# Understanding the Amplifier Offset Voltage and Output Swing in Resistive Current Sensing

## How does an amplifier's input offset voltage impact the accuracy of a current sense resistor's measurement? And how does amplifier output swing affect your shunt resistor value? Learn this and more in this technical article.

In previous articles, we discussed that op-amp-based amplifiers, as well as purpose-built current sense amplifiers, can be used to gain up the voltage across a current sense resistor. These amplification stages are not ideal and can introduce an error in our measurement.

In this article, we’ll discuss how the amplifier input offset voltage can affect the accuracy of our measurement. We’ll also see that the amplifier output swing can be a key factor when choosing the shunt resistor value.

### Input Offset Voltage

Ideally, the output of a differential amplifier should be at zero volts when its both inputs are exactly at the same potential. However, in reality, the output goes to zero volts when there is a small voltage difference between the inputs. This non-ideality originates from the inherent mismatch between the components inside the amplifier.

The voltage that must be applied between the input terminals of the amplifier to bring its output voltage to zero volts is called the input offset voltage. To model this non-ideal effect, we can place a voltage source in series with one of the amplifier inputs and assume that the amplifier is ideal and has zero offset voltage. This is shown in Figure 1.

**Figure 1**

**Figure 1**

In Figure 1, the circuitry inside the gray box represents an integrated current sense amplifier. Since the offset voltage is modeled by the external voltage source V_{offset}, the circuitry inside the gray box is assumed to have an offset voltage of 0V, i.e. with V_{A} = V_{B}** _{ }**we have V

_{out}= 0. The polarity of the offset voltage can be either positive or negative.

### Error From the Amplifier Offset Voltage

Consider the high-side current sensing diagram shown in Figure 2.

**Figure 2**

**Figure 2**

The amplifier actually senses the voltage between nodes A and B. Assuming a differential gain of A_{d} for the amplifier, we have:

\[V_{out} = A_d(V_A - V_B) = A_d(V_{offset} + V_{shunt})\]

If V_{offset} is comparable with V_{shunt}, the error can be significant. For example, with V_{offset} = V_{shunt}, the error is 100%, as calculated below:

\[Percentage~Error = \frac{V_{sensed} - V_{shunt}}{V_{shunt}} \times 100\% = \frac{(V_{offset} + V_{shunt}) - V_{shunt}}{V_{shunt}} \times 100\% = 100\%\]

The error is maximized when V_{shunt} is at its minimum. Hence, in order to find the worst-case error from the offset voltage, we should consider the minimum value of the load current I_{load}.

How can we reduce this error?

For a given load current range, we can either increase the shunt resistor value to have a larger V_{shunt} and/or use an amplifier with a smaller offset voltage.

However, it should be noted that increasing the shunt resistor value increases the power dissipated by the resistor. Additionally, higher wattage resistors are more expensive and require a larger board area.

### Determining the Value of the Shunt Resistor

As discussed above, there is a trade-off between the accuracy on the low end of the measurement range and the power dissipated by the shunt resistor. The accuracy can be maximized by choosing a large shunt resistor.

However, in a high-current precision application, the power dissipated by the current sense resistor can limit the maximum resistor value that can be used. In this case, we can choose the maximum value of R_{shunt} based on its maximum allowable power dissipation. If the maximum power dissipation that is budgeted for the shunt resistor is P_{max}, the maximum shunt value can be calculated by:

\[R_{shunt} \leq \frac{P_{max}}{I_{load,~max}^2}\]

The maximum value of R_{shunt} can also be limited by the amplifier output swing to the positive rail. The swing limitation of an amplifier depends on the supply voltage used to power the amplifier as well as the output stage of the amplifier.

Although the output of a rail-to-rail amplifier can get very close to the supply rails, it cannot actually reach them. Even with a rail-to-rail amplifier, the output swing might be limited to hundreds of millivolts away from supply rails, depending on the technology.

The output swing limitations of an amplifier also depend on the current level flowing through the output stage. The amplifier datasheet must be consulted to determine the available swing at the amplifier output.

If V_{out, max} is the maximum voltage at which the amplifier output stage remains in its linear region of operation, then the maximum shunt value can be calculated by:

\[R_{shunt} \leq \frac{V_{out,~max}}{I_{load,~max}A_d}\]

**Equation 1**

**Equation 1**

We can also find the lower limit of R_{shunt} by taking into account the output swing to the negative rail. If V_{out, min} is the minimum output voltage that guarantees the amplifier to be in its linear region of operation, we can find the minimum value of R_{shunt} as:

\[R_{shunt} \geq \frac{V_{out,~min}}{I_{load,~min}A_d}\]

*Equation 2*

*Equation 2*

#### Example Shunt Resistor Value Calculation

Assume that we need to monitor a load current between 40 mA and 1 A. The amplifier has a gain of 50, V_{out, max} = 4.9 V and V_{out, min }= 100 mV. Assuming that the application has not limited the power dissipated by the shunt resistor, what value of R_{shunt} do we need?

