Case 4: VOUT = –mVIN – b

The circuit shown in Figure 4–19 yields a solution for Case 4. The circuit equation is obtained by using superposition to calculate the response to each input. The individual responses to VIN and VREF are added to obtain Equation 4–56.

Figure 4–19. Schematic for Case 4: VOUT = –mVIN – b

 

Comparing terms in Equations 4–56 and 4–16 enables the extraction of m and b.

The design specifications for an example circuit are: VOUT = 1 V @ VIN = –0.1 V, VOUT = 5 V @ VIN =– 0.3 V, VREF = VCC = 5 V, RL = 10 kΩ, and 5% resistor tolerances. The simultaneous Equations 4–59 and 4–60, are written below.

From these equations we find that b = –1 and m = –20. Setting the magnitude of m equal to Equation 4–57 yields Equation 4–61.

Let RG1 = 1 kΩ, and then RF = 20 kΩ.

The final equation for this example is given in Equation 4–63.

The final circuit is shown in Figure 4–20 and the measured transfer curve for this circuit is shown in Figure 4–21.

Figure 4–20. Case 4 Example Circuit

Figure 4–21. Case 4 Example Circuit Measured Transfer Curve

The TLV247X was used to build the test circuit because of its wide dynamic range. The transfer curve plots very close to the theoretical curve, and this results from using a high performance op amp.

As long as the circuit works normally there are no problems handling the negative voltage input to the circuit because the inverting lead of the TLV247X is at a positive voltage. The positive op amp input lead is grounded, and normal op amp operation keeps the inverting op amp input lead at ground because of the assumption that the error voltage is zero. When VCC is powered down while there is a negative voltage on the inverting op amp input lead.

The most prudent solution is to connect the diode, D1, with its cathode on the inverting op amp input lead and its anode at ground. If a negative voltage gets on the inverting op amp input lead it is clamped to ground by the diode. Select the diode type as germanium or Schottky so the voltage drop across the diode is about 200 mV; this small voltage does not harm most op amp inputs. RG2 is split into two resistors (RG2A = RG2B = 51 kΩ) with a capacitor inserted at the junction of the two resistors. This places a power supply filter in series with VCC.