Case 3: VOUT = –mVIN + b

The circuit shown in Figure 4–16 yields the transfer function desired for Case 3.

Figure 4–16. Schematic for Case 3: VOUT = –mVIN + b

The circuit equation is obtained with superposition.

 

Comparing terms between Equations 4–45 and 4–15 enables the extraction of m and b.

 

The design specifications for an example circuit are: VOUT = 1 V @ VIN = -0.1 V, VOUT = 6 V @ VIN = -1 V, VREF = VCC = 10 V, RL = 100 Ω, and 5% resistor tolerances. The supply voltage available for this circuit is 10 V, and this exceeds the maximum allowable supply voltage for the TLV247X. Also, this circuit must drive a back-terminated cable that looks like two 50-Ω resistors connected in series, thus the op amp must be able to drive 6/100 = 60 mA. The stringent op amp selection criteria limits the choice to relatively new Sop amps if ideal op amp equations are going to be used. The TLC07X has excellent singlesupply input performance coupled with high output current drive capability, so it is selected for this circuit. The simultaneous equations (Equations 4–49 and 4–50), are written below.

 

From these equations we find that b = 0.444 and m = –5.6.

 

Let RG = 10 kΩ, and then RF = 56.6 kΩ, which is not a standard 5% value, hence RF is selected as 56 kΩ.

 

The final equation for the example is given below

 

Select R1 = 2 kΩ and R2 = 295.28 kΩ. Since 295.28 kΩ is not a standard 5% resistor value, R1 is selected as 300 kΩ. The difference between the selected and calculated value of R1 has a nearly insignificant effect on b. The final circuit is shown in Figure 4–17, and the measured transfer curve for this circuit is shown in Figure 4–18.

 

Figure 4–17. Case 3 Example Circuit

Figure 4–18. Case 3 Example Circuit Measured Transfer Curve

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 TLC07X is at a positive voltage. The positive op amp input lead is at a voltage of approximately 65 mV, and normal op amp operation keeps the inverting op amp input lead at the same voltage because of the assumption that the error voltage is zero. When VCC is powered down while there is a negative voltage on the input circuit, most of the negative voltage appears 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. As a further precaution, RG can be split into two resistors with the diode inserted at the junction of the two resistors. This places a current limiting resistor between the diode and the inverting op amp input lead.