To minimize the error from the amplifier offset voltage, we should choose the maximum value for R_{shunt}. Applying Equation 1, we obtain:

\[R_{shunt} \leq \frac{V_{out,~max}}{I_{load,~max}A_d} = \frac{4.9}{1 \times 50} = 98~m\Omega\]

Using Equation 2, we can find the minimum value of Rshunt as:

\[R_{shunt} \geq \frac{V_{out,~min}}{I_{load,~min}A_d} = \frac{0.1}{0.04 \times 50} = 50~m\Omega\]

The largest standard value in this range should be chosen to minimize the offset error. After choosing the appropriate value of Rshunt, we can evaluate the offset error by applying the following equation:

\[Percentage~Error = \frac{(V_{offset} + V_{shunt}) - V_{shunt}}{V_{shunt}} \times 100\% = \frac{V_{offset}}{R_{shunt} \times I_{load,~min}} \times 100\%\]

To reduce the error to the desired level, we need to choose an amplifier with sufficiently low input offset voltage.

I would like to attract your attention to a subtle point here. Considering the output swing limitation of an amplifier, is there a contradiction in the offset voltage definition?

The offset voltage is defined as the differential voltage that must be applied between the input terminals of the amplifier to bring its output voltage to zero volts; however, the output of a single-supply amplifier cannot really swing to the ground potential.

These details can be important when measuring very small currents.

### Determining the Offset Voltage of a Single-Supply Amplifier

As discussed above, when the negative rail of an amplifier is tied to ground, its output can only get close to the ground potential. This is illustrated in Figure 3.

**Figure 3**

**Figure 3**

In this figure, the blue curve shows the amplifier output versus the differential voltage applied to the input.

For very small values of the differential input, the output voltage reaches V_{out, min}. Since the output doesn’t go below V_{out, min}, we cannot directly measure V_{offset}.

Instead, we can fit a line to the transfer function curve in the linear region of operation and consider the intersection of this line with the horizontal axis as the input offset voltage of the amplifier. You can imagine that if the amplifier was ideal, the dotted line would go through the origin.

We can now find the equation of the dotted line and determine the offset voltage. If the output is at V_{out1} and V_{out2} when the input is respectively at V_{diff1} and V_{diff2}, we can find V_{offset} as:

\[V_{offset} = V_{diff1} - \frac{(V_{diff2} - V_{diff1})}{(V_{out2} - V_{out1})} \times V_{out1}\]

### Can We Use a Single-Supply Amplifier for Measuring Very Small Currents?

Using a single-supply amplifier to measure near-zero load currents can introduce an unacceptable error due to the output swing limitations of the amplifier. This is shown in Figure 4.

**Figure 4. **Image courtesy of Texas Instruments.

**Figure 4.**Image courtesy of Texas Instruments.

To circumvent this problem, the reference design “0-1A, Single-Supply, Low-Side, Current Sensing Solution Reference Design” from Texas Instruments uses the LM7705 inverting charge pump to generate a -0.23V supply for the negative rail of the amplifier. This reference design is based on a discrete solution that uses an op-amp along with external gain-setting resistors as shown in Figure 5.

**Figure 5. **Image courtesy of Texas Instruments.

**Figure 5.**Image courtesy of Texas Instruments.

According to the mentioned reference, the negative supply must be at least 100 mV less than the system ground to make sure that the amplifier exhibits a linear output down to 0 V.

Another technique that enables measuring small currents is shown in Figure 6.

**Figure 6**

**Figure 6**

In this case, V_{ref} is used to add a DC value to the output voltage. The transfer function will change to:

\[V_{out} = A_d(V_A - V_B) + V_{ref}\]

This technique can also be used to sense bidirectional load currents (positive and negative load currents) by means of a single-supply amplifier. Depending on the desired current range, the appropriate value for V_{ref} should be chosen.

### Conclusion

The amplifier input offset voltage affects the accuracy on the low end of the measurement range. To reduce this error, we should maximize the shunt resistor value.

However, the shunt value can be limited by the budgeted power and the available swing at the amplifier output. Output swing limitations can also introduce error when measuring near-zero load currents.

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1 CommentIt took me a hot minute to figure out the current Iload was being multiplied by the gain in equations 1 & 2! I was like that result should be divided by the gain…until I realized it was. In case anyone else is confused, it’s because (Vout / iLoad) / Ag simplifies to Vout/(iLoad x Ad). Just a bit confusing at first glance.