2 cascaded 555 monostable problem

Discussion in 'General Electronics Chat' started by hazim, Jun 18, 2009.

  1. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Hi.

    Here I have 2 555 monostable circuits, the first with time 2x and the second with time x, connected in cascaded form, the output of the first timer is the voltage supply of the second one.. I'm using this circuit with a relay to let a dc motor run for time x in a direction and then it turns in the opposite direction for time x...etc.

    It could not work because the circuit's supply voltage is 6V, the output of the first timer is 5 with no load, when it is connected to the other part of the circuit as a supply voltage, it drops to about 3V or less, this makes the second 555 timer not working properly, and its output useless...

    I tried using transistors such as BC337 and BC548 (common emitter and common collector) with base connected to the output, collector/emitter connected to 6V/load... but this didn't rise the voltage as it should be

    Any idea
     
  2. Wendy

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    Mar 24, 2008
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    Show us a schematic of what you tried. Instead of using #1 as the power supply, it would be better to use it for the reset input (pin 4).

    Do you want both directions, and no still motion? Or are you wanting it both directions and a stop?

    The reason I ask is the 555 sort of has ½ of a H Bridge, but not a very good one.

    We'll also need the voltage/current specs of the motor.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    The output of a BJT 555 timer when high is about 1.3v lower than Vcc, even with no load. As you've discovered, it drops quite a bit more under load.

    Use the output of the 1st timer to control the reset pin (4) of the 2nd timer. Use a resistor to limit the current from pin 3 of the 1st timer to pin 4 of the 2nd timer; 470 Ohms would be a likely value.
     
  4. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    In the schematic I drew the relay (its pins).... I don't want it to stop, no still motion is needed... the relay is 6V or 5V. the motor is 6V and 0.2A to 0.4A..
     
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  5. SgtWookie

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    It's not going to work like that.

    If you want the motor to reverse, you'll need to either use an "H" bridge circuit, or a double-pole double-throw relay.

    Also, it's VERY hard on the motor to simply reverse it while it's running full speed.
     
  6. Wendy

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    Mar 24, 2008
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    OK, I think this will do what you want, Figure 1 is for illustration only. It may work, but I suspect the 555's won't last long. Figure 2 shows a true H Bridge using BJTs, and should be OK.

    [​IMG]

    D1-4 ... Red LEDs
    R5,R7 ...180Ω
    R6,R8 ...100Ω
    Q1,Q3 ..2N2222
    Q2,Q4 ..2N2907
    C1 ......100µF
    C2 ......0.1µF

    You may want to use beefier transistors, but not Darlingtons.
     
    Last edited: Jun 19, 2009
  7. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    ok Bill, your circuit may help me but I'll now explain what I want to do exactly by this circuit so you could understand the idea...
    In a project I'm doing, a line following robot will come along a line to a garage. The garage door will open when the car comes toward it. Here I used infrared transmitted from the car and will be detected when the car-the line following robot- becomes near the garage. When IR is detected will switch a relay in the garage just for the time the IR is detected at which the car is less than 15cm far from the garage and needs about 2 to 5s to enter the garage... the relay will switch back off when the car is just about 7cm far from the door..

    So what I have is a single 6V pulse taken from the relay that should start a circuit at the instant this pulse starts (when the car is 15cm far from the door). This circuit will start a circuit to run a motor to open the door of the garage, and after a certain duration (for example 5 seconds) the motor will run in reverse to close the door...
    Bill, how does the motor runs for Figure 2?
     
    Last edited: Jun 19, 2009
  8. Wendy

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    Mar 24, 2008
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    Either Q1 or Q2 is on for one side, and Q3 or Q4 is on for the other. Think of the transistors as switches, if Q1 and Q4 is on, then current flows one way, if Q2 and Q3 is on current flows the other way. The reason this is called an H bridge is the same configurations can be drawn as so...

    [​IMG]

    The switches can be transistors, MOSFETS, relays, whatever.

    Wookie said it first, you need to allow for a dead time to let the motor settle. Try to reverse a motor too fast and there will be a current surge.

    The configurations of the transistors (both common emitter) means there is no losses. The LEDs voltage drop keep both of them from being on at the same time, creating a dead short with predictable results.
     
  9. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I have the H-bridge L293 and I need to have inputs as the picture attached shows. Maybe 2 555 timers should be used to generate these two inputs (monostable)... I don't know much about timing using 555 timers... If you could give me a circuit that gives 2 outputs as attached then I'll be very thankful...
     
  10. Wendy

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    No problem, but it will be later than sooner. Think 3 monostable 555's. How do you want to trigger them?
     
  11. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I don't know... just assume I have 6V supply voltage for the circuit that will produce the 2 outputs in the attached file. What is important in the circuit is how it starts, the same as the attached file, it could be periodic-astable or monostable, since the 6V supply will be timed and adjusted till all the operation ends-when the first cycle ends
     
  12. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Don't forget me Bill:) I need that circuit as soon as possible since it's a part from my senior project and I'll present the senior Thursday..
    After Thursday I'll present the senior here in the forum, I'll put video.
     
  13. Wendy

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    OK, takes time to draw stuff.

    [​IMG]

    Just put the H bridge where the LEDs are attached.
     
  14. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Thank you
    Can you shortly explain how the circuit will function please
     
  15. Wendy

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    A basic 555 Monostable, which triggers on the negitive edge, followed by another. The 1st one trips the second, which is your null time, and it trips the 3rd. Thinking about it, I can eliminate the transistor. DOH!

    [​IMG]
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    The L293 has four inputs, and they need to be TTL level.

    TTL is 4.5v to 5.5v. 6v is too high. You can use 6v for the 555 supply if you're using bjt 555's, as the outputs won't go higher than around 4.7v (Vcc-1.3v) - however the logic portion of the L293 needs to be supplied with 5v.
     
  17. Wendy

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    There is also a 558, which I haven't used before. It is 4 timers, similar to 555's but not quite the same, in a 16 pin package.
     
  18. Wendy

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    BTW, the drivers I've show, using Q1,Q2,D1,D2,R5, and R6 are inverting. I have been handling a lot of 555 threads lately, I thought this drawing was there.

    [​IMG]

    Attach the drivers to the 555s, they are inverting power amps. The LEDs (red) will help prevent shootthrough, where both transistors are on at the same time and can blow themselves out. Merge the previous 555 circuit with this one.
     
  19. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    No, no need for the last circuit, that of the motor driver, I'll use the L293 driver with two of its inputs
     
  20. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    R2=r4=r8, c2=c5=c8???
     
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