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WEBENCH® Tools/VFC320: voltage to frequency convertor

Part Number: VFC320
Other Parts Discussed in Thread: OPA197

Tool/software: WEBENCH® Design Tools

HOW does the circuit in VFC 320, fig. 14, VFC operate with bipolar input, how is the feed back on the negative cycle.

  • Hello James,

    It appears from the VFC320 datasheet Electrical Characteristics table that the input can be positive > 0, negative -10 V to 0 V, or a differential input voltage (e1, e2). In either case, the internal operation of the VFC320 remains essentially the same and follows the sequence of events described in the datasheet's Theory of Operation section. 

    The case when is e1 is a positive voltage being applied to the input resistor of the input integrator and e2 is at ground is well described in the Theory of Operation. However, when the e1 is at ground and e2 is being driven by a 0 to -10 V input the operation details are not provided. The only difference between the two cases is that when e2 is 0 V to -10 V, the integrator acts as a non-inverting integrator. When e2 is negative it appears as a common-mode voltage at the non-inverting input and the inverting input, the op amp's virtual ground, will track that e2 voltage. Since e2 is negative the output will integrate negative just as it does for e1 positive case.

    The VFC320 datasheet FIGURE 4, Functional Block Diagram, states in the diagram block "VIN: for Differental Input Voltages use e1 and e2." Even though differential voltage can be accommodated it appears e1 would have to be positive relative to e2, for the circuit to operate as described in the Theory of Operation section.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • Hello Thomas,

    Thanks for you reply.

    I could understand the theory part of the tracking, what I cannot understand for is:-

    R1=11K(eleven) and R2=100K, R3=40.2K & R4=8.66K

    for Positive input lets assume +100mV, (e1, 3510B +ve input) then at the junction of R3 and R4=+177mV (considering gain as 10).

    if Negative input is same in Magnitude and opposite in Polarity i.e. -100mV, then the junction for R3 and R4 should be at -177mV, am I wrong and missing something, I am trying to calculate and get it correct.

    Kindly guide me where I am going wrong.

    Regards James

  • Hello James,

    The legacy 3510 op amp being applied in Figure 14 is intended to provide an absolute value (precision full wave rectifier) function. A voltage applied to the 3510 non-inverting input, positive or negative in polarity, but having the same magnitude would produce the same positive output voltage. For a +100 mv, or -100 mV input voltage, the intended output voltage would be the same +1.0 V. That voltage would be reduced to +177 mV at the junction of R3 and R4 in the Figure 14 circuit.

    When I evaluate the circuit with +100 mV applied, the output voltage at the 3510 output is +1 V, and +177 mV at R3, R4. This all works because D1 becomes forward biased under these conditions and the feedback loop is closed as needed. However, when the input voltage polarity is reversed and -100 mV is applied to the op amp input, D1 becomes reverse biased and the feedback loop is essentially opened. This no longer results in the needed absolute value behavior.

    I checked the circuit behavior with some TINA sims and the op amp output is slammed against the negative swing rail. When I compare it to the many absolute value/ precision rectifier circuit seen on line.It seems that something is missing from the Fig. 14 circuit. I thought that possibly the addition of the Q1 sign bit circuit might be a contributor in that circuit's operation, but simulations verified it doesn't fix the problems.

    My thought is to use one of the proven absolute value circuits and a modern op amp in place of the 3510. The 3510 hasn't been available from TI for many years. There is a classic absolute value circuit paper from TI/Burr Brown that proves some useful absolute value circuits:

    A modern precision op amp such as the OPA197 should work well in one of these circuits.

    Regards, Thomas

    Precision Amplifiers Applications Engineering