| Datenblatt-Suchmaschine für elektronische Bauteile |
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ML4819CS Datenblatt(PDF) 11 Page - Micro Linear Corporation |
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ML4819CS Datenblatt(HTML) 11 Page - Micro Linear Corporation |
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11 / 15 page ![]() ML4819 11 The value of R9 (pin 3) depends on the selection of R2 (pin 6). R V ImA k IN MAX PEAK SINE PEAK 2 260 1 414 072 510 == × = () () . . Ω (12) R VR V k k CLAMP IN MIN PEAK 9 2 4 8 510 80 1 414 22 > × = × × = () . . Ω Ω (13) Choose R9 = 27k W The peak of the inductor current can be found approximately by: I P V A LPEAK OUT IN MIN RMS = × = × = 1 414 1 414 200 90 314 . . . () (14) Next select NC, which depends on the maximum switch current. Assume 4A for this example. NC is 80 turns. R VN I CLAMP C LPEAK 11 49 80 4 100 = × = × =Ω . (15) Where R11 is the sense resistor, and VCLAMP is the current clamp at the inverting input of the PWM comparator. This clamp is internally set to 5V. In actual application it is a good idea to assume a value less than 5V to avoid unwanted current limiting action due to component tolerances. In this application VCLAMP was chosen as 4.8V. Having calculated R11 the value SPWM and of R18 can now be calculated: S V mH Vs PWM = − ×= µ 380 20 2 100 80 0 225 ./ R R AS R C R k knF k SC PWM T T 18 9 18 6 25 2 5 28 8 0 7 0 225 10 14 1 30 = × ×× × = × ×× × × ≅ . .. .( . ) Ω Ω (16) Choose R18 = 33k W The following values were used in the calculation: R9 = 27kW ASC = 0.7 RT = 14kW CT = 1nF VOLTAGE REGULATION COMPONENTS The values of the voltage regulation loop components are calculated based on the operating output voltage. Note that voltage safety regulations require the use of sense resistors that have adequate voltage rating. As a rule of thumb if 1/4W through-hole resistors are used, two of them should be put in series. The input bias current of the error amplifier is approximately 0.5µA, therefore the current available from the voltage sense resistors should be significantly higher than this value. Since two 1/4W resistors have to be used the total power rating is 1/2W. The operating power is set to be 0.4W then with 380V output voltage the value can be calculated as follows: RV W k 5 2 380 0 4 360 == () / . Ω (17) Choose two 178k W, 1% connected in series. Then R6 can be calculated using the formula below: R VR VV Vk VV k REF B REF 6 5 5 356 380 5 4 747 = × − = × − = Ω Ω . (18) Choose 4.75k W, 1%. One more critical component in the voltage regulation loop is the feedback capacitor for the error amplifier. The voltage loop bandwidth should be set such that it rejects the 120Hz ripple which is present at the output. If this ripple is not adequately attenuated it will cause distortion on the input current waveform. Typical bandwidths range anywhere from a few Hertz to 15Hz. The main compromise is between transient response and distortion. The feedback capacitor can be calculated using the following formula: C RBW C kHz F 8 5 8 1 3 142 1 3 142 356 2 044 = ×× = ×× =µ . . . Ω (19) |
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