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IX9915 датащи(PDF) 8 Page - IXYS Corporation |
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IX9915 датащи(HTML) 8 Page - IXYS Corporation |
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8 / 10 page ![]() INTEGRATED CIRCUITS DIVISION IX9915 8 www.ixysic.com R01 always into pin FB). This error causes the regulated output voltage to increase which increases the current through R1 by an amount equal to IIB, thereby restoring the current through R2 to its original value. Reducing the VREG error created by the input bias current to less than 1% is accomplished by setting the value of R1 using the following formula: Where: 2.2 Compensation The dominate pole of the error amplifier is around 13kHz. In a typical system with a low-bandwidth requirement, it doesn't need any external compensation. Frequency response of the system can be optimized for the specific application by placing a compensation network between the OC and FB pins of the IX9915. For designs with more critical bandwidth requirements, measurement of the loop response must be made and compensation adjusted as necessary. 2.3 Design Example A design example for the bleeder circuitry in LED lamp exhibits the detailed steps. In this example, it will target the predetermined voltage VLINE-TH=25V and maximum bleeding current IH-MAX=25mA. In order to flow the maximum bleeding current IH-MAX through the Darlington transistor: If taking RE=100: In fact, the components in the dashed rectangle function as a comparator, its gain: If taking R0=40k, the gain of the comparator is around 82dB. That is to say, once the error amplifier starts to regulate, the Darlington transistor will be shut off by this comparator. So, IOC can be ignored for affecting the predetermined voltage: Almost full power supply voltage will cross over R3, taking R3=100k to minimize its power consumption: Substituting: • IQ=75A, • I1=VREF / R2, • VLINE-TH=25V into formula (2): R 1 V REG 50 A ------------- 50 A 100 I IB MAX = V REG I HMAX – R E V BE + = (1) V REG I HMAX – R E V BE + = 25mA 100 1.5V + = 4V = A R 0 R 2 g m R 1 R 2 + ------------------------------ = R 2 R 1 R 2 + ------------------ V REF V REG ------------- 1.299V 4V ----------------- 0.325 == = g m 1S (typical) = V LINE-TH V REG I 1 I Q + R 3 + (2) P V rms 2 100k ------------------- = R 2 9.6k R 1 R 2 V REG V REF ------------- 1 – = R 1 20k |
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