Select resistors to minimize grounding load current source error

Operational amplifiers are commonly used to generate high performance current sources in a variety of applications, including industrial process control, scientific instrumentation, and medical devices. The "Single Amplifier Current Source" published in Analog Dialogue, Volume 1, Number 1, 1967, describes several current source circuits that provide constant current through a floating load or a grounded load. In industrial applications such as pressure transmitters and gas detectors, these circuits are widely used to provide currents from 4-mA to 20-mA or from 0-mA to 20-mA.

The improved Howland current source shown in Figure 1 is very popular because it can drive grounded loads. Transistors that allow relatively high currents can be replaced with MOSFETs to achieve higher currents. For low-cost, low-current applications, transistors can be removed, as described in Analog Dialogue, Vol. 43, No. 3, 2009, "The Heart of Precision Current Sources: Differential Amplifiers".

The accuracy of this current source depends on the amplifier and resistor. This article describes how to select an external resistor to minimize the error.

Figure 1. Improved Howland current source drives the grounded load.

Figure 1. Improved Howland current source drives the grounded load.

By analyzing the improved Howland current source, the transfer function can be derived:

By analyzing the improved Howland current source, the transfer function can be derived: (1)

Tip 1: Set R2 + R5 = R4

In Equation 1, the load resistance affects the output current, but if we set R1 = R3 and R2 + R5 = R4, the equation is reduced to:

In Equation 1, the load resistance affects the output current, but if we set R1 = R3 and R2 + R5 = R4, the equation is reduced to: (2)

The output current here is only a function of R3, R4 and R5. If there is an ideal amplifier, the resistor tolerance will determine the accuracy of the output current.

Tip 2: Set RL = n &TImes; R5

To reduce the total number of resistors in the device bank, set R1 = R2 = R3 = R4. Now Equation 1 is simplified to:

To reduce the total number of resistors in the device bank, set R1 = R2 = R3 = R4. Now Equation 1 is simplified to: (3)

If R5 = RL, the formula is further simplified to:

If R5 = RL, the formula is further simplified to: (4)

The output current here depends only on the resistor R5.

In some cases, the input signal may need to be attenuated. For example, with a 10 V input signal and R5 = 100 Ω, the output current is 100 mA. To get an output current of 20 mA, set R1 = R3 = 5R2 = 5R4. Now Equation 1 is simplified to:

Now Equation 1 is simplified to:

If RL = 5R5 = 500 Ω, then:

If RL = 5R5 = 500 Ω, then: (5)

Tip 3: The value of R1/R2/R3/R4 is large, which can improve current accuracy.

In most cases, R1 = R2 = R3 = R4, but RL ≠ R5, so the output current is shown in Equation 3. For example, in the case of R5 = 100 Ω and RL = 500 Ω, Figure 2 shows the relationship between resistor R1 and current accuracy. To achieve 0.5% current accuracy, R1 must be at least 40 kΩ.

Figure 2. Relationship between R1 and output current accuracy.

Figure 2. Relationship between R1 and output current accuracy.

Introduction


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