance observer is attached.

1. Read the actual probe attenuation, compensation range, input resistance/capacitance and voltage ratings.
2. Set or verify the oscilloscope probe factor so displayed amplitude is correctly scaled.
3. Connect the probe to the instrument’s specified compensation/reference signal or another appropriate known edge.
4. Use the shortest practical reference connection for the compensation check.
5. Adjust only the intended compensation element until the step is neither rounded nor excessively peaked.
6. Then estimate DUT loading using the probe’s tip impedance over the frequency range of interest.

1. For R_i=1 MΩ, what R_p gives a 10× low-frequency divider?
2. If C_i=18 pF and R_p=9R_i, what C_p satisfies first-order compensation?
3. Under that idealized compensation, what are R_tip and C_tip?
4. What qualitative step shape indicates τ_p<τ_i?
5. Why can a compensated probe still ring on a fast edge?
6. A source has high resistance and modest signal amplitude. Give one reason to prefer 10× and one reason 1× might still be attractive.
7. If a second identical probe changes the first trace, what conclusion is justified?
8. Why is the actual probe specification required even after learning the ideal compensation equations?

Chapter 03 will move from the analog input path to the acquisition problem: sampling, record length, time base and trigger are different functions and must not be conflated.