Reducing Voltage Offsets in Precision Amplifiers

Precision amplifiers have voltage offset errors caused in part by input bias current. This article analyzes this problem and provides solutions based on resistor networks, both discrete and integrated. The analysis shows that integrated resistors have better performance than more expensive discrete solutions. For precision electronics , amplifier circuits must meet the accuracy requirements of the design specifications. One problem faced when designing these amplifiers is the voltage offset caused by the current flowing into the amplifier input. In this article, we first analyze the cause of the offset and provide a corresponding solution based on integrated resistor networks.

Problem Analysis

Before we attempt to solve the problem, we need to understand where the problem comes from. Therefore, we first consider the simplified block diagram of an ideal op amp (Figure 1). The analysis of this circuit will be very familiar to many first-year students (assuming that the amplifier input current is zero):

Figure 1. Simplified block diagram of an ideal op amp.

Introducing finite input impedance can make the analysis result closer to the actual situation. At this time, the operational amplifier will have a certain input bias current. We add a current source to each input terminal of the ideal operational amplifier to simulate this effect and build a model (Figure 2).

Figure 2. Current source model of the ideal op amp in Figure 1, simulating input bias current.

To analyze the effect of each current source, assume VIN = 0V. Assuming that the VIN impedance is smaller than the other impedances in the equation, IBIAS+ will be bypassed to ground and will not have any effect. Since VIN = 0V, V- is also equal to 0V. In addition, since the potential across R1 is the same, it is 0V, so it can be ignored in the analysis. In this way, we can easily get the output offset (VOUT) caused by the input bias current (IBIAS-) and the feedback resistor (R2):

VOUT = IBIAS- x R2

Solving the problem

To improve the circuit, we add a resistor (R3 in Figure 3). We need to verify the effect of this additional resistor, which will introduce a negative bias voltage at the non-inverting input: IBIAS + x R3. Therefore, we can adjust R3 to eliminate the effect of the bias current on the inverting input. Of course, a reasonable choice is to adjust the bias current of the non-inverting and inverting inputs to be approximately equal.

When VIN = 0V, notice that we have superimposed a voltage in the circuit, and we can easily get VOUT, that is, the output voltage is equal to the voltage at the non-inverting terminal multiplied by the voltage gain, plus the offset due to the input leakage current of the inverting terminal. Because VIN = 0V, any voltage applied to the non-inverting terminal is the leakage current of this terminal and R3:

If R3 is equal to the parallel resistance of R1 and R2, the voltage generated by the input bias current will be cancelled. For precision applications that often use this technique, the resistors should be selected according to the following principles:

The R2/R1 ratio must be accurate to set the gain with high precision.

R3 and the parallel resistors R1/R2 need to be kept exactly equal to compensate for the error introduced by the input bias current.

These resistors should maintain the same temperature characteristics.

The precision op amp in Figure 3 can use either integrated or discrete resistors.

Figure 3. A compensation resistor (R3) is added to the circuit in Figure 2 to offset the effects of input bias current.

The integrated resistor MAX5421 (as an example) has a built-in 15kΩ resistor and operates from either +5V or -5V; a similar device, the MAX5431, has a built-in 57kΩ resistor and operates from either +15V or -15V. These devices not only include precision integrated resistors, but also allow switching between different resistors. When using resistors to set the gain of an op amp, the gain can be set between 1, 2, 4, and 8.

The data sheets of the devices show that they have a constant resistance at the node for resistor pairs with a ratio of 2, 4, and 8. At a ratio of 1, the node is just a low resistance. Therefore, the matching resistor should be equal to the wiper resistor for all ratios (Table 1). The resistor tolerances are shown in Table 2.

Note that these tolerances are guaranteed maximum values ​​over the full operating temperature range of -40°C to +85°C, thus ensuring high precision gain tolerances. Figure 4 shows a typical integrated resistor design (a precision amplifier). The main technical advantage of the MAX5421 or MAX5431 integrated resistor chips is the matching and consistent temperature characteristics between the resistors. The required system gain can be selected by electronically switching between the gain-setting resistors.

Figure 4. This precision amplifier consists of precision resistors (the MAX5421 IC) and a general-purpose rail-to-rail op amp (the MAX4493).

The absolute resistance values ​​of integrated resistors have large errors, but this has no effect in these circuits because the gain value is determined by the accuracy of the resistor ratio, which can be guaranteed to be within ±0.025%. If external resistors are used for matching, it is difficult to get the appropriate resistance value, but integrated resistors are easy to match. Integrated resistors can be factory trimmed to ensure that the gain setting resistors have consistent temperature characteristics.

The error in R1 and R2 also affects R3, which should be the same value as the parallel combination of R1 and R2. If R3 is not required in the system, the system cost can be reduced by using the digitally programmable precision resistor dividers MAX5420 and MAX5430. These devices have the same performance as the MAX5421 and MAX5431, but do not include matching resistors. For fixed-gain applications, the MAX5490, MAX5491, and MAX5492 resistor dividers can be used. This family of devices includes only one fixed-gain resistor pair and does not include matching resistors.

Discrete Resistor Solution

We now turn to the solution of setting the gain with discrete components and analyze that solution. The discrete resistor pair must not only have a ratiometric tolerance of ±0.025%, but must also maintain the rate of change within tolerance over the entire temperature range. In practice, this means that each resistor must have a tolerance of 0.0125%. The resistor data sheet usually gives the initial error and the temperature coefficient. From this we can calculate the maximum error over the entire temperature range. The example given below is based on very high precision discrete resistors with low temperature coefficients:

Initial error: 0.005%

Temperature coefficient: 2ppm

Operating temperature range: -40°C to +85°C

Therefore, the resistor tolerance over the entire operating range is:

To achieve the same gain accuracy as the integrated resistor op amp solution, the ultra-high precision resistors mentioned above must be used. Although such discrete resistors are available, they are very expensive, costing around a few dollars each. Even if the input offset matching requirements are reduced, the cost of discrete components is difficult to accept in order to achieve performance close to that of the integrated resistor solution.

The cost of a pair of resistors is much higher than that of devices such as the MAX542x or MAX543x, which integrate the resistors required for the four gain settings, as well as matching resistors and all the switches and logic required to switch the gain settings. Conclusion We have analyzed the voltage offset error caused by input bias current. By comparing the discrete and integrated resistor solutions, it can be seen that using integrated resistors can achieve better performance than the more expensive discrete solutions.