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Calculate the Output High Voltage Error for a Current Sense Amplifier

Introduction
When using a current sense amplifier, it is important to know how much error is present in the output so an accurate interpretation of the data can be obtained.

Output error can be measured by the Output High Voltage parameter of the amplifier which is defined as the difference in voltage between a VRS- pin and an Output pin. The typical error that can be expected for any current sense amp will be defined in the Electrical Characteristics (EC) table found on its datasheet. Although this rate is what one can expect for most cases, it is still important to know how this parameter is calculated based upon the amplifier's circuit design. To do this, we will demonstrate how to do an error budget analysis calculation.

Calculation demonstration
For the purposes of this demonstration, we are using the Touchstone TS1100 current-sense amplifier although the process can be applied to any current sense amplifier. In this case, the amplifier has a voltage power supply of 3.6V and a gain of 100V/V. First, the full scale voltage across the RSENSE resistor must be found. The equation for the full scale VSENSE is:

VSENSE(FS) = VSUPPLY /GAIN [1]

Substituting the 3.6V power supply and the 100V/V gain option into equation [1], the resulting full scale VSENSE is shown below.

VSENSE(FS) = 0.036 V

Next, the gain across the amplifier's entire temperature range (-40°C to +105°C) must be found. The maximum gain error across the entire temperature range must be calculated. The maximum Gain Error (GE) across temperature can also found in the EC table on the datasheet, and shown below.

GE = ±0.6%

Considering the plus-minus error specification, a maximum and minimum swing for the gain across temperature is found. The inequality is defined below.

GAIN⋅(1-GE) ≤ GAIN ≤ GAIN⋅(1+GE) [2]

Solving the inequality results in the amplifier exhibiting the following gain over the entire temperature range.

99.4V/V ≤ GAIN ≤ 101.6V/V

Now, the voltage output can be determined by calculating the effects from the input offset voltage over temperature. The voltage output equation to solve for is:

VOUT = [GAIN⋅(1±GE)⋅VSENSE(FS) ] ± [GAIN(1±GE)⋅VOS ] [3]

The Input Offset Voltage parameter is also listed in the EC table. For this demonstration, the maximum and minimum input offset voltage over the entire temperature range has been accounted for. Note that over the entire temperature range the input offset voltage will vary from a negative value to a positive value:

-200μV ≤ VOS ≤ +200μV

The voltage output equation has been rewritten to define the absolute maximum and minimum voltage outputs:

VOUT(MAX) = [GAIN(MAX) ⋅VSENSE(FS) ] + (GAIN(MAX) ⋅VOS(MAX) ) [4]

VOUT(MIN) = [GAIN(MIN) ⋅ VSENSE(FS) ] + (GAIN (MIN) ⋅VOS(MIN) ) [5]

Next is the absolute minimum voltage output equation. Due to the negative value of the absolute minimum input offset voltage, the equation was rewritten to accommodate the negative value so it would not drop out. Substituting the previously defined values into equations [4] and [5], the absolute voltage output is calculated to have the following swing over the entire temperature range:

3.558V ≤ VOUT ≤ 3.642V

Please note, a 3.6V power supply was selected for this demonstration and obtaining a voltage output higher than the power supply is unrealistic. Therefore, the absolute maximum voltage output value that was calculated is disregarded. The 3.6V power supply is inserted into the voltage output inequality as the maximum possible value. The actual absolute voltage output across temperature is defined below.

3.558V ≤ VOUT ≤ 3.6V

Lastly, the Output High Voltage parameter (VOH ) is calculated. The parameter is also defined in the EC table.

VOH = VRS- – VOUT [6]

Note that the voltage on the load side of the RSENSE resistor is equal to the voltage on the supply side of RSENSE , and subsequently equal to the power supply voltage. Solving equation [6], the selected amplifier, with an input offset voltage of 200μV and a gain error of ±0.6% over temperature, will exhibit the following Output High Voltage range:

0V ≤ VOH ≤ 0.042V

This calculation was performed specifically using the 100V/V gain option, a 3.6V power supply and the maximum possible error specifications. If the same calculations are carried out with the same errors and 3.6V power supply, but varying the gain option, the ranges over temperature can be calculated for the Output High Voltage, as shown below.

Conclusion
The Output High Voltage (VOH ) parameter of the amplifier can be compared with the electrical characteristics table on the datasheet. When the two are compared, the TS1100 current sense amp shows a maximum possible value of 0.2V. This calculation walk-through accounted for the maximum possible error specifications across temperature. Thus, the Output High Voltage for all of the gain options of the amplifier is still well below the maximum spec listed in the datasheet. This demonstration shows that under high error scenarios this amplifier performed at a very high level.

7 comments on “Calculate the Output High Voltage Error for a Current Sense Amplifier

  1. Andy Pienkowski
    April 25, 2013

    A diagram for this article would not go amiss.  It looks interesting but it might take a little longer than necessary to work out your arguments because you don't have any annotated diagrams at all.

  2. Davidled
    April 25, 2013

    All equations are much formula. If interface circuit with Touchstone TS1100 current-sensing amplifier is described, it could be more beneficial for understanding equation. I am wondering what type criteria are used to implement this type of formulation. Formulation may consider any factor related to PCB board level including noise.

  3. Daniel Honniball
    April 26, 2013

    Thank you for the suggestion Andy. I will take note of that, and try to incorporate a diagram next time. 

  4. Daniel Honniball
    April 26, 2013

    DaeJ,

    Thank you for your comments. Although a circuit schematic is drawn in the CSA's datasheet, I should have shown the typical application circuit for the CSA. A power supply and a load are assumed for the interfacing circuits.  The only factors considered for the write up, were the errors of the CSA, and the effects temperature has on those errors. Errors exhibited from the board layout and interfacing circuits, are ignored.

  5. Andy Pienkowski
    April 27, 2013

    The time available to digest an articicle such as this is almost non-existent during the working day so that it is important to help the reader in every way possible; diagrams can help considerably.

  6. Brad Albing
    April 27, 2013

    Hi Andy – I'll work with Daniel – and everyone else – next time to make sure we get a schematic included to make processing the info go a bit more quickly.

  7. Davidled
    April 27, 2013

     I am curious how all equations are derived, without any interface circuit.  May be, I am missing something else. When checking TS1100 part, there is no similar information regarding on any equation as indicated in blog. Secondly, without considering any noise with high frequency interference in the board level or other module, it sound like this math formulation is more closely in simulation level.

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