空白PIC12F1822的电流消耗以及60V的低功耗5V电源

我正在设计一个电池管理系统（BMS），它可能需要监视四个12V的SLA电池，总电压高达56 VDC，并且我需要向电路提供5个VDC，这将使用PIC和其他可以吸引多达5毫安的组件。我想把电池组的电流降到小于1毫安。所以我想使用降压开关稳压器。我已经模拟了使用PIC12F1822（或其他设备）驱动的P沟道MOSFET的设计，但是我需要制作一个5V的稳压器来提供它的功率在56伏包的高端。我用两个白色LED和249K电阻器构建了一个电路，它提供5.13伏直流电压和40V输入（140 UA），但是PIC就位，电压下降到大约2.35 VDC，这不足以驱动PMOS晶体管。PIC12F1822应该只画出30-60uA，内部时钟为32千赫。一个空白（擦除）设备默认为外部HS时钟，但我认为它将绘制没有时钟的最小电流。所以我想知道如果我把它编程为一个500千赫的MFiToSc，它在5 VDC上只画280个UA（大约与3V的L版本相同）。我计划使用25 kHz或更小的PWM，但我需要的占空比（根据模拟）是500毫秒，这将需要更高的时钟速率。在4 MHz的HFIFToC通常绘制700 UA。我不知道使用两个白色LED的并联调节器是否足够用于此目的。我想出了另一个选择，应该持有约4.3 VDC。这里是电路：[图像拒绝访问]：（i）可以在突发模式下运行Boost转换器，为BMS电路充电大电容，然后关闭Boost转换器，并在睡眠模式下运行BMS，并每秒唤醒一次，以节省功率。有没有其他设计方案？谢谢！



      以下为原文

    I am designing a battery management system (BMS) that may need to monitor four 12V SLA batteries with a total voltage as high as 56 VDC, and I need to provide 5 VDC to the circuitry, which will use a PIC and other components that may draw as much as 5 mA. I want to minimize the current draw from the battery pack to less than 1 mA. So I want to use a buck switching regulator. I have simulated a design that uses a P channel MOSFET driven by a PIC12F1822 (or other device), but I need to make a 5V regulator to supply its power on the high side of the 56 volt pack. I built a circuit using two white LEDs and a 249k resistor, which provides 5.13 VDC with 40V input (140 uA), but with the PIC in place the voltage drops to about 2.35 VDC, which is not enough to drive the PMOS transistor.

The PIC12F1822 is supposed to draw only 30-60 uA with a 32 kHz internal clock. A blank (erased) device defaults to an external HS clock, but I thought it would draw minimal current without a clock. So I wonder if it will draw less current if I program it for a MFINTOSC of 500 kHz where it should draw only 280 uA at 5 VDC (about the same as the L version at 3V). I plan to use PWM of 25 kHz or less, but the duty cycle I need (according to the simulation) is 500 nSec, which would require a higher clock rate. The HFINTOSC at 4 MHz draws typically 700 uA.

I don't know if a shunt regulator using two white LEDs is sufficient for this purpose. I came up with another option that should hold about 4.3 VDC. Here is the circuit:

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I might be able to run the boost converter in burst mode to charge a large capacitor for the BMS circuit, and then shut down the boost converter and run the BMS in sleep mode and wake up every second or so to save power. Any ideas for alternate designs? Thanks!