How do you turn on an NPN BJT Transistor?

Discussion in 'General Electronics Chat' started by Guest3123, Oct 8, 2016.

  1. Guest3123

    Thread Starter Member

    Oct 28, 2014
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    Alright, with an n-Channel MOSFET, you apply positive voltage to the Gate, to saturate the gate. This voltage is known as Vgs(th) on the datasheets.

    The IRLB8721PBF n-Channel MOSFET has a Vgs(th) voltage rating of 1.8Vdc. That means the gate needs at least 1.8Vdc to allow voltage and current to flow from the Drain to the Source. Simple right? I want someone to explain to me, using the NPN BJT Transistor I'm going to specifically list, how the same thing happens, using the data in the data sheet I'm going to list. After I explain my understanding how an n-Channel MOSFET works, and how it's wired up in a very simple circuit.

    So here's my drawing I just did for the n-Channel MOSFET I've listed, used as a Solid State Relay (SSR) to protect the switch. I must've worked with this particular n-Channel MOSFET several times, doing all kinds of things, so hopefully I got the circuit correct.

    Feel free to correct me if I'm wrong, but I'm pretty sure I'm not, especially since I have videos to show me dropping the gate current down to as low at 20uA in circuits to control mains voltage, and stuff with my Arduino Uno I got for $3 from China. So.. here it is.

    Hunted001186.jpg

    Now.. Let's talk about the NPN BJT Transistor. I would like to know what the turn on, or forward biasing voltage is called in the specifications on the data sheet for the KSP05TA NPN BJT Transistor.


    Where Vgs(th) is the gate voltage, or gate source voltage to turn on the n-Channel MOSFET, and the _______________ is the voltage to turn on the KSP05TA NPN BJT Transistor.

    Where Id is the MAX drain current going from the Drain & Source on the n-Channel MOSFET, the ______________ is the same thing on the KSP05TA NPN BJT Transistor.

    I will need to protect my (uC) micro controller by connecting a pull down resistor between the gate & source on the n-Channel MOSFET, is the same to connect a pull down resistor between the ___________ and ____________ on the KSP05TA NPN BJT Transistor.

    Now I'll try to guess. This doesn't mean I don't need an answer to my three questions I've asked. I'm going to show effort, because god only knows, I've searched online, and watched several videos showing how to wire these NPN BJT Transistors up, nobody explains what the data in the datasheet does or whatever.

    So.. here it goes. I'm going to try give my hypothesis, as to how these NPN BJT Transistors work.

    The Vebo (Emitter Base Voltage) is the positive voltage to be applied to the Base, to turn the NPN BJT Transistor on, connecting ground to Emitter.
    The Vceo (Collector Emitter Voltage) is the Maximum voltage I can have as my load (Collector to Emitter).
    The Ic (Collector current) is the Maximum current I can switch. (Collector to Emitter).

    Is that correct? If not, please explain.


    This is a top secret circuit I'm working on.. It using the NPN BJT Transistor I've mentioned.

    Here's my top secret drawing, I'm trying to learn about Logic Gates, and stuff before I tackle Bit counters, and frequency dividers, etc. Which I've posted earlier about, but not ready to tackle yet. Learning the basics first. So here's my top secret drawing. It drives an LED, for starters. I'm sure I can use an n-Channel MOSFET to drive other things. This type of circuit is useful for (backup power supplies, and a fun Laser Tripwire). But I'm sure it can be used for much more technical things, like CPU chips, etc.

    Hunted001187.jpg
     
  2. hp1729

    Well-Known Member

    Nov 23, 2015
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    NPN and PNP devices are current activated devices, not voltage operated like FETs. Your basic circuit is good for the base circuit but when the transistor turns on the LED needs to be in series with the collector resistor.
    Drawing current through the base lead turns on the transistor (lowering resistance between emitter and collector). Your resistors are good values.
    Gate to source resistor? Base to emitter resistor.. If the input does not drive to a good low you need the base-emitter resistor to keep the NPN transistor off.
     
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  3. shteii01

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    Apply 0.7 volts to BE junction.
     
  4. blocco a spirale

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    Before you start another thread, please read this:
    https://en.wikipedia.org/wiki/Concision
     
    Last edited: Oct 8, 2016
  5. Papabravo

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    Who knew, a guy whose first language is probably not English, could be so erudite. Bravo Blocco!
     
  6. shortbus

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    Using the Vgs(th) as a turn on point is wrong. For what your trying, don't even look or consider the Vgs(th) value. Use that low of a voltage to turn the mosfet you chose will cause the mosfet to heat up very fast. Look at and use the same voltage as shown in the line of the data sheet called "Id", this is the fully turned on, low resistance point, gate voltage of a mosfet.

