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LTC6990CS6PBF Datasheet(PDF) 17 Page - Linear Technology |
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LTC6990CS6PBF Datasheet(HTML) 17 Page - Linear Technology |
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17 / 28 page  LTC6990 17 6990f Figure 12. VCO Transfer Function VCTRL (V) 1 100 80 40 20 60 0 2 6990 F12 4 3 APPLICATIONS INFORMATION The accuracy of KVCO does depend on VSET because the output frequency is controlled by the ratio of VCTRL to VSET. The frequency error (in Hertz) due to ΔVSET is ap- proximated by: ΔfOUT ≅KVCO •VCTRL • ΔVSET VSET As the equation indicates, the potential for error in output frequency due to VSET error increases with KVCO and is at its largest when VCTRL is at its maximum. Recall that when VCTRL is at its maximum, the output frequency is at its minimum. With the maximum absolute frequency error (in Hertz) occurring at the lowest output frequency, the relative frequency error (in percent) can be significant. VSET is nominally 1.0V with a maximum error of ±30mV for at most a ±3% error term. However, this ±3% potential error term is multiplied by both VCTRL and KVCO. Wide fre- quency range applications (high KVCO) can have frequency errors greater than ±50% at the highest VCTRL voltage (lowest fOUT). For this reason the simple, two resistor VCO circuit must be used with caution for applications where the frequency range is greater than 4:1. Restricting the range to 4:1 typically keeps the frequency error due to VSET variation below 10%. For wide frequency range applications, the non-inverting VCO circuit shown in Figure 13 is preferred because the maximum frequency error occurs when the frequency is highest, keeping the relative error (in percent) much smaller. Example: Design a VCO with the Following Parameters fOUT(MAX) = 100kHz at VCTRL(MIN) = 1V fOUT(MIN) = 10kHz at VCTRL(MAX) = 4V Step 1: Select the NDIV Value First, choose an NDIV that meets the requirements of Equation (3a). 6.25 ≤ NDIV ≤ 10 The application’s desired frequency range is 10:1, which isn’t always possible. However, in this case NDIV = 8 meets both requirements of Equation (3). Step 2: Calculate KVCO and f(0V) Next, calculate the intermediate values KVCO and f(0V) using Equations (3b) and (3c). KVCO = 100kHz − 10kHz 4V − 1V = 30kHz/V f(0V) =100kHz + 30kHz/V •1V =130kHz Step 3: Calculate and Select RVCO The next step is to use Equation (3d) to calculate the cor- rect value for RVCO. RVCO = 1MHz • 50k 8•1V • 30kHz/V = 208.333k Select RVCO = 210k. Step 4: Calculate and Select RSET The final step is to calculate the correct value for RSET using Equation (3e). RSET = 1MHz • 50k 8 • 130kHz − 1V • 30kHz/V ( ) = 62.5k Select RSET = 61.9k In this design example, with its wide 10:1 frequency range, the potential output frequency error due to VSET error alone ranges from less than ±1% when VCTRL is at its minimum up to ±36% when VCTRL is at its maximum. This error must be accounted for in the system design. |
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