BJT delay timer problem?

Discussion in 'General Electronics Chat' started by samy555, Oct 2, 2015.

  1. samy555

    Thread Starter Active Member

    May 24, 2010
    I've built the following circuit:
    Voltage at node (2) is about 2V. If I disconnect (R3) at node (3) and the circuit is powered ON then the voltage at the top of cap (C1) is gradually increasing from 0 to 2 volt. But if (R3) is reconnected then the voltage at node (3) never increase more than 0.64 V. The problem in that the transistor never saturate, VCE = approximately 10V and a 12V relay connected to the collector will never activate.
    I need two things:
    FIRST: How can I make the circuit operates a 12V relay?
    SECOND: Where to connect a diode to discharge C1 when the circuit is OFF?
    Thank you very much.

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    Last edited by a moderator: Oct 3, 2015
  2. dl324

    Distinguished Member

    Mar 30, 2015
    The combination of R2+R3 is too large to provide enough current to saturate the transistor.

    Regarding how to switch a relay and how to discharge C1; need you to describe how you plan to control the relay and why you need C1.
  3. ronv

    AAC Fanatic!

    Nov 12, 2008
    You might find it easier with a FET instead of a BJT,

    It's not to say it can't be done with the BJT it is just that R2 will need to be smaller, so C1 bigger for the same timing. To figure the size divide the relay current by the beta of the transistor and make sure R2 can supply that base current. The transistor won't be saturated but pretty low drop. It would also help to use a higher gain transistor like the BC547C.
    Last edited: Oct 2, 2015
    samy555 likes this.
  4. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
    Swap R1 and Led 1, the capacitor will charge upto 10v, better still use a fet.
    samy555 likes this.
  5. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    That is incorrect. Node 4, the base, is 0.64 V. Node 3 can be considered the center of a voltage divider between two batteries, 2 V on the left and 0.64 V on the right. This is a simple Ohm's Law calculation.

    You have stumbled into a classic timing circuit problem, one that everyone here has dealt with at some time. As mentioned above, the total resistance between the 2 V source at node 2 and the base at node 4 is over 100 K, way too large to allow enough base current to saturate the transistor. This is a common problem with long time delays and bipolar transistors - when the timing resistor is large enough for the desired delay, it is too large to do anything else. Changing to a MOSFET solves that problem because it needs no current to operate so the series timing resistor can be very large. However, it's turn on characteristics are much less precise, so the amount of time delay will vary significantly from one device to the next, and with temperature. Also, a MOSFET turns on more slowly as the input voltage ramps up. Notice in Ron's simulation how long it takes for the output to get down near 0 V.

    There are circuit techniques to solve all of this with more discrete components, or you can change to a device designed for timing applications like the 555.

    Last edited: Oct 3, 2015
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  6. dannyf

    Well-Known Member

    Sep 13, 2015
    R1 and led form a 2v voltage source. That, plus r2, forms a constant current source to charge up the capacitor, because r2's value is very high.

    once r3 and the bjt are brought in, the quickly discharge the cap.

    two solutions. One, use a mosfet. Two use a high beta bjt.
  7. crutschow


    Mar 14, 2008
    What's the resistance of the relay coil you want to control?
    That will determine the minimum values of R2 and R3.
  8. samy555

    Thread Starter Active Member

    May 24, 2010
    I wanted that when electrical power is applied to a circuit, the LED does not illuminate immediately, but takes a few seconds before it shines (Later I'll replace LED with 12V relay).
    The transistor is off until the voltage on C1 is >= 0.7V.

    Thank you
  9. samy555

    Thread Starter Active Member

    May 24, 2010
    Thank you very much
    Does D3 (in your schematic) to discharge the capacitor?
  10. samy555

    Thread Starter Active Member

    May 24, 2010
    it is a very good point
    thank you very much
  11. dl324

    Distinguished Member

    Mar 30, 2015
    You could add another transistor to make Q1 a split darlington; and add a signal diode in series with LED1 if the current R's and C give you the delay you want. The diodes and resistors in the circuit should discharge C1 before the next relay activation.
  12. Bordodynov

    Active Member

    May 20, 2015