    The mosfet you picked is not a good one for use with a micro controller, you should chose one that is known as a "logic level mosfet".
     
  7. Guest3123

    Thread Starter Member

    Oct 28, 2014
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    So if the NPN BJT Transistor is a current controlled device, then how is the amount of base current determined to turn on the NPN BJT Transistor? What values on the datasheet determine this?

    Using the data found in the NPN BJT KSP05TA Datasheet.
     
    Last edited: Oct 8, 2016
  8. hp1729

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    hFE (current gain) is stated as 50 (minimum). collector current is 10 mA. So base current is 0.0002 Amps for this condition.
    Fig 1, shows DC current gain at given collector currents. Collector current divided by gain gives you base currents.
    Usually there is a better graph for this but I don't see one on that data sheet. Best experience ... transistor exercise. Plug it into a breadboard. choose a collector resistor to give you a maximum of 10 6mA (or what ever example is shown on the data sheet). Select various base resistors, starting high. Measure base voltage, collector voltage. Calculate base and collector currents. Calculate gain. Go to the next smaller resistor and repeat the process. When the transistor reaches saturation select a different collector resistor and repeat the whole process.
    Notice how much gain changes. Graph base voltage, collector voltage, currents, gain ...
    Excel is handy for this.
    Very educational. Would you like an example? I have to convert the Excel files to PDF to post them here.
     
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  9. Guest3123

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    Oct 28, 2014
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    If you would like to show me an example, that would be good.
     
    Last edited: Oct 8, 2016
  10. GopherT

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    I was hoping the Wikipedia entry would have been more detailed.
     
  11. Guest3123

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    Oct 28, 2014
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    Honestly.. I think I'm showing effort here.Here's my final circuit, to see if I now understand the NPN BJT Transistor, and how it can be turned on.
    I think I nailed it. Also here : http://i.imgur.com/xQjWt10.jpg

    So if the KSP05TA NPN BJT Transistor has a a maximum Collector - Emitter Current (Ic) of 500mA, and the HFE is 50, then the MAX Base current is.. 0.5A (Ic) x 50 HFE = 0.01A (Ib) Correct?

    NPN BJT TRANSISOTR CIRCUIT FINAL.jpg
     
    Last edited: Oct 8, 2016
  12. GopherT

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    @Guest3123
    Sorry but your in for a redo but you are getting there.

    E is connected to ground so it is at 0 volts.

    If voltage is applied to the base (B) it will behave like a diode and as voltage at the base is increased, it will essentially stop increasing at B once 0.7 V is reached so, in this case, B will be at 0.7 V.

    Now that you know those two points, try again.
     
  13. hp1729

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    The circuit is simple. An NPN transistor (2N3904) with a collector resistor (5,100) and a base resistor that gets changed.
     
  14. Guest3123

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    Oct 28, 2014
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    I was taught by numerous sources online that the emitter has a voltage drop (diode) of usually around 0.6 - 0.7v. That's why I thought that.
    That's why 5v supply minus 0.7v E Vd = 4.3v Ve.

    So it doesn't matter what voltage I put into the base? It comes to 0.7v..? How's that possible?

    This is where I was learning how to do the circuit I did.
    5DoYuxDzczQ?t=2m32s
     
  15. GopherT

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    The "diode" is from base to emitter. It is represented on the BJT schematic symbol with the arrow, just like a diode. One diode drop.
     
  16. Guest3123

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    I'm truly lost again. I don't know what the hell is what now. Sorta.

    NPN CIRCUIT try 2.jpg
     
  17. GopherT

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    Do you have a box of real parts and battery with a volt meter?
     
  18. hp1729

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    Welcome to the real world exercise. The base voltage does change as current through it changes. Just like a diode. Previous to this the student did an exercise on diodes where this is discovered. These exercises are from a closed technical school, Nevada Institute of Technology, Las Vegas.
     
  19. GopherT

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    It is a very small change. Lets get to the details of the B to E drop other than 0.7V as a second refinement of his knowledge.
     
  20. hp1729

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    Agree, the change is not linear. But it is not a constant 0.7 Volts.
    The suggested diode exercise is just a forward biased diode with varying resistors. You start with a high value resistor, measure the diode voltage, calculate the current and the effective resistance of the diode. The objective is to learn the characteristics of various diodes. By the end of the exercise you should be able to measure the forward voltage of a diode in a circuit and know about how much current is going through it. This was a component level technician's troubleshooting class. Far short of engineering though. Simple designing is taught but engineering goes much further than the class covered.
    The objective was slot machine repair.
     
    Last edited: Oct 8, 2016
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