M8 · A sense chain that does not clip

Goal

• Work out a sense chain’s burden, its offset-equivalent current and its filter corner.
• Find why the example misses its error target at the smallest current, and fix it.
• Keep the output below the amplifier’s largest output and the ADC’s full scale at the largest current, in the worst case of the offset and the pad resistance.

Before you start

• Read the sense chain and current sensing.
• Know the two gotchas this mission is about: pad and solder resistance and a current-output amplifier’s headroom.

Steps

1. The chain

Open the sense chain with the current-output example:

▶ Open in the sense chain: Example sense chain: current-output amplifier, two-terminal shunt
Preset from the example sense-current

It measures up to 2 A with a shunt of 10 mΩ on pads that add 500 µΩ, and must read 200 mA within 0.01.

2. Burden and offset

Work out the sense voltage at the largest current (sense voltage) and the current the amplifier’s offset looks like (offset-equivalent current). Enter both below. A larger shunt lowers the second and raises the first [Texas Instruments 2023].

3. Why the low end fails

At the smallest current, add the pad error (pad error) and the offset (relative error). Say which of the two is larger, and why a Kelvin connection removes one of them [O'Sullivan 2012].

4. The fix, and what it costs

Change the chain until the tool reports the error at the smallest current within the target: the connection, the shunt or the amplifier’s offset. Then check the largest current: the worst-case output must stay below the amplifier’s largest output and the ADC’s full scale. If it does not, lower the gain. Write down what your fix costs in burden and dissipation.

5. The filter

Work out the filter’s corner (R-C corner frequency) and compare it with the Nyquist frequency of the ADC’s sampling rate (Nyquist frequency). Enter the corner below.

Acceptance criteria

• The sense voltage at the largest current, the burden on the measured circuit.
  Your value (V)Check
  Show the example's value

  20 mV (from sense-current: Example sense chain: current-output amplifier, two-terminal shunt)

• The current the amplifier's offset looks like.
  Your value (A)Check
  Show the example's value

  10 mA (from sense-current: Example sense chain: current-output amplifier, two-terminal shunt)

• Which of the pad error and the offset is larger at the smallest current, and a change that brings the error within the target.
• At the largest current, the worst-case output below the amplifier's largest output and the ADC's full scale.
• What your fix costs in burden and dissipation.
• The filter's corner frequency, below the Nyquist frequency.
  Your value (Hz)Check
  Show the example's value

  1.447 kHz (from sense-current: Example sense chain: current-output amplifier, two-terminal shunt)

Clear this mission

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Gotchas

• A current-output amplifier’s output cannot swing to its supply. Its headroom sets the largest output whatever the gain (headroom).
• The data sheet gives the offset’s largest magnitude, not its sign: check both signs, the one that raises the output at the top and the one that lowers the reading at the bottom.
• With a current-output amplifier, the filter’s capacitor sees $R_\mathrm{OUT}$ in series with $R_f$ (the filter's resistance), so the corner is lower than $R_f$ alone gives.
• A single pole below the Nyquist frequency still passes part of the signal there. Read the filter’s gain at the Nyquist frequency, and at the switching frequency if you enter it.

Go deeper

• Technical article System Trade-offs for High- and Low-side Current Measurements (SSZTA51) — Texas Instruments · retrieved 2026-09-24
  A short comparison of placing the shunt in the ground return or on the supply side.
• Technical article Optimize High-Current Sensing Accuracy by Improving Pad Layout of Low-Value Shunt Resistors (Analog Dialogue 46-06) — M. O'Sullivan, Analog Devices · retrieved 2026-09-24
  How the solder of a low-value shunt enters the reading, and the four-pad layout that keeps it out.

Quiz

1. A 5 mΩ shunt carries 4 A. What sense voltage, and burden, does it give?

   1. 20 mV
   2. 2 mV
   3. 200 mV
   4. 0.8 mV
   Answer and explanation

   Correct answer: A Ohm's law: 4 A times 5 mΩ is 20 mV, which the measured circuit loses across the shunt.

2. An amplifier with an offset of 50 µV reads that shunt. What current does the offset look like?

   1. 1 mA
   2. 10 mA
   3. 0.25 mA
   4. 100 mA
   Answer and explanation

   Correct answer: B The offset over the shunt: 50 µV over 5 mΩ is 10 mA, the same at every current.

3. The pads add 0.5 mΩ to that shunt, sensed beyond the pads. By what fraction does the reading run high?

   1. 1 %
   2. 10 %
   3. 5 %
   4. 0.5 %
   Answer and explanation

   Correct answer: B The amplifier sees the current through the shunt and the pads: 0.5 over 5 is a tenth, at every current.

4. What removes the pad error?

   1. A larger gain
   2. A Kelvin connection: the sense traces meet the shunt's own terminals, away from the load current's pads
   3. A lower sampling rate
   4. A larger filter capacitor
   Answer and explanation

   Correct answer: B With a four-terminal shunt, or a four-pad layout for a two-terminal one, the pads carry the load current but are no longer in the sense path.

5. Raising the shunt tenfold lowers the offset-equivalent current tenfold. What does it cost?

   1. Nothing
   2. Ten times the burden and the dissipation at the same current
   3. A tenth of the bandwidth
   4. A higher offset
   Answer and explanation

   Correct answer: B The sense voltage and the shunt's loss both grow with its resistance; the output may also clip, so the gain must come down.

6. A current-output amplifier's output reaches 4.5 V at full current, on a 5 V supply whose output swings to about 1 V below it. What happens?

   1. Nothing: the gain sets the output
   2. The output clips before full current: lower the gain
   3. The offset doubles
   4. The shunt overheats
   Answer and explanation

   Correct answer: B The output cannot exceed its headroom whatever the gain, so the top of the range is lost unless the gain comes down.