Time Delay Circuit isn't working

Discussion in 'General Electronics Chat' started by jcarver1112, Jul 11, 2015.

  1. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    Hello Everyone,

    I wanted to start by saying I'm new here and somewhat new to electronics in general. I'm kind of self educating at the moment and I'm working on transistors. I had a little project in mind to use one and help me learn but I'm having no luck. I'm wondering if i'm not using the wrong type of transistor.

    I believe I have replicated the attached circuit but I don't see a delay. The LED just turns on right away.

    The values of my resistors and cap are a bit different but i dont think they should impact it this way.
    1K resistor to LED (same as drawing)
    270k resistor to base of transistor (instead of 33k)
    Additional 10k resistor from+ side to the capacitor. Right about where the push button is in the drawing.
    Capacitor is only 470uF(instead of 1000uF) *this was the reason for the 10k to slow the charge down some

    Some of the transistors I've tried are
    TIP3055
    PN2222
    PN4393

    I've attached the schematic as well as a picture of it on my breadboard. Any advice is greatly appreciated.

    circuit.PNG unnamed.jpg
     
  2. absf

    Senior Member

    Dec 29, 2010
    1,493
    372
    It should work. You are expecting a delayed ON but the circuit is a delayed OFF setup. The light will remain ON after you release the button.

    A delayed ON would need a different setup.

    Allen
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    For a delayed on, add a resistor in series with the switch.
     
  4. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    The 10k is in series with the switch. I thought with that it would be both delay on and off.

    In the picture you can see the yellow wire (12v), then 10k resistor, then + of the cap. That should be delaying the turn on correct?
     
  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    It is a p.ss poor timing circuit...

    Note that the addition of R3 slows down the turn-on just a fraction of a second, but you have to wait 4 min for the Led to go completely out.
    245.gif

    Swagatam is a noob.
     
  6. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    Wow mike, now that is a HUGE help! I thought I was going crazy. Ive seen this circuit draen several times as a delay on, but as far as i could tell the transistors were far to sensitive for that. So as I see it, there really isn't a way using this circuit to accomplish a several second delay without using a very high resistor in place of the 10k. But then draining would take forever. Less than idea. So, how about a push in the right direction? Is there a such thing as a transistor or MOSFET that has a high turn on voltage. Like 5-9v?

    Also what program do you have there? That could save some serious head scratching lol
     
  7. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The original circuit you posted is not a "delay on", it is a pulse-stretcher, meaning a momentary push of the button charges up the capacitor, and then the LED stays on for a preset time, and then goes off. The posted circuit is a very poor design, typical of the crap posted on the internet by folks who are better at playing with html than designing circuits...

    I "fixed" the pulse stretcher to make it conform to best practices. The simulation is posted below:

    First, I added R2, not to make the circuit a "delay on", but to protect the switch contacts from the inrush current that would otherwise flow into the uncharged capacitor. If the circuit is powered with a MN1604 like 9V battery, it probably has a high-enough internal resistance not to weld the switch contacts, but if V1 is a high-current power supply and C1 is more than a few hundred uF, then R2 is needed to prevent damage to the switch.

    Second, I added a second transistor connected in a Darlington configuration to reduce the impedance on node rc, to get a much longer time delay per uF of capacitance in C1. Note that with this connection, I am getting about 1sec of delay per uF in C1. 1000uF would give a delay of 1000sec.

    Third, I modified the timing network by adding R1 and greatly increasing R3. R1 speeds up the rate of discharge of node rc right as the transistor(s) turns off, giving a much cleaner turn-off. Note that the LED current I(D1) stays at ~10mA, and then turns off much more abruptly than in the original circuit.

    Note that the timing sequence actually begins when the switch S1 is released.

    91.gif

    Simulation is done with LTSpice...

    Are you looking to make a "delay on" circuit rather than a "pulse stretcher"?
     
    Last edited: Jul 12, 2015
    absf likes this.
  8. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    Mike, that was absolutely great.

    Looking though the changes you suggested, all of them make perfect sense.

    -Using the Darlington configuration allowed the used of a much higher resistance on R3 while still supplying a higher current path for the load (LED). This also slowed the discharge of C1 though the base of Q1 after its off, is that correct?
    - With that you were able to add R1 allowing a faster and more controlled discharge, so the light turns off much more abruptly. The original design on the bread board did "fade out" the led, which was simulated well on your post #5.
    - As for the addition of R2, your right, I hadn't given it any consideration but a large capacitor connected to a high current source like a car battery or other high current power supply, you would just be asking to melt that switch the moment it made contact.
    -And the reduction of C1 to 10uF makes sense now that your using the Darlington for the higher gain.

