Transistor Circuit - Base and Collector current.

Discussion in 'General Electronics Chat' started by Dr.killjoy, May 6, 2016.

  1. Dr.killjoy

    Thread Starter Well-Known Member

    Apr 28, 2013
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    The cicuit I using is a simple 2N2222 led switch.. What I don't understand is that 2n2222 is current driven and we use a resistor to limit the current but doesn't the voltage drop cause of the reisitor ??Sorry for the noob question but still try to learn.
     
  2. wayneh

    Expert

    Sep 9, 2010
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    The base-emitter current and the collector-emitter current need to both be limited. Often the load limits the collector current. For a typical switch, you would control the base current to 5-10% of the current in the load.
     
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  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    There are two resistors, one to limit the collector or emitter current through the LED, and one to limit the current into the base.
    If you post a schematic or sketch of your circuit, we probably can explain it. Be sure to label each part.
    R1, R2, etc. for resistors
    C1, C2, etc. for capacitors
    Q1, Q2, etc. for transistors
    D1, D2, etc. for diodes, including LEDs

    ak
     
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  4. Dr.killjoy

    Thread Starter Well-Known Member

    Apr 28, 2013
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    Here is what I am playing with

    20160507_091724.jpg
    Mods Note:
    Please compress the file size to a small and clear resolution as 800x600 or 1024x768.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    If the load (R3 and LED) is connected between emitter and ground the base resistor (R2) and the collector resistor (R1) is not needed and you can remove it from the circuit (replace this resistor with a piece of wire). Because in this case the BJT work as a emitter follower. And in this case R3 (Ie) will "set" the base and collector current.
    Ie = Ib + Ic = Ib + β*Ic = (β + 1)*Ib ---> Ib = Ie/(β + 1) and Ic = β/(β + 1)*Ie
    And for this circuit you do not need the BJT use only a switch and R3+LED
    http://forum.allaboutcircuits.com/threads/emitter-follower-analysis.83287/#post-593745
    http://forum.allaboutcircuits.com/threads/transistor-biasing.37227/#post-235682
    http://forum.allaboutcircuits.com/threads/which-method-npn.97963/#post-731172
     
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  6. wayneh

    Expert

    Sep 9, 2010
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    That's called an emitter-follower configuration (as opposed to a low-side switch) because the base and emitter voltage rise to whatever is at the top of the load. You could eliminate either R1 or R3.

    Gaaa, Jony130 beat me to it, with a more thorough answer to boot. I need more coffee!
     
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  7. BR-549

    Well-Known Member

    Sep 22, 2013
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    The purpose of a resistor is to limit current, not set voltage. Now we can set a voltage with a resistor, but we do that by applying and setting current.

    A resistor is current device, not a voltage device. A resistor is useless and non-functional without current.
     
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  8. Dr.killjoy

    Thread Starter Well-Known Member

    Apr 28, 2013
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    But doesn't the resistor change the voltage also ..
     
  9. BR-549

    Well-Known Member

    Sep 22, 2013
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    It will depend on configuration. For instance, a battery and one resistor circuit. Or battery and parallel resistors.

    The voltage across a resistor does not come from the power supply. Even if you put a resistor across the supply. It HAS to be INDUCED by current flow