Hot!Current draw of blank PIC12F1822, and low power 5V supply from 60V

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2018/02/11 01:03:52 (permalink)
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Current draw of blank PIC12F1822, and low power 5V supply from 60V

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:
 
[Access denied for image :( ]
 
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!

 
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    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 01:04:36 (permalink)
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    #2
    NKurzman
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 01:08:29 (permalink)
    +1 (1)
    Why not use a low quiescent, current high Efficiency dedicated regulator? TI has tools that will do most of the work for you.
    #3
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 03:08:37 (permalink)
    +1 (1)
    I did find some that might do the job:
     
    https://www.mouser.com/ds/2/256/MAX5033-104528.pdf 76Vin 5Vout 270uA quiescent 10uA shutdown $3.50
     
    https://www.mouser.com/ds/2/256/MAX15062-257927.pdf 60Vin 5Vout 65% efficiency at 5 mA out 48Vin $2.50
     
    http://www.ti.com/lit/ds/symlink/lm5165-q1.pdf 65Vin, 5Vout, 205 uA active, 5uA shutdown, 75% efficiency at 5 mA out $3.25
     
    I was able to get a simulation with less than 1 mA draw from the 60V rail for 5V 5mA output by using a 15mH inductor and 2 kHz PWM with 12uSec duty cycle. Only 50% efficiency. I'd also have to provide some means of reading the output voltage to adjust the PWM.
     
    Perhaps best overall efficiency would be to run the regulator in bursts to charge a capacitor that can power the circuit in sleep mode between samples.
     
    This BMS will be used on 12V, 12 A-h SLA batteries, so a 1mA constant draw would drain 10% of charge in 1000 hours, or over a month. That's probably similar to self-discharge.
     
    Thanks.

     
    #4
    Antipodean
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 06:29:46 (permalink)
    +1 (1)
    Are those regulators linear ones or switchmode ones?
     
    My take on this would be to use an automotive specification switching regulator to drop from 60V to 5V. Linear Technology have some and it is a case of plug numbers onto the formula they give, and use their suggested board layout and you will have a reliable regulator with minimal components. You should be able to get much higher efficiency than you are quoting above.
     
    There should also be some National Semiconductor available from TI (the ones with LM part numbers) with similar ability and efficiency. 

    Do not use my alias in your message body when replying, your message will disappear ...

    Alan
    #5
    NKurzman
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 09:12:19 (permalink)
    +1 (1)
    I f you go to TIs web site they have a selection tool. That will take you in too their web based design and simulation tools. You may be able to find aunt tweek a better match. A discontinued mode pwm regulator may be a good choice.
    #6
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 13:25:26 (permalink)
    +1 (1)
    I'll probably use the MAX5033. It comes in an SOIC-8 and PDIP-8 package which will be a lot easier to solder on the prototype. The others come in tiny 2mm-3mm square packages with 0.5mm lead spacing. It seems crazy to put an IC in such a tiny package when the ancillary components are 10 times bigger.
     
    These are all switching regulators, and most of them use resonant ZV PWM switching techniques along with pulse-skipping (PSM) for light loads. I wonder just how they are able to achieve such low quiescent current and high efficiency, but there are probably many components that can be utilized in a chip, that suffer when packaged as discrete devices and mounted on a PCB. I think I will continue to make a prototype of my own design just to learn a bit more about high efficiency micropower switching regulators. Might come in handy if I need something for special applications, such as perhaps higher input voltages as seen in EV battery packs or line voltage (up to 350 VDC).
     
    Another method I considered was to use an opto-isolator as the control element, so there would be no practical limitation to the high side bus voltage except for the PMOS transistor. Much of the power dissipation comes from resistive connections from the lower 5VDC supply to the high-side switching components, and high values of resistance slow things down because of capacitance. I chose an H11L1 with Schmitt trigger which has 2 uSec response time and 100nSec rise/fall time to drive the PMOS gate. The LED must be driven with at least 5mA but since the duty cycle is less than 1% it is an average of just 50uA. But it also needs a regulated power supply on the high side, and that's why I had to come up with an efficient regulator. Also, the low-side PIC would need to powered before it could drive the high side switching circuit.
     
