Current Source Built with a Negative Resistance

Posted on 13-March-2011 by mathscinotes

Introduction

I have seen quite a few schematics lately that are using an operational amplifier (opamp) configured as a negative resistor. I thought it would be interesting to analyze this circuit and show how it can provide an economical solution to a common sensor interface scenario.

Background

Figure 1 illustrates a common sensor interface scenario. This scenario consists of:

  • Sensor
    The sensor has a resistance characteristic that varies according to some external parameter, e.g. temperature, ambient light level, gas concentration.
  • Current Source
    A current source drive can be desirable for a number of reasons: (1) The sensor may be calibrated for a specific current level, (2) The sensor may have a linear response for a given current drive, (3) The sensor output voltage range needs to be restricted in some way.
  • Amplifier
    The amplifier is often needed to provide isolation from the monitoring circuitry and to scale the output for further signal processing.
Figure 1: Common Sensor Interface Scenario.

Figure 1: Common Sensor Interface Scenario.

Negative Resistance Opamp Circuit

Figure 2 shows an operational amplifier connected as a negative resistance.

Figure 2: Negative Resistance Circuit.

Figure 2: Negative Resistance Circuit.

Equation 1 shows the derivation of the negative resistance equation.

Eq. 1 {{v}_{I}}={{v}_{O}}+i\cdot {{R}_{F}}
{{v}_{I}}={{v}_{O}}\cdot \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}
{{v}_{I}}={{v}_{I}}\cdot \frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}}+i\cdot {{R}_{F}}
-{{v}_{I}}\cdot \frac{{{R}_{2}}}{{{R}_{1}}}=i\cdot {{R}_{F}}\Rightarrow {{Z}_{IN}}\triangleq \frac{{{v}_{I}}}{i}=-{{R}_{F}}\cdot \frac{{{R}_{1}}}{{{R}_{2}}}

Equation 2 shows the derivation of the voltage gain equation.

Eq. 2 {{v}_{I}}={{v}_{O}}\cdot \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}\Rightarrow {{A}_{v}}\triangleq \frac{{{v}_{O}}}{{{v}_{I}}}=\frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}}

Introduction

My application example is shown in Figure 4.

Figure 3: Negative Resistance Example Application.

Figure 3: Negative Resistance Example Application.


The basic idea here is to use negative input resistance of the opamp circuit to cancel out the resistance of RS. Figure 4 illustrates this source transformation.
Figure 4: Source Transformation.

Figure 4: Source Transformation.


The parameters of this case are the following:

  • V_{CC} = 10 V
  • I = 1\text{mA}
  • A_v = 10

Figure 4 shows my solution in Mathcad.

Figure 5: Example Solution in Mathcad.

Figure 5: Example Solution in Mathcad.

Conclusion

I worked through some basic design equations for the application of a negative resistance circuit to a sensor interface example. I have been seeing this circuit quite a bit lately and it is an interesting application of basic linear circuit analysis.

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