The circuit shown in Figure 1 is a robust and flexible loop-powered current transducer that converts the differential voltage output of a pressure sensor into a 4 mA to 20 mA current output.

The design is optimized for a variety of bridge-type voltage or current-driven pressure sensors, using only four active components and achieving a total unadjusted error of less than 1%. The loop supply voltage range is 12 V to 36 V.

The inputs of this circuit are ESD protected and provide voltage protection above the supply rails, making it ideal for industrial applications.

This design provides a complete 4 mA to 20 mA transmitter pressure sensor detection solution, with the entire circuit powered by the loop. There are three important circuit stages: sensor excitation driver, sensor output amplifier, and voltage-to-current converter.

The total current required by the circuit is 1.82 mA (maximum), as shown in Table 1. Therefore, the bridge can be used to drive pressure sensors with currents up to 2 mA without exceeding the maximum available loop current of 4 mA.

Sensor excitation driver

Either voltage drive or current drive is required, depending on the pressure sensor selected. This circuit uses half of the ADA4091-2 (U2A) and supports one of two options with different configurations selected via switch S1. Switch S1 provides one of the drive options.

Excitation: Voltage Drive Configuration

Figure 2 shows the voltage drive configuration of S1, which is located on the PCB marked VOLTAGEDRIVE (For the complete circuit layout and schematic, see the CN0289 Design Support Package: http://www.analog.com/CN0289-DesignSupport )

The voltage drive circuit is usually configured for a 10 V bridge drive voltage. In this mode, the minimum allowed bridge resistance is:

For bridge resistances below 5 kΩ, the drive voltage can be reduced to 5 V by removing R6 and using a buffer configuration.

Other values ​​of the driving voltage can be obtained by selecting the appropriate R6 as follows:

in:

Note that the loop voltage V _LOOP should be at least 0.2 V higher than the bridge drive voltage to give U2A enough headroom.

Excitation: Current Drive Configuration

By moving S1 to the position marked CURRENT DRIVE on the PCB, the circuit can be switched to the current drive configuration shown in Figure 3.

In current drive mode, the maximum allowed bridge drive current of 2 mA must be maintained. The circuit configuration is R4 = 2.49 kΩ and I _DRIVE = 2 mA. Use the following equation to select the R4 value to obtain a lower I _DRIVE value:

_{The driving voltage V DRIVE} can be calculated by the following formula :

The U2A power supply requires 0.2 V headroom, therefore:

In Figure 3, R _BRIDGE = 3 kΩ, I _DRIVE = 2 mA, V _DRIVE = 11 V, V _LOOP ≥ 11.2 V.

The ADA4091-2 op amp was chosen for this circuit because of its low power consumption (250 μA per amplifier), low offset voltage (250 μV), and rail-to-rail input and output characteristics.

Bridge output instrumentation amplifier and gain and offset resistor selection

The bridge output is filtered with a common-mode filter with a bandwidth of 39.6 kHz (4.02 kΩ, 1 nF) and a differential-mode filter with a bandwidth of 2 kHz (8.04 kΩ, 10 nF).

The AD8226 is an ideal instrumentation amplifier choice because of its low gain error (0.1%, Class B), low offset (58 μV at G = 50, Class B; 112 μV at G = 50, Class A), and excellent gain nonlinearity (75 ppm = 0.0075%) and rail-to-rail output characteristics.

The AD8226 instrumentation amplifier amplifies the 100 mV FS signal by a factor of 50 V to 5 V with gain setting resistor R3 = 1.008 kΩ. The relationship between gain G and R3 is as follows:

Where G = 50, R3 = 1008 Ω.

Output zero loop current I _LO = 4 mA:

Since the ratio of R10 to R8 is 100:1

Combining the last two equations we get:

When I _LO = 4 mA, the AD8226 output is 0 V; the offset resistor R12 can be calculated as follows:

If V _OUT = 5.00 V, then the output loop current I _LH = 20 mA, therefore:

The current flowing through R12 is:

The current flowing through R9 is:

The R9 value can be calculated by the following formula:

In actual use, the calculated values ​​of R3, R12 and R9 will not be provided as standard values, so the actual values ​​used in the circuit will have a fixed error. These errors can be calculated by the following equation. Gain, offset, and total error measurements resulting from resistors R3, R9, and R12, expressed in %FSR (where FSR = 16 mA):

The total error at zero-scale output (4 mA) is not affected by gain error.

The total error at full-scale output (20 mA) can be calculated as follows:

Full-scale error = gain error + offset error

In the actual circuit, the 0.1% resistor must be selected closest to the EIA standard, so the fixed gain and offset errors mentioned earlier are obtained. A combination of two 0.1% resistors can be used to get closer to the calculated value. For example, the following series combination of 0.1% resistors is very close to the calculated value:

• R3 = 1 kΩ + 8.06 Ω = 1008.06 Ω (calculated = 1008 Ω)
• R9 = 30.9 kΩ + 655 Ω = 31.555k Ω (calculated = 31.56 kΩ)
• R12 = 124 kΩ + 2.26 kΩ = 126.26 Ω (calculated = 126.25 Ω)

The error for these combinations is calculated as follows:

• Offset error = −0.008% FSR
• Gain error = +0.010% FSR
• Full scale error =

However, in some cases, the resistor supplier cannot even provide the standard 0.1% resistor value, so a substitution is required.