    I do like the idea of a linear regulator using white LEDs as shunt regulators. Two in series provides 5.13VDC at 140uA and 4.88V at 12uA, but the PIC seems to load them down to get only 2.35V. However, adding a transistor provides a solid 4.5VDC, and a single LED would provide 1.8VDC for low power devices.

     
    #7
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 14:43:59 (permalink)
    +1 (1)
    I programmed the PIC for lowest power (I think), but there was still only 2.35 VDC on the Vdd and Vss of the device, with the LEDs and a 249k resistor to 40 VDC. I would think it should draw only 20 uA typical and 40 uA maximum, which corresponds to a resistor of 5/40 = 125k so that would be about 13 volts as a voltage divider. But perhaps the LEDs have a significant leakage current below threshold. They drop 4.88V at 12 uA which is equivalent to 406k, and 5.13V at 140 uA, which is 128k.
     
    Here is my code (borrowed from another project):
    // CONFIG1
    #pragma config FOSC = INTOSC    // Oscillator Selection (INTOSC oscillator: I/O function on CLKIN pin)
    #pragma config WDTE = OFF       // Watchdog Timer Enable (WDT disabled)
    #pragma config PWRTE = OFF       // Power-up Timer Enable (PWRT enabled)
    #pragma config MCLRE = OFF      // MCLR Pin Function Select (MCLR/VPP pin function is digital input)
    #pragma config CP = OFF         // Flash Program Memory Code Protection (Program memory code protection is disabled)
    #pragma config CPD = OFF        // Data Memory Code Protection (Data memory code protection is disabled)
    #pragma config BOREN = OFF       // Brown-out Reset Enable (Brown-out Reset enabled)
    #pragma config CLKOUTEN = OFF   // Clock Out Enable (CLKOUT function is disabled. I/O or oscillator function on the CLKOUT pin)
    #pragma config IESO = ON        // Internal/External Switchover (Internal/External Switchover mode is enabled)
    #pragma config FCMEN = OFF      // Fail-Safe Clock Monitor Enable (Fail-Safe Clock Monitor is disabled)

    // CONFIG2
    #pragma config WRT = OFF        // Flash Memory Self-Write Protection (Write protection off)
    #pragma config PLLEN = OFF      // PLL Enable (4x PLL disabled)
    #pragma config STVREN = ON      // Stack Overflow/Underflow Reset Enable (Stack Overflow or Underflow will cause a Reset)
    #pragma config BORV = LO        // Brown-out Reset Voltage Selection (Brown-out Reset Voltage (Vbor), low trip point selected.)
    #pragma config LVP = OFF        // Low-Voltage Programming Enable (High-voltage on MCLR/VPP must be used for programming)

    // #pragma config statements should precede project file includes.
    // Use project enums instead of #define for ON and OFF.

    #include <xc.h>

    #include <stdio.h>
    #include <stdlib.h>

    #define DS3         LATAbits.LATA5  //pin2 GP5
    #define DS2         LATAbits.LATA2  //pin5 GP2
    #define TAP         ANSA4           //pin3 GP4
    #define _XTAL_FREQ  4000000
    #define ERROR       51              // 3.2V/20=160 mV
    /*
     *
     */
    void main(void) {
        int     ADCvalue = 0;
        
        OSCCON = 0b00000000;    // 32 kHz LF
        TRISA = 0;  // 0b00011011;     //GP2, GP5 output
    //    ANSELA = 0b00010000;    //AN3 GP4
        LATA = 0b00000000;
    //    ADCON0 = 0b00001101;    // AN3 ADC ON
    //    ADCON1 = 0b10001000;    // Right Just, Osc/8, Vref=Vdd
    //    __delay_ms(1000);
        while (1) {
            sleep();
            ADCON0bits.ADGO=1;
            while(ADCON0bits.nDONE);
            ADCvalue = 256*ADRESH+ADRESL;
            if(ADCvalue > 512+ERROR) {
                LATAbits.LATA2 = 1;
                __delay_ms(50);            
                }
            else if(ADCvalue < 512-ERROR) {
                LATAbits.LATA5 = 1;
                __delay_ms(50);
                }
            LATA = 0;
            __delay_ms(1000);
        }
            
