555 Timer overheating when adding power transistor

Discussion in 'The Projects Forum' started by Ziplock9000, May 4, 2010.

  1. Ziplock9000

    Thread Starter New Member

    Apr 14, 2009
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    I have a 555 timer circuit running at around 20Khz/50% duty cycle using a 10v input supply. Everything runs smoothly.

    I then added directly to the output a power transistor, it's OWN power supply (30v+) and a coil based load.

    At this point, everything is still ok, the load on the second power supply behaves.

    Everything seems fine until I start to ramp the voltage up on the LOAD power supply. At that point the 555 seems to get very hot and I have to shut down.

    I'm not sure which of the following is happening.
    - Some of the LOAD voltage is getting back into the 555 circuit via the power transistors.
    - There is some back EMF that is doing the same
    - My idea that two seperate power supplies can be used each side of a power transistor is flawed (like a relay).

    Anyone have any explainations before I need to upload a circuit diagram?

    Thanks.
     
  2. gerty

    AAC Fanatic!

    Aug 30, 2007
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    We need to see how you have everything connected. Do you have a resistor from the 555 pin 3 to the base of the transistor, you didn't mention it, and this is one of the reasons why we need to see what you have.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Posting a complete circuit diagram (schematic) is mandatory before you will receive useful comments.

    Please include all values for resistors/capacitors, part numbers for the transistors, resistance or current rating of your coil, the coil's voltage rating, and the inductance of your coil (if available).
     
  4. Ziplock9000

    Thread Starter New Member

    Apr 14, 2009
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    No I'm not using a resistor on pin 3 before it leads to the power transistor.. Here is my circuit (minus the internals of the 555 circuit part)

    sorry for the bad image quality

    [​IMG]
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Big problem, a resistor on the base is mandatory. This is true for both BJTs and MOSFETs (gate there). If you ran this circuit for any duration you have probably blown the 555 and the BJT.

    You also need a kickback swamping diode on the coil. Basically the diode is back biased on the coil, and absorbs the voltage generated by the coil when it is turned off that would damage other components.

    What is the power supply voltage? If it is 12V then a MOSFET is a better option.

    Have you read this...

    LEDs, 555s, Flashers, and Light Chasers

    The 555 Projects

    Bill's Index
     
  6. n1ist

    Active Member

    Mar 8, 2009
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    The transistor should be wired with the emitter to ground, base to 555 thru a resistor, and collector to the coil. The power supply is connected to the other side of the coil (positive) and ground (negative), and a flyback diode should be across the coil, connected with the cathode to the supply end.
    /mike
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    Something like this...

    [​IMG]

    The difference between a motor and a relay isn't that big.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    I just took another look at your schematic. The diode is still needed as shown, but the fact the load is on the emitter should eliminate a base resistor. You will experience large losses though, the 555 drops 1.2VDC from Vcc due to Darlington transistor Base Emitter drop, and then you loose another 0.6VDC in addition to the other losses.

    What kind of current is going through this coil?

    Here are several other points, the 555 has a max Vcc of 15V, if you fed the 555 with more than that it fries. You didn't mention the 555 power supply voltage. I have designed circuits that will use higher voltages.

    If you have 15VDC feeding the 555, which in turn feeds the coil, you will have 12.9VDC going to the coil.

    Here is my answer for another gentleman with a similar problem.

    [​IMG]

    You'll note this design does not have the 1.2VDC drop on the output. Here is why the 555 does...

    This is from LEDs, 555s, Flashers, and Light Chasers .

    [​IMG]

    This from the 555 Schmitt Trigger article.

    [​IMG]
     
    Last edited: May 7, 2010
  9. debjit625

    Well-Known Member

    Apr 17, 2010
    790
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    Ok a standard 555/556 timer IC can source or sink up to ~200mA(it also depends on different vendors) and you are using the pin 3 (as source) directly with the power BJT's base. Power BJT's use a large amount of base current to operate so for that you have to use a buffer section between the IC and the Power BJT.Ya in case of MOSFET they are not like this.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    BJT 555's can source or sink up to 200mA if they are being powered by 15v.

    It is less when the Vcc is less.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Actually they will handle the full 200ma at any voltage. BJTs will smoke where CMOS won't, CMOS can be self limiting but BJTs are more like switches. Running anywhere near max will shorten the lifespan though, and I feel uncomfortable using 100ma, that is a lot of current for chip design this old.

    I'll run the experiments and document if you'd like, I'm pretty sure of this

    MOSFETs have a different set of problems, but overall they are superior for this kind of duty.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    I cordially invite you to try this. :)

    Well, sort of. The BJT versions have that pesky Darlington follower for the high side, which is annoying. I'm wondering why nobody has come out with a version that swaps the NPN Darlington for a PNP.
    Yep, it is. As always, de-rating by 50% is always a good practice.

    Absolutely!

    Well, the CMOS versions have nearly rail-to-rail output, but the current source ability is really quite low. They can sink about 10x the current that they can source. However, the BJT version can sink about 20x as much as the CMOS versions can, and source about 400x as much.
     
  13. Wendy

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    Mar 24, 2008
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    Last edited: May 9, 2010
  14. debjit625

    Well-Known Member

    Apr 17, 2010
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    Ok its seems we are going out of topic,in the case of BJT (BIPOLAR JUNCTION TRANSISTOR) I reffered it for the power transistor in the circuit 2SC3320 ,its not about the internal circuit component of 555 timer IC.Now the 200mA is offcourse the max ratting but I have used a aprox symbol before it(~200mA) it should be noticed.
    Now for switching purpose Power MOSFET are far more superior than power transistor,the reason behind it is on the name ,the transistor use a base current to operate and in case of power transistor its quit large amount of current,on the other hand the MOSFET use a electric field i.e few voltage is applied to the gate to operate the mosfet and the current is very much less than a base current of a power bjt.Normaly in UPS or Inveter you find MOSFET rather than BJT for switching for power efficiency offcourse the frequency factors so in this case BJT is ok, but use some driving stuff for the larger base current like a Darlington pair.

    Now a simply explanation what is wrong in this circuit,suppose we connect a battary's positive and a negative with a wire and let this wire can withhold the power
    without getting fused after sometime we will found that the battary has gone too hot.In case of this battary it will get discharged and becomes cool but in case of 555 IC it burns up.So thats what we called the power factor or wattage.

    Good Luck
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    Actually it wasn't off topic, since how much current a 555 will source is critical. Look up emitter follower BJT transistor designs, AKA common collector. The current through the base is the emitter current divided by the transistors gain. I made the same mistake from a casual glance. A Darlington or Sziklai Pair will reduce this current even more since their beta gain is fantastically high, and may also solve the OPs problem. We need to know what the coil is pulling, and are waiting for the OP.

    With a beta of 100, the 555 will feed 200ma to a 2A load through the transistor. At 200ma the 555 will dissipate around 1/3W though, so it will get quite warm.
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    OK, here are some revised schematics, including the limited one the OP provided, that might fix the problem.

    [​IMG]

    Since we don't have the full schematics there might still be something that will adversely effect the design.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    Better to use a transistor as a saturated switch, if possible. Power dissipation will be much less. In your examples, you are using voltage followers on the 555's voltage followers. You already lose ~1.3v from the internal Darlington follower, and for every PN junction between that and the load, you'll lose another .63v or more.

    If the OP ever comes back, maybe they'll explain more about their application. Until then, we're dead in the water.
     
  18. debjit625

    Well-Known Member

    Apr 17, 2010
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    No I think its not clear yet,what I said is to use a darlington pair to operate the Power Transistor not the coil ,its goes like this the 555 operates the darlington pair and the darlington pair operates the power transistor ,the darlington pair is used here as a buffer as it needs a small amount of current from 555 which will not heat it up.
     
  19. Wendy

    Moderator

    Mar 24, 2008
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    Ultimately the power still comes from the 555, we are talking about gain (as mentioned before). A simple power darlington will do the job, though there are lots of ways to do something like this.

    I drew my idea, feel free to draw yours. :)

    Wookie is probably right, we don't know how the OP is powering the circuit, there are more questions than answers at the moment. Like the fact was mentioned the transistor was powered by 30V, and the 555 can only take 15V.
     
    Last edited: May 9, 2010
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