For example, the EVAL-CN0289-EB1Z evaluation board provides the following resistor values:

• R3 = 1000 Ω (calculated value = 1008 Ω)
• R9 = 31.6 kΩ (calculated = 31.56 kΩ)
• R12 = 124 kΩ (calculated = 126.25 kΩ)

Based on the values ​​provided by the evaluation board, the error caused by the resistor value can be calculated as follows:

• Offset error = +0.45% FSR
• Gain error = +0.66% FSR
• Full scale error = +1.11% FSR

The reference voltage

Use the ADR02 5 V reference voltage to set the drive voltage or current of the bridge and set the 4 mA zero-scale offset. Its initial accuracy is 0.1% (Grade A), 0.06% (Grade B), and has 10 μV pp voltage noise. Additionally, it will operate from supply voltages up to 36 V and consume only 1 mA (max), making it ideal for loop-powered applications.

The reference voltage

By forcing a current equal to the sum of the signal component (I ₉ ) and the offset component (I ₁₂ ). to flow through R10, an output current of 4 mA to 20 mA is produced. The current I ₁₀ = I ₉ + I ₁₂ ) produces a voltage across R10, which is applied to the sensing resistor R10 through U2B and Q1. The current flowing through R8 is 100 times the current flowing through R10. Therefore, the loop current I _LOOP can be calculated by the following formula:

Choose values ​​for R8 (10 Ω) and R10 (1 kΩ) that allow you to easily obtain 0.1% tolerance.

For the circuit to work properly, the circuit current II _CIRCUIT  is generated by the bipolar NPN transistor controlled by the U2B output and the gain should be at least 300 to minimize linearity errors. Its breakdown voltage should be at least 50 V.

Output transistor Q1 is a 50 V NPN power transistor dissipating 1.1 W at 25°C. The circuit has worst-case power dissipation with 20 mA output current into a 0 Ω loop load resistor and a V _CC supply of 36 V. Q1 power dissipation under these conditions is 0.68 W.

_{The supply voltage V LOOP} of the drive circuit board depends on the loop supply V _{LOOP_SUPPLY,} the loop load R7 and the loop current I _LOOP . The relationship between these values ​​is as follows:

For the circuit to operate properly, the supply voltage V _LOOP must be greater than 7 V to provide sufficient headroom for the ADR02 reference.

therefore,

For 20 mA maximum loop current and R7 = 250 Ω

The minimum loop supply voltage also depends on the bridge drive circuit configuration. In the voltage drive mode with V _DRIVE = 10 V, the supply voltage V _LOOP must be greater than 10.2 V so that U2A can maintain sufficient margin (see Figure 2).

In current drive mode, the supply voltage V _LOOP must be greater than 11.2 V so that U2A can maintain sufficient headroom (see Figure 3).

The loop supply voltage is limited to 36 V (maximum).

Error Analysis of Active Components

Table 2 and Table 3 respectively represent the A and B level maximum errors and rss errors of AD8226 and ADR02 caused by active components in the system. Note that the ADA4091-2 op amp is only available in one grade level.

Total circuit accuracy

A reasonable approximation of the total error due to resistor tolerance is to assume that each critical resistor contributes equally to the total error. The five critical resistors are R3, R8, R9, R10, and R12. The worst-case tolerance due to the 0.1% resistor can result in a maximum total resistor error of 0.5%. If the rss error is assumed, the total rss error is 0.1√5 = 0.224%.

Due to errors due to active devices (Class A), a 0.5% worst-case resistor tolerance error needs to be added to the previous worst-case error:

• Offset error = 0.29% +0.5% = 0.79%
• Gain error = 0.15% + 0.5% = 0.65%
• Full scale error = 0.44% + 0.5% = 0.94%

These errors assume ideal resistors, so the errors arise only from their tolerances.

Although the circuit is allowed to have a total error of 1% or less, if better accuracy is required, the circuit needs to have offset and gain adjustment capabilities. For a 4 mA output and zero-scale input, the offset is calibrated by adjusting R12, and then for a full-scale 100 mV input, full scale is adjusted by changing R9. These two adjustments are independent of each other; offset calibration is assumed to occur first.

The actual error data of the circuit is shown in Figure 4. The total output error (%FSR) is calculated by dividing the difference between the ideal and measured output current by the FSR (16 mA) and multiplying the result by 100.

Note that the error between the 0 mV and 1 mV inputs is caused by the AD8226 output stage saturation voltage, and the error range of the circuit under load is 20 mV to 100 mV. All rail-to-rail output stages are limited by their ability to reach the supply rails via saturation voltage (bipolar outputs) or on-resistance (CMOS outputs).

If errors due to the output saturation voltage cause problems, the input signal from the bridge can be biased by connecting an appropriate resistor between the +5 V reference voltage and one side of the bridge output.