    }


     
    #8
    qɥb
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 14:47:27 (permalink)
    +1 (1)
    The comment on this line is wrong.
    Does it make any difference if you DO enable BOR?
    #pragma config BOREN = OFF       // Brown-out Reset Enable (Brown-out Reset enabled)

     

    PicForum "it just works"
    #9
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/11 15:19:16 (permalink)
    +1 (1)
    The comments were not changed from the previous project. I just tried a different device of same type, and it also shows only 2.35 volts across the power pins. If I measure each LED separately, I read 0.97 and 0.60 volts, showing the effect of the multimeter. That seems to indicate that the LED current is minimal, probably less than 1 uA.
     
    [edit] I measured current draw of the device with a 5VDC supply and it is 925 uA, and at 2.5VDC it's 720 uA.
     
    [edit2] I tried a blank PIC16F1825 and it draws 300 uA at 5 volts and 150 uA at 2.5 volts.
    post edited by PStechPaul - 2018/02/11 15:59:14

     
    #10
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/12 01:06:09 (permalink)
    +1 (1)
    OOPS! I just found out that I had a different main.c file in my project, so my changes did not apply. After fixing that, the PIC12F1822 draws just 68 uA at 5V (where the LEDs barely glowed, so drawing current), and at 4 VDC the current is just 49 uA. With the shunt regulator and the 249k resistor in series with the LEDs, it draws 4.12 mA. The LEDs glow moderately bright, and regulate Vdd to 5.12 VDC. That's 34.8 VDC across the 249k resistor, so actual current should be 1.40 mA. My super duper lab grade Harbor Freight free multimeter measures the resistor as 205k in one direction and 260k in the other. I think I need to clean my PCB!
     
    Cleaning the PCB did wonders. The resistor now reads 245-250k. The PIC draws 65 uA at 5 volts and 39 uA at 4 volts. Also, I forgot that I have a 10k resistor across the input of the circuit, which itself draws 4 mA at 40 volts. The circuit itself draws 200 uA at 40 VDC.
     
    Yabba-Dabba-Doo!
    post edited by PStechPaul - 2018/02/12 01:46:06

     
    #11
    al_bin
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/12 03:23:05 (permalink)
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    Why you use this topology? vs  controller on low side, powered from output.
    In second case shunt can be turned off after startup(lower quiescent current)
    and ground is common for all circuits.
     
    Albert
    #12
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/12 15:29:51 (permalink)
    +1 (1)
    There will be a controller on the low side to perform the BMS functions, but I need to provide 5 VDC power from the high side bus, which might be anything from 12 VDC to 60 VDC. A linear regulator for 5mA at 5V (25mW) would draw at least 5mA (300mW) from the high bus, which is undesirable. A buck switching supply should be able to do that with as little as 1mA at 60V (60mW 42% efficiency) or 2.5mA at 12V (30mW 83% efficiency). As noted above, there are dedicated switching controllers that will meet this specification.
     
    By enabling the switching supply for only 100mSec for every 1 second sample, the effective current drain can be reduced to 1/10. So even a linear supply could function with an average of 500 uA pack draw. The two white LEDs could be left on continuously at 50uA at which they will glow noticeably and provide a good 5V reference. The low side controller could disable the buffered output of the supply, but with the device in sleep mode it will draw only about 50 uA more.
     
    I will probably continue making a prototype based on this design. Some of the circuitry and code may be useful for various purposes, even if a dedicated switching supply is more practical for this specific application.

     
    #13
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/12 17:13:55 (permalink)
    +1 (1)
    Here is a simulation of a simple linear power supply that can work on 30V to 60V pack voltage (or 12V to 130V with different values for R1). It assumes a PIC on the low side that operates at about 5mA for 100 mSec out of every second. The supply voltage varies from 4.12 to 4.69 volts and average current is well under 1mA:
     

     

     
    #14
    al_bin
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/13 01:32:34 (permalink)
    +2 (2)
    OK. So you traded the difficulty of controlling the upper MOSFET
    for the controller's linear power supply and lack of common ground.
    Your choice, quite clever.
    I am trying to compare the profits and losses of such a solution, and not to undermine it.
    I apologize for the lack of precision of my English.
     
    Albert
    #15
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/13 03:03:38 (permalink)
    +1 (1)
    I appreciate all the comments, and they help me look differently at the design with an eye to improvement. Don't worry about your English - it is quite adequate and I think I understood what you meant. It opened the door to my more comprehensive analysis and better design decisions. Thanks.
     
    [edit] Here are some actual data points for the voltage regulator, powering the PIC12F1822, in sleep mode. I am using an MMBTA06 transistor, 249k resistor, 1 uF capacitor across the PIC, and two white LEDs in series, which visibly glow with as little as 6 VDC applied:
     
    Vin   Vout   Iin(uA)
     6.0  4.35   21.5
    12.0  4.50   44.0
    20.0  4.57   76.8
    30.0  4.62  115.4
    40.0  4.65  153.6
    50.0  4.68  194.4
     
    post edited by PStechPaul - 2018/02/13 04:04:24

     
    #16
    al_bin
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/13 04:24:09 (permalink)
    +1 (1)
    Consider to use depletion mode MOSFET instead R1(in #13)  for wide range input voltage as in
    https://www.infineon.com/dgdl/Infineon-Application_Note_Applications_for_Depletion_MOSFETs-AN-v01_00-EN.pdf?fileId=5546d4624cb7f111014cd63d1a197d94
    Figure 5 ?
    Zener diode may be LED or upper side as well.
     
    Regards Albert
    #17
    uc1
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/13 08:01:01 (permalink)
    0
    Hi Paul,
    The reason you have low PS is due to limited base current to the PNP. 
    If you want to achieve low cost, you can search the circuits " shunt voltage regulator using transistor ", but you have to consider the power dissipation on transistor.
    #18
    PStechPaul
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/13 16:44:17 (permalink)
    +1 (1)
    It is not necessary to have 5.0 volts for the PIC. It's only 10% low at 12V in and holds that to within 0.18 volts up to 50V. The MMBTA06 has a beta of at least 100 so it should regulate well enough when the PIC and other circuitry draw 5 mA. The maximum power will be 60*5 = 300 mW and it's rated for 350 mW. Besides, the 10% duty cycle results in just 30 mW average.
     
    The BSP179 is $0.83 in quantity 10, while the MMBTA06 is $0.14.
     
    I don't really need a 400V device, but if I did the BSP179 might be a good choice. The MMBTA06 is rated 80V and 500mA, which is plenty.
     
    But thanks for the info. I was not very familiar with depletion mode MOSFETs. They are similar to JFETs, like the 2N3819, which I have used many years ago (early 1970s!).

     
    #19
    al_bin
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    Re: Current draw of blank PIC12F1822, and low power 5V supply from 60V 2018/02/14 01:37:59 (permalink)
    +1 (1)
    You're right, they're very similar to JFETs, but less known, so I dared to put a link.
    Your calculations and price data are completely correct, as always.
    But I did not mean voltage ranges.
    Looking at the data from your post # 16 you only need 4uA ((6- (4.35 + 0.7)) V / 249kohm) flowing through the LED, but you consume 180uA ((50- (4.68 + 0.7)) V / 249kohm) at 50V.
    By using the current source that I suggested, you can set the current optimally (eg 10uA or whatever) regardless of the power supply voltage.
    I am also interested in how this power source will behave when the output MOSFET capacity will be charged/discharged.
     
    PS.
    Check price BSS169, BSS159 etc. or Microchip (Supertex) products
     
    Albert
    post edited by al_bin - 2018/02/14 01:53:54
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