2N3904 and finder relay

Discussion in 'The Projects Forum' started by jm-a, Jan 22, 2011.

  1. jm-a

    Thread Starter Active Member

    Oct 20, 2010
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    Hi all,

    After reading advices of members of this forum -thanks a lot guys for my previous post about SSR -, i decided the best solution would be to use a finder relay.

    It'a a plug-in module, in case of failure, remove clip and put an other one, no need to desolder like a SSR.....

    As usual, queries, if someone could reply, would be really great.

    Are values for these resistors correct:
    R3 Collector resistor
    Rb Base resistor
    Rl Led Resistor

    I thought to add test points , but finally a green Led will do the job.

    - 2N3904 Ic Maxi is about 200mA : Right use for this project, is it a power transistor or must be changed?

    -About Led : Iled choice: 10mA for Vf:2,1 V


    -I = 90 mA for the coil + 10 mA for the led = 100 mA

    -Ic=100 mA with hfe=30 ( mini ) Vcesat=0,3V Ib=3,3mA

    -Rc value (15-12-0,3)/ 100 mA= 27 Ohm Rc=27 Ohm

    -Rb value for Vbe sat= 0,7 V Rb=( 15-0,7) 3,3mA Rb=4,3KOhm

    -R led (12-2,1) / 10 mA R led=1MOhm

    Are these basic math correct?

    I made a mistake on the schematic, it is not Nominal Output , but must read Power Coil!!!!!!



    Thanks a lot.

    jm
     
    Last edited: Jan 22, 2011
  2. CDRIVE

    Senior Member

    Jul 1, 2008
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    You have a few errors and a number of issues in your print, like unneeded components. Is this a school assignment? Also, what is a "Finder" relay?
     
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  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,773
    931
    Try 1Kohm for Rled, and not 1Mohm.

    You should reduce the base resistor just a little more, so the switch saturates quicker. 3.5Kohm perhaps.

    Lastly, a 2222 would be a better choice, so the current draw will stay under 25% of max. With the 3904, you are at 50% or more of its max.
     
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  4. jm-a

    Thread Starter Active Member

    Oct 20, 2010
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  5. SgtWookie

    Expert

    Jul 17, 2007
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    You're using 10k for R1 as part of the R1/C1 timing, but R1 is also in the current path for the base of the transistor - and you also have a pull-down resistor and diode on the left.

    This will prevent sufficient base current; the transistor won't go into saturation.

    What kind of time delay are you hoping to achieve from when the power is first applied?
     
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  6. jm-a

    Thread Starter Active Member

    Oct 20, 2010
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  7. Audioguru

    New Member

    Dec 20, 2007
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    The hFE of a transistor is used only when it has plenty of collector-emitter voltage (5V for the 2N3904 transistor) so it is not saturated. The datasheet for the 2N3904 shows that its base current must be 1/10th the collector current when saturated like it must be here. Its saturation voltage loss is not shown for a collector current as high as 100ma but is probably 0.5V max. Then the voltage to the relay and series resistor is only 14.5V, not 15V. But the 27 ohms series resistor is fine.

    The base current must be 10mA so the base resistor must be re-calculated.
     
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  8. SgtWookie

    Expert

    Jul 17, 2007
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    I've drawn a new schematic for you that should prove to be more suitable for your application.

    It uses a reasonably common LM311 single comparator IC with timing components to determine when the relay should be energized and the relay turn on. Note that pin 6 of this comparator should be left disconnected.

    See the attached schematic and simulation.

    R1 and R2 establish a voltage that is approximately 2/3 your 15v input. Note that R10 provides what is called "hysteresis"; it prevents the output from toggling rapidly when the two inputs are nearly equal. The yellow (B) trace on the plot below demonstrates that the voltage drops slightly when the output changes states.

    R3 and C1 create the RC delay time. D1 provides a discharge path for C1 when power is removed. Note that this is displayed as the cyan (A) trace in the plot.

    The output of the comparator is displayed as the green (C) trace on the plot below. R4 limits the output sink current to about 5mA. This means you can use other common comparators, such as an LM2903 or LM393; just be certain to ground the inputs to the unused comparator channel.

    R5 ensures that Q1 is turned OFF when the comparator is not sinking current.
    Q1 is a common PNP transistor rated for up to 200mA collector current, but we're only using a fraction of that; around 14mA.

    R6 is the base current limiting resistor for Q2; limiting the base current to about 14mA. You could increase R6 to 1.4k to limit current to 10mA; I just used 1k as a typical value.

    R7 makes certain that Q2 is off when Q1 is not conducting.

    R8 is retained from your original schematic. It should be rated 1/2 Watt or higher.

    The remainder with the LED, relay coil, and reverse-EMF diode are electrically the same as your schematic.
     
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  9. SgtWookie

    Expert

    Jul 17, 2007
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    The PN2222 you linked to above is equivalent to the 2N2222 transistor. It will work just fine.
     
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  10. jm-a

    Thread Starter Active Member

    Oct 20, 2010
    105
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    Hi,

    Thanks SgtWookie, but I've just read datasheet of 2N3904 to find Rb value.

    And is my math good?

    I found:


    VBE(sat) Base-Emitter Saturation Voltage IC = 10 mA, IB = 1.0 mA 0.65 0.85 V
    IC = 50 mA, IB = 5.0 mA 0.95 V

    So i choose , Vbe sat of 1V . Right or bad idea?

    Then (15 -1 ) / 10 mA = 1,4 KOhm

    Rb= 1,4 KOhm

    Your schematic is excellent, i take it.

    Thanks a lot.

    jm
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, that is all good. :)

    I am glad that you like it. All of the components are necessary to make the circuit do what you need it to do.

    It might be constructed a little more simply by using a 555 timer and components, but I will leave it up to someone else to do that.
     
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  12. jm-a

    Thread Starter Active Member

    Oct 20, 2010
    105
    2
    Hi,

    One more query, i think it's the last of course.

    I see on your schematic, on the left side , square wave... , but i use 7815 output with dc current ; so can i use it with your circuit?

    Thanks a lot .

    jm
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Certainly. I simply used a signal generator source as a way to turn on and off the 15v in the simulation.

    Please do not forget to use the required capacitors from IN to GND (>= 0.33uF) and OUT to GND (0.1uF to 10uF) on the 7815, otherwise you will likely have problems.
     
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  14. jm-a

    Thread Starter Active Member

    Oct 20, 2010
    105
    2
    Hi all,

    I reply to myself !!!!!

    After reading good advices on this forum, I post a very basic schematic .....

    Of course, i'm going to take SgtWookie answer :)

    I'm only curious to know if my 2 basics drafts should work.

    I've removed a diode, replaced 2N3904 with 2N2222, changed R2 value and also a few basics maths.

    From 2N2222 datasheet : Ib=15 mA Ic= 150 mA Vbe sat Maxi 1,3 V


    Schematic-V1

    Changes: R1 to 1 KOhm and R2 to 10 KOhm

    C1 1500 µF

    Req : 1Kohm // 10 Kohm about 1 KOhm

    Maths for Rb: Ve = 15 x ( 10 / 11 ) Ve= 13,7 Volts

    Choice for Ib: 10 mA 13,7 -1,3 = ( Req +Rb ) x 10 mA

    Rb = 240 Ohm



    Schematic-V2

    Diode added to discharge C1 .

    Removed 20 KOhm resistor

    Added C1 1500 µF

    Now for Rb 15 -1,3 = ( Rb x 10 mA) Rb= 1,4 KOhm






    Thanks a lot.

    jm
     
    Last edited: Jan 23, 2011
  15. Audioguru

    New Member

    Dec 20, 2007
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    Both circuits will have a very short delay because the high base current of the transistor through R1 quickly charges the capacitor.
    You need a delay circuit to drive the transistor.

    The circuit on the right side has a base current that is half of what is needed.
    It has a useless extra diode on its left side.
    It does not have a resistor to discharge the capacitor when the power is turned off.
     
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  16. jm-a

    Thread Starter Active Member

    Oct 20, 2010
    105
    2
    Hi,

    Thanks Audioguru, it is the right answer i waited for....

    About useless diode, i agree, someone told me that idea - not a member of this forum -; i was suspicious about this, as you too.

    About delay, i need only roughly one second.

    1500 µF x 1KOhm = 1,5 second, it's what I only want.

    Aside short delay time, should this circuit ( V1 ) work?

    Thanks a lot.


    jm
     
  17. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    Could it be?
    Whoever originally spiced the original circuit didn't use a battery as a 15V source. Instead a pulse gen was employed to simulate power being turned on and off. If this is the case, that mystery diode would rapidly discharge the timing cap when the gen went low, as a Gen, (as apposed to a simple switch) would represent near zero ohms to the V+ rail when low.
     
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  18. SgtWookie

    Expert

    Jul 17, 2007
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    OK, decided to see if your circuit could be fixed.

    Have a look at the attached.

    It's similar to what you had before, but I've added an LED in the current path to the base; it could be a blue or a white LED rated for 20mA.

    Before, when the capacitor started charging, the transistor would be turned on when the capacitor voltage was only ~0.7v. This would require a very large capacitor - large and expensive.

    Adding the LED in series with the base allows the capacitor to charge to about 4v before the transistor starts to turn on. As it is, it takes about 340mS from when the transistor starts to turn on until it is saturated; during that time power dissipation in the transistor reaches ~380mW, which is within the 625mw package limit (about 61% of maximum).

    As shown, it takes about 1.1 seconds for the voltage across the coil to reach 9.6v. Increasing C1 to 4700uF would give about 1.4 seconds delay. 6800uF would give almost a 2 second delay.

    D2 discharges the capacitor when power is removed. You should also use a diode across your 7815 regulator, cathode connected to IN, anode to OUT. The power switch to your 7815 should connect the IN to your voltage supply for ON, and to ground via a small resistor (~100 Ohms) or light bulb for OFF.

    As before, R3 should be rated for 1/2 Watt or more. (I made a typo in the value; it should still be 27 Ohms) The other resistors can be 1/4W.
     
    Last edited: Jan 23, 2011
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  19. jm-a

    Thread Starter Active Member

    Oct 20, 2010
    105
    2
    Hi SgtWookie,

    Thanks a lot for your helpfull post.

    Best regards.

    jm
     
  20. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    I can't seem to resist the challenge of discrete circuits, so here's yet another more complex design that guarantees saturation. The theory of operation is described in the attachment. This circuit enables use of much smaller values of C1. Increasing C1 to 500uF will produce a 5 second delay.

    Longer delays:
    Because of the feedback nature of this circuit there is still quite a bit of headroom left for increasing the values of R1 and R2. I spiced this circuit to a to over 15 seconds with C1 = 500uF R1= 200k and R2 = 100k, but this assumes a perfect Cap with no leakage, which is unrealistic for an Electrolytic. It also assumes Q1 beta >=215 & Q2 beta >=277. For certain, the circuit is operating at extreme limits with R1 & R2 values this high.

    EDIT!! Ron H has correctly pointed out (in post 22.) that Q3 (PNP) is vulnerable to Vbe breakdown.
    http://forum.allaboutcircuits.com/showpost.php?p=324235&postcount=22

    A modified circuit will be posted later in this thread that will replace Q3 (PNP) with two NPN's.
     
    Last edited: Jan 24, 2011
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