    Also thank you for that simulation software. I see that its freeware so I'm going to download it and play with it a bit.

    So back to the original topic, the goal was to make a delay on, the not pulse stretcher, though this exercise alone has considerably helped me understand how to effectively utilize a npn transistor (or 2) in a simple circuit. One thing I haven't tried yet, that you may be able to answer easily; could the LED be placed on the other side of the transistor? In every drawing I've see, its been on the collector side.
     
  9. radiohead

    Active Member

    May 28, 2009
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    These images are similar to the one you uploaded. These capacitor-discharge timers should only be used as a teaching tool to demonstrate that an NPN will conduct using the current from the capacitor when connected to the base. Also, the PNP will not conduct when current from the capacitor is applied to the PNP base...as the capacitor drains, the light will gradually get brighter. If you want a better timer, I would suggest using a 555 or any other precision timer.
     
  10. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Let's use LTSpice to answer the question:

    91iv.gif

    Which one do you like?
     
  11. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is how I would do it. Here, putting the LED in the emitter is an advantage. Getting about 0.5sec of delay per uF of timing capacitance.

    Note that the switch must remain closed beyond the delay duration. Note how I discharge the timing capacitor for the next timing cycle.

    91d.gif
     
  12. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    Very nice, it does make a considerable difference with both the position of the led and the position of the resistor, however I don't understand why it does. I'm semi familiar with how an npn transistor works, at least the academic explanation of it but I'm just not making the connection as to why it responds in such a way.

    For what its worth, when I say I'm building something to learn how, this is what I mean. I have a desire to understand all aspects of how something works as well as how each component interacts with the others to produce a "system". As well I'm a very hands on so creating a simulation, understanding what is happening, then taking it to the breadboard is what helps me learn the best. The help I've received here so far as a new member has been second to none, so I just wanted to say thank you to you and the other contributors.
     
  13. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    Ok let me first start by seeing if I understand the simulated results here.

    V(on), represents the closing of the switch applying power from the circuit, correct? And it is supplying about 4.5mA of current?

    V(rc) is the charged voltage of C1 (or the point at rc technically)? Why does it stop at 3.4v?



    I have to be honest I'm pretty lost on whats happening in this one.
     
  14. jcarver1112

    Thread Starter New Member

    Jul 11, 2015
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    Also I've started messing with LTspice, but as a newbie I'm still struggling lol.

    Just a few general observations from yours.
    #1 My voltage supply looks different, but no idea why.
    #2 My switch looks different. The +/- labels are on the wrong side and I have no idea how you made those little flags for on/off
    #3 I have no idea what you've done in that box to the left titled Stimulus.
    Capture.PNG
     
  15. ian field

    Distinguished Member

    Oct 27, 2012
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    You need to add a charging resistor in series with the push button.

    A fully discharged 1000uF looks pretty much like a dead short at the instant of pressing the button, so a current limiting resistor will make the push button last longer as well as giving you a CR turn on time constant.
     
  16. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The red trace V(on) is "the voltage at node on as a function of time". Read the voltage on the left Y axis (units in Volts). Time is the x-axis, because the directive .tran 10 tells the simulator to simulate vs. time for 10seconds. The default behavior of the LTSpice voltage-controlled switch is that it is on when its control voltage is >0V; and it is off when its control voltage is <oV. The default can be overridden; read the Help file.

    I did not plot the current that flows through the switch, so you are misinterpreting the 4.5mA. The only current I plotted is the Orange trace, I(D1). To read the current, look at the right Y-axis, units of mA. This only applies to the LED current.

    Yes, the Yellow trace is V(rc), or the time-history of the voltage at node rc. The units are Volts, so read it against the left y-axis. It stops at ~3.4V because is is clamped there by 2ea. Vbe drops (about 0.6V each) plus the forward voltage of a Red LED (about 1.9V).

    When the switch closes, C1 charges through the switch and R3. D2 is blocking. The LED begins turning on when V(rc) exceeds about 3.1V. This makes the delay before the LED turns on...
    When the switch opens, C1 discharges through D2 and R1, making the timing network ready for another cycle.
     
  17. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Work through one of the many LTSpice tutorials. Start with the one on the Linear.com website... Working the tutorial will answer most of your three questions... If not, post them again...
     
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