Quick TTL current question

Discussion in 'The Projects Forum' started by Peace Frog, Apr 28, 2009.

  1. Peace Frog

    Thread Starter Member

    Apr 13, 2009
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    Alright, I'm looking at using this gate to drive one of two transistors. My question is what kind of current output can I count on? There are three current ratings and I don't know which one I'm looking at.

    Regards,
    Ben
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Post the circuit that you're considering.
     
  3. thingmaker3

    Retired Moderator

    May 16, 2005
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    You want the "Iout" and "Icc" lines taken together. The chip is rated to source or sink up to 25mA on any given output pin, but not more than 50mA total. So if you are driving one transistor as stated, you can use the full 25mA.
     
  4. Peace Frog

    Thread Starter Member

    Apr 13, 2009
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    1
    Per the SgtWookie's request... (Ignore part numbers, that's just the first thing I came up with with PSPICE)

    [​IMG]

    Ohh, and I forgot to tie the emitters to ground.
    So if I use a Vcc=5v and a BJT here I'd just use a limiting resistor to get the proper ic.

    ic*R+Vto=5V

    where ic is not to exceed 25mA?

    Regards,
    Ben
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    How much collector current will your transistors be drawing?
     
  6. Peace Frog

    Thread Starter Member

    Apr 13, 2009
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    The transistors aren't set yet. I'm finalizing the schematic before I make a digikey order.
     
  7. leftyretro

    Active Member

    Nov 25, 2008
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    Don't you need to include series resistors to the base connection of the transistors to limit the current to the desired amount?

    Lefty
     
  8. Peace Frog

    Thread Starter Member

    Apr 13, 2009
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    Yes, the schematic is the idea I'm working off. As stated above I'm asking if the value of R should be given by:

    R*Ic+Vto=5
     
  9. thingmaker3

    Retired Moderator

    May 16, 2005
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    First, subtract 0.7V dropped across the EB junction. Voltage across the base resistor will therefore be 4.7 Volts.

    E=I*R therefore E/I=R

    4.7V / 25mA = 188Ω

    Next size up is 191Ω
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The collector current isn't set by the transistors. When a transistor is saturated, the collector-emitter voltage is nearly zero, so the current (Ic) is determined by your load resistance and the voltage across it. From there, you generally want to make the base current Ib=Ic/10.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    Of course. The 74HC08 high-speed Cmos IC (it is not a 7408 TTL) will try to supply up to 60mA to the base of the transistor which is more than the 25mA that is its max allowed current.
     
  12. Peace Frog

    Thread Starter Member

    Apr 13, 2009
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    Yes, that's what I thought.

    Ron H, I haven't heard the Ib=Ic/10 recommendation before. Why?
     
  13. Peace Frog

    Thread Starter Member

    Apr 13, 2009
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    Are you getting these numbers from the datasheet?
     
  14. CDRIVE

    Senior Member

    Jul 1, 2008
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    There isn't much an issue here. One output pin on those chips could drive the base of quite a few 2N2222's. Less than a mA of Ibe will saturate it with a Rc of 100Ω.
     
  15. thingmaker3

    Retired Moderator

    May 16, 2005
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    He missed the bit about your 2N2222's being place-holders, and he's using the datasheet for the 2N2222. (Or possibly he's going from memory on the 2N2222 stats. I thought it was 50mA, but I'm definately going from memory. :rolleyes:)
     
  16. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Many noobs think that it should be Ib=Ic/β, where β is beta, the DC current gain. Beta is generally specified in the active region, e.g., Vce=5V. When the transistor is saturated, Vce is on the order of 0.1V, and the beta is much lower. You can usually get away with base current lower than Ic/10, but the saturation characteristics of transistors is almost invariably specified at a forced beta of 10. Therefore it is good design practice to design to this spec.
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    It has nothing to do with the transistor datasheet, and everything to do with the HC08 specs and characteristics. You are correct about Icc(max) being 50mA. Audioguru was pointing out that a shorted output could draw as much as 60mA, which will exceed the Icc(max) spec.
     
  18. thingmaker3

    Retired Moderator

    May 16, 2005
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    Okay. I'm lost now. Why would a short max out at 60mA?
     
  19. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Why not? It all depends on driver transistor geometry.
    The Icc(max) spec is the current above which the device might be damaged. It is not a short circuit current spec.
    I have no idea where Audioguru got the 60mA number.:confused: I couldn't find output device V-I curves in the Philips family characteristics. Perhaps it is from bench experience.
     
  20. thingmaker3

    Retired Moderator

    May 16, 2005
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    Makes sense. Thank you!
     
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