工业控制 | 能源技术 | 汽车电子 | 通信网络 | 安防监控 | 智能电网 | 移动手持 | 无线技术 | 家用电器 | 数字广播 | 消费电子 | 应用软件 | 其他方案

电路设计->电源电路图->其他电源电路图->直流电源稳压电路

直流电源稳压电路

作者:dolphin时间:2012-07-30

电路能力测试-2:直流电源稳压电路

Circuit Knowledge Test-2: DC-DC LDO Regulator

2006年马槟厂设计工程师试题-2 - 爱碎碎念的老翁 - weng3309 的博客

  1. What will be the DC voltage at pin-1 of IC LM7805?

  1. What will be the supply current for DV 5V regulated output on this drawing?

  1. If 78M05 with DPAK package dissipation 60°C/W, then what surface temperature will be for IC 78M05 at 25°C room temperature?

  1. Same condition as the Question-3: If DC 5V output will supply 150 mA for all circuits, what surface temperature will be for IC 78M05 at 25°C room temperature?

  1. What minimum input voltage will be allowed for 5V stabilized output if the minimum dropout voltage for LM78M05 is 1.5 Volt?

  1. What will be the suitable resistance value and power rating of RP92 RP93 for lowest LM78M05 surface temperature theoretically at 100 mA supply current and 10 volt low line input?

试题解答

Circuit Knowledge Test-2: DC-DC LDO Regulator

ANS-1:

Vin = 12 – 34 * 0.084 = 12 – 2.86 = 9.14 Volt.

ANS-2:

Same as the input current 84 mA.

ANS-3:

Power loss = (9.14 – 5.0) * 0.084 = 0.35 W. then Tht = 0.35 * 60 = 21 °C

Final junction temperature Tj = 25 + 21 = 46 °C

ANS-4:

Vin = 12 – 34 * 0.15 = 6.9 Volt.

Power loss = (6.9 – 5.0) * 0.15 = 0.285 W. then Tht = 0.285 * 60 = 17 °C

Final junction temperature Tj = 25 + 17 = 42 °C

ANS-5:

Vmin = 34 * 0.084 + 1.5 + 5.0 = 9.36 Volt.

ANS-6:

Vdrop = 10 – 5.0 – 1.5 = 3.5 then R = 3.5/0.1 = 35 Ohm.

RP92 = RP93 = 68 Ohm.



评论

技术专区