Cut power in very short regular intervals

Discussion in 'General Electronics Chat' started by Scottm355, Aug 1, 2016.

  1. Scottm355

    Thread Starter New Member

    Aug 1, 2016
    9
    0
    Hello,

    I am after some help designing a device that will cut DC power in very short regular intervals. I would say that a a momentary power interruption every 30 to 60 seconds would be sufficient. The power only needs to be interrupted for a fraction of a second.

    I need the device for a digital slot car track. I have carried out a modification allowing analog cars to operate on the digital track however the autonomous function doesn't work in analog mode. The digital chips in each car use the momentary power interruptions caused by the cars going over a lane change section as an indication that the Car is still moving and not stuck. The modification I have carried out to allow analog cars to operate on the digital track involves powering the rails of the slot car track with two digital chips out if cars. Problem is that the power supply is constant so in Autonomous mode the chips "shut down" after about two minutes of running. Hope this makes sense. Any help would be appreciated.
     
  2. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,676
    899
    How about a PWM driver with a duty cycle of 99.7% (100 - (0.2 s/60 s))? Or, trigger a 0.2 second monostable every 60 seconds and use that pulse to cut the power (various ways to do that).

    If you reset the cars' timers more frequently, say once every 2 seconds for 0.02 seconds, that would be a 99% duty cycle and might be easier to create.

    John
     
  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    Period - 30-60 seconds - check.
    Off time - "momentary" - ?
    Voltage - "slot car" - ?
    Peak current - ?

    I'm not a big fan of using a 555 for long (tens of seconds) period timing, so I'd go with an oscillator/divider driving an output pulse driver.

    IF the system runs on +18 V or less
    THEN the circuit is one CD4060 and one P-channel power MOSFET (plus timing and protection components).

    ak
     
    atferrari likes this.
  4. Scottm355

    Thread Starter New Member

    Aug 1, 2016
    9
    0

    The off time required could be understood by considering the length of time a slot car experiences power interruptions in normal operation when passing over a lane change section. Considering the cars speed is variable I don't believe the off time duration is critical. The shorter the duration the less chance it will be noticed by the the person controlling the analog car so perhaps go with 1/10 of a second for now.

    Voltage is 14vdc

    Peak current I have never actually measured. I would estimate each car could be capable of drawing between 1 and 2 amps so two cars could be 2 to 4 amps. The power supply to the circuit I want momentarily interrupt is protected by a 5 amp fuse and has never blown. Can check accurately if required.
     
  5. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,676
    899
    Maybe you didn't understand my comment. Many of the analog ways to create an asymmetric PWM (e.g., 99.9% on time) don't work well at greater than 99% on time. Ninety eight to 99% on time works better/easier. So, if you allow 0.1 seconds off time, 99% allows a 10-second on time. I doubt the cars can tell the difference between 99% on time and 99.7% on time.

    Now, if you do it digitally and have an 8-bit byte resolution, then 99.6% becomes somewhat feasible; however, Microchip's PIC devices easily allow 1024 (10-bit) resolution (99.9%) on time.

    Please answer this: Does the car's program preclude a more frequent resetting than every 30 seconds or so?

    John
     
  6. Scottm355

    Thread Starter New Member

    Aug 1, 2016
    9
    0
    image.jpeg
    I have measured more accurately how long the digital chips will function for in Autonomous mode before they "shut down" and the time is 45 seconds. In reality when in Autonomous mode each car is likely to pass over a lane change second at least every 10 seconds. If I am trying to design a circuit to mimic this behaviour these are the kind of frequencies to target. I have attached a picture to give you an idea of my modification. The circuit boards you see normally are located in a digital car. The input rail voltage is constant and carries digital data to each car. The output is directly to the cars motor. My analog modification has the rail supply (and digital data) provided directly to each chip and the output from each chip powers the rails.
     
  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    First pass. R3 adjusts the off time from approx. 0.1 s to 0.5 s. R1-C1 set the oscillator frequency such that the Q14 output has a period of approx. 30 s. After power-on the output will pulse off in around 15 seconds, then every 30 seconds after that. Decrease R1 for a shorter interval. Q1 can be any P-channel power MOSFET that is rated for at least 8 A at 25 V. Since it is on most of the time it might need a heatsink, depending on its Rdson.

    Schematic updated 1608.04

    ak
    PowerInterruptPulser-1-c.gif
     
    Last edited: Aug 4, 2016
  8. Scottm355

    Thread Starter New Member

    Aug 1, 2016
    9
    0
    Thanks AK

    Might need a little help understanding a few components of the circuit but I am confident I can build it. Can you confirm the following for me please.

    Is R3 a variable resistor?
    I am assuming that D on the Q1 MOFSET is the power supply to my digital chips?
    What is the little C3 circuit drawn on the lower right of the schematic?
     
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    R3 is a variable resistor, either a full size potentiometer or a trimpot. Its value is supposed to give an off-time adjustment range of approx. 0.1 s to 0.5 s.
    D on the MOSFET is the drain pin. It is the switched output that is turned off periodically. The MOSFET source (S) is connected to the 14 V system power supply. (My library does not have a 14 V symbol.)
    C3 is a decoupling capacitor. The circuit will not work reliably without it. It should be placed as close as possible to the CD4060 power and ground pins, Vdd and Vss.

    Note - it is common practice in digital logic schematics that the chip power and ground connections are *not* shown. This prevents a lot of clutter in the drawing. Refer to the CD4060 datasheet for these connections.

    If you want the delay period between off-times to be adjustable, replace R1 with a 10 K fixed resistor and 50 K pot in series.

    ak
     
  10. #12

    Expert

    Nov 30, 2010
    16,257
    6,762
    I question your choice of a mosfet with Rds of 0.28 ohms. You will lose most of a volt at 4 amps with a 14 volt Vgs and the transistor will be, "on" for ~99% of the time (about 3 watts in the mosfet).
    As a nit-pick, your drawing shows the gate drawn backwards.
     
  11. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    Nice catch on the symbol; I knew something was off but I was rushing.

    As I said, "Q1 can be any P-channel power MOSFET that is rated for at least 8 A at 25 V." Probably should be at least 30 V.

    ak
     
    #12 likes this.
  12. #12

    Expert

    Nov 30, 2010
    16,257
    6,762
    Sorry. I missed that.:oops:
     
  13. dannyf

    Well-Known Member

    Sep 13, 2015
    1,775
    360
    Two approach to it.

    1 a time base generator that triggers a switch on the edge. It can be done with a timer like 555, or oscillator plus divider. Or a MCU.

    2. A pwm generator with very high duty cycle. A MCU is a more practical solution.

    If you are familiar with arduino or pic, it is simple. Otherwise, take the oscillator approach.
     
  14. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    Updated schematic in post #7.

    ak
     
  15. Kjeldgaard

    Member

    Apr 7, 2016
    71
    16
    I have a modification to AnalogKid's schematic:
    PowerInterruptPulser-1-c_Edit.jpg
    When I did not like having too long switching times of Q1's gate, I have taken U1 Reset in use and modified C1 to 47nF.

    From reset, Q14 goes high after 28.6 seconds. and Q6 go high again 0.11 seconds later and resets U1.
    This should provide a nominal time period of 28.7 seconds and an off time of 0.11 seconds for Q1.
     
    AnalogKid likes this.
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    * * * *

    I've done a similar trick in many designs, some of them posted on this site, but I completely missed it on this one. Nice catch.

    Note that some off-time duration adjustment is possible. Moving the reset diode to Q7 and Q8 gives off-times of 0.22 s and 0.44 s.

    Also, I'm not sure the 100 ohm gate resistor is needed. The output resistance of a 4060 is approx. 250 ohms at 10 V, more than enough to prevent oscillation.

    ak
     
  17. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    Updated design per Kjeldgaard, tweaked for common components. On time = 27 s. Off time = 0.11 s.

    ak
    PowerInterruptPulser-2-c.gif
     
  18. Scottm355

    Thread Starter New Member

    Aug 1, 2016
    9
    0
    For this small circuit or any other
    Reasonably small circuit built on a PCB, what would the process be to have a PCB developed to suit the design? The little research I have done leads me to thinking that the PCB isn't that expensive to have printed but to have the drawing file developed to give the printer can cost a bit.
     
  19. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    @AnalogKid:
    There was no any resistor or diode between Q14 and Vg, and Vg will be bonded by Q14 all the time, does the function match the TS want?
     
  20. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    The gate is driven by Q14 all the time. It is not a tri-state output, so there is no need for a pull up or pull down resistor. The 100 ohm resistor in series with the gate usually is there in high frequency switching circuits like power supplies to prevent resonant ringing caused by circuit inductance and the MOSFET gate capacitance. The 4060 does not have a particularly fast output transition, especially when driving a capacitance, so I don't think the gate series resistor does anything. I think the schematic in post #17 captures everything in posts #1 and #4 (and #15).

    ak
     
    ScottWang likes this.
Loading...