My transistor gets very hot. Where should I be putting a resistor?

Discussion in 'General Electronics Chat' started by samoz, Jun 15, 2009.

  1. samoz

    Thread Starter New Member

    Jun 10, 2009
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    I am wiring up a very simple circuit just to experiment with transistors. This consists of the emitter pin connected to +5V, the collector pin connected a resistor and LED which are in turn connected to ground, and the base pin to a wire which I attach to either +5V or 0V.

    I'm using NPN transistors which are rated for 1.8W and PNP transistors rated for 300 mW.

    After I leave my transistor connected for a while, I find that it gets very hot. Why is this and how can I fix it? Should I put a resistor on the base pin? Could I possibly have the transistor hooked up backwards and would this matter?

    Edit: Here is a quick schematic I made of the circuit.

    [​IMG]
     
    Last edited: Jun 15, 2009
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Post a schematic into this thread so that we can see how you have hooked things up.

    hgmjr
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Add a 470 Ohm resistor between the switch and the base of the transistor.

    This will limit the base current to about 10mA.

    Right now, you have the NPN transistor configured as an emitter follower. With this configuration, the emitter will be about 0.7v lower than the voltage on the base.

    However, if you put a PNP transistor in the circuit (emitter towards +5V) then it would be a common emitter configuration, and maximum current will flow through the base. Without a resistor limiting the current, the transistor will get very hot.

    Here is a schematic of two circuits; one LED driven by an NPN transistor, one by a PNP transistor. The 470 Ohm resistors limit the base current to around 10mA. Note that with the switch positions as they are now, the LEDs would be illuminated.
    [​IMG]
     
  4. creakndale

    Active Member

    Mar 13, 2009
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    For the K.I.S.S. principle:

    You should connect one side of the 220 ohm resistor to 5V.
    The other side of the 220 ohm resistor would connect to the anode of the LED.
    The cathode of the LED would connect to the Collector of the NPN transistor.
    The Emitter of the NPN transistor connects to ground.

    Assuming the NPN transistor has an HFE (gain) of 40, you would connect a 10k resistor to base of the transistor with the other end of the 10K switchable to 5V or ground as you've shown.

    creakndale
     
  5. DrNick

    Active Member

    Dec 13, 2006
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    None of these solutions explain why the transistor is running so hot.

    Limiting the base current to 10 mA doesn't really do much for you. As it is when the circuit is on, the transistor is configured as a diode, not an emitter follower, as the base and collector are shorted. In this configuration (assuming the LED has about a 1-2V drop when conducting...dont know what color it is) the emitter current will only be about 15 mA.

    Even if the transistor was connected incorrectly it should not get hot. Perhaps your device is bad. The other option is that the resistor in your circuit is not what you think it is.

    Lastly it is possible that the thermal resistance from your case to ambient is very high, making your circuit naturally susceptible to getting hot.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    It shouldn't get hot, so I suspect a wiring error. Not using a base resistor with this configuration is completely OK, the base current will be in the neighborhood of 0.1ma, give or take. With a 2.1V drop Vf, and .6V BE, the current through the LED is 10ma.

    Likely you have the collector and emitter reversed, and the BE is carrying a lot of current, in which case the transistor is blown. A base resistor would have prevented that, but like I said, if it is wired correctly it isn't necessary.

    Transistor wattage is BE drop X current, or .6V X .01A = 6mw

    What is the case style of this transistor?
     
  7. samoz

    Thread Starter New Member

    Jun 10, 2009
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    Alright, that makes more sense now. In a circuit I'm designing, I'd like to have several transistors all use the same wire for their base pin (so that they all are either on or off). Would I put a larger resistor (10K?) between the switch and the wire that connects to all the base pins? Would that work?
     
  8. axeman22

    Active Member

    Jun 8, 2009
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    Ok.. my first post in reply.. so hoping I am correct!! (knowing I' m probably not though :) )

    Ref my attached pic

    here's my take on it all..

    - I'm using the BC548 (referenced here) - you'll not that a 5mA Base current(emitter to base) will saturate it such that it will pass a 100mA Collector current.
    - I'm also using a fairly standard LED where it requires 1.6V and 20mA for full brightness.

    From my calcs the transistor has a gain of 20, i.e. the collector current will be 20 times the base current. So I use the logic for base current that it will be 1/20th of the desired collector corrent, which in this case is 20mA(because that's all you need for full brightness on the LED).

    I look at the simple circuit from TP5(ground) to TP1(Vcc) (via TP4 then TP2)and see that the transistor will drop 1.4v from Emitter to Collector(I'm actually not so sure on that but I work on it being 2 x PN junctions at 0.7 each), then the LED will drop 1.6V and the desired current flow is 20mA up this path - so I need to whack in a resistor which will drop the remaining voltage and work with the current calculations I have. The resistor needed to drop 6V and pass 20mA I calculate to be a 300Ohm resistor. So this is what I would refer to as the main load path - i.e. from TP5 to TP1 - via TP4 and TP2

    Then there is what I'd refer to as the switching path. TP5 to TP1, via TP3 when the switch is made to that position. In this case I know that the emitter to base of the transistor will drop 0.7V and it will need to pass 1mA of current in order to allow the transistor to pass 20mA of current on the colllector circuit to drive the LED. So I then plan that if we have 9V to lose via this path and the transistor drops just 0.7V then I need a resistor between the Base and Vcc (R2) to drop the remaining voltage and keep the current to 1mA - ohm's laws there produces a result of R2 needing to be a 8.3KOhm resistor - (8.3/.001)

    then when you switch the switch to the OFF position the base voltage will be the same as the emitter therefore no current flow emitter to base and as a result no current flow emitter to collector and therefore also no current flow through your LED.

    Hope that makes sense.. Please don't assume I am the oracle on this.. this is just what I understand so far and wanted to share my knowledge as I've taken much from this forum and feel that I can give back here.
     
  9. axeman22

    Active Member

    Jun 8, 2009
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    Whilst on the subject of transistor theory..

    1. when fully saturated - i.e. iEB = Max and iEC = MAX also.. what is the voltage drop accross Emitter to Collector ? it is as simple as 1.4Volts as there are two PN junctions or ..?


    2. Is this correct - The Emitter to Base Vdrop should never be more than 0.7Volts. You should plan your supporting circuit such that the vEB is only ever 0.7V and with the desired current flow you build you suporting circuit. To connect the negative of a 9V battery to the emitter and the positive of the battery to the base with no other circuitry would force the Emitter to Base PN junction to dissipate 9Volts.. assuming the transistor didn't blow up this would be dissipated as heat, you would measure 9Vdrop from E to B. ..???

    Thanks!
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    No, you will need a separate resistor for each transistor's base.

    Each transistor is slightly different. If you try to share current from one resistor, one transistor will get more of the current, causing it to warm up, which drops the Vbe even more, which causes it to get even more current.
     
    Last edited: Jun 15, 2009
  11. creakndale

    Active Member

    Mar 13, 2009
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    Axeman22,
    To comment on a couple of things...
    1. When the transistor is fully saturated (used as a switch) you use the Vce (sat) spec from the datasheet. It's typically 0.1V to 0.2V (not 1.4V).
    2. You are correct in that Vbe should not be more than 0.7V. You need a base resistor or... poof.
    3) The DC current gain (hFE from the datasheet) is around 100 for the BC548 so you only need about 200uA of base current to saturate. Not sure where you got a gain of 20 but 1mA of base current won't hurt the transistor. It will just saturate the it a bit harder and make Vce smaller.

    creakndale
     
  12. axeman22

    Active Member

    Jun 8, 2009
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    well here in lies the interesting part.. practical versus theory!

    take a look at the circuit below, I built this and took all the actual measurements listed.

    I'd like to understand..

    1. why there is such a high Voltage drop between emitter to base - 3.15Volts - is this simply because I have such a low resistance value on the base circuit. I know I should have about 1.7kΩ but just playing, was surprised to see it so high. I calculated that for a BC548 where 5mA Ibe equals saturation and 0.7Vbe then R2 should be 1660Ω (to drop 8.3V with 5mA)

    2. the Vdrop over the Emitter to Collector Junction is what really baffles me.. I jammed it hard on to full saturation with the 11mA running through base so expected to see around 0.2v Emitter to Collector but it is 2.38V !

    3. Any suggestions on what values I should have for R2 given that I just want to jam on the transistor, effectively an on/off switch. - and with that resistor you suggest what should be Vec?

    Thanks!
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    Was your switch floating or something? I suspect your transistor might be blown, there is no way you should measure more than 0.7V on a forward biased BE junction, 3V don't fly.

    It might also be possible that you have a sever oscillation, with can screw up the readings. I tend towards the blown transistor theory myself.

    BTW, the original schematic you showed, as opposed to the verbal description, would have worked. A common collector design is a high impedance input. You showed common collector, but described common emitter.
     
    Last edited: Jun 16, 2009
  14. axeman22

    Active Member

    Jun 8, 2009
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    sorry.. the switch actually doesn't exist.. it is hard wired from Base to TP3. what's the best way to check if a transistor is blown? (which it could be!) .. I checked it on the multimeter and getting hFE of 131.6.... hmm??
     
  15. Wendy

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    Mar 24, 2008
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    I asked this once before, what is the case style? It is very easy to mix up which lead is which on some transistors.
     
  16. axeman22

    Active Member

    Jun 8, 2009
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    It's a Z1308 from dse.com.au - a BC548 - TO92 pack. refer wiki

    The DSE catalog has 3 variants of configurations for pins listed against the TO-92 package and there is no way of telling in the catalog which one is which!

    I'm working on the assumption that when you're LOOKING at the FLAT face then moving left to right it is C - B - E.

    ?
     
  17. Wendy

    Moderator

    Mar 24, 2008
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    I can't say as to your part, but this is a 2N2222A plastic transistor pinout, top view.
     
  18. axeman22

    Active Member

    Jun 8, 2009
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    have a look at the latest readings.. every bit of wording I could find supported the way I was using it.. but just for fun I swapped the other way around as you have shown and now the voltages seem to make a lot more sense, I think you are right!.. I was surprised how well it seemed to work in reverse wiring..? what's what that..?

    so as I have it now, when you're LOOKING at the face, from Left to right it goes E-B-C
     
  19. axeman22

    Active Member

    Jun 8, 2009
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    Ahhhrk, no I'm second guessing myself.. I just took the transistor out and measured the voltage drop accross the Base to Emitter and Base to Collector.. I got higher over what I thought to be the Base to Collector. Research tells be Base to Emitter is always higher. so perhaps I actually had it in the right way before! ... damn this is killing me. How can one tell for sure what lead is C or E.. seem such a fine difference..

    plus if I go back to my original thinking the thing measure a hFE of 143 whereas if I swap it around it measures hFE of about 3 only..
     
  20. SgtWookie

    Expert

    Jul 17, 2007
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    Read through this Wiki article: http://en.wikipedia.org/wiki/Bipolar_junction_transistor
    Note that the emitter is more heavily doped than the collector. If you try to swap the emitter and collector, you'll be operating the transistor in the reverse of what it's optimized for, and performance will be very poor.

    An hFE of 143 is in the ballpark of what's to be expected (note that hFE is for small signal gain, not for operation in saturation).
     
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