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CS5112EDWF24 数据表(PDF) 8 Page - Cherry Semiconductor Corporation |
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CS5112EDWF24 数据表(HTML) 8 Page - Cherry Semiconductor Corporation |
8 / 10 page 8 Application Notes: continued Step 4 Determine the maximum on time at the minimum oscilla- tor frequency and VIN. For discontinuous operation, all of the stored energy in the inductor is transferred to the load prior to the next cycle. Since the current through the inductor cannot change instantaneously and the induc- tance is constant, a volt-second balance exists between the on time and off time. The voltage across the inductor dur- ing the on cycle is VIN and the voltage across the inductor during the off cycle is VOUT - VIN. Therefore: VINton = (VOUT -VIN)toff (4a) where the maximum on time is: ton(max) Å . (4b) Step 5 Calculate the maximum inductance allowed for discontin- uous operation: L(max) = (5) where h = efficiency. Usually h = 0.75 is a good starting point. The ICÕs power dissipation should be calculated after the peak current has been determined in Step 6. If the efficiency is less than originally assumed, decrease the efficiency and recalculate the maximum inductance and peak current. Step 6 Determine the peak inductor current at the minimum inductance, minimum VIN and maximum on time to make sure the inductor current doesnÕt exceed 1.4A. Ipk = (6) Step 7 Determine the minimum output capacitance and maxi- mum ESR based on the allowable output voltage ripple. COUT(min) = (7a) ESR(min) = (7b) In practice, it is normally necessary to use a larger capaci- tance value to obtain a low ESR. By placing capacitors in parallel, the equivalent ESR can be reduced. Step 8 Compensate the feedback loop to guarantee stability under all operating conditions. To do this, we calculate the modulator gain and the feedback resistor network attenu- ation and set the gain of the error amplifier so that the overall loop gain is 0dB at the crossover frequency, fCO. In addition, the gain slope should be -20dB/decade at the crossover frequency. The low frequency gain of the modulator (i.e. error ampli- fier output to output voltage) is: = Ã , (8a) where Ipk(max) = = =2.3A. The VOUT/VEA transfer function has a pole at: fp = 1/(¹RLoadCOUT) , (8b) and a zero due to the output capacitorÕs ESR at: fz = 1/(2¹ESR COUT). (8c) Since the error amplifier reference voltage is 1.25V, the output voltage must be divided down or attenuated before being applied to the input of the error amplifier. The feedback resistor divider attenuation is: . The error amplifier in the CS5112 is an operational transcon- ductance amplifier (OTA), with a gain given by: GOTA = gmZOUT (8d) where: gm = . (8e) For the CS5112, gm = 2700µA/V typical. One possible error amplifier compensation scheme is shown in Figure 9. This gives the error amplifier a gain plot as shown in Figure 10. For the error amplifier gain shown in Figure 10, a low fre- quency pole is generated by the error amplifier output impedance and C1. This is shown by the line AB with a - 20dB/decade slope in Figure 12. The slope changes to zero at point B due to the zero at: fz = 1/(2¹R4C1). (8f) Figure 9. RC network used to compensate the error amplifier (OTA). VOUT VFB1 VFB2 M U X SELECT Error Amplifier 1.25V + Ð C1 R4 C2 R1 R2 R3 ÆIOUT ÆVIN 1.25V VOUT (2.4V)/(7) 150m½ VEA(max)/GCSA RS RLoad Lf 2 Ipk(max) VEA(max) ÆVOUT ÆVEA ÆVripple Ipk Ipk 8fÆVripple VIN(min) ton(max) L(min) fSW(min) VIN2(min) ton2(max) 2 POUT/h ] 1 fSW(min) [] 1 - VIN(min) VOUT(max) [ |
类似零件编号 - CS5112EDWF24 |
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类似说明 - CS5112EDWF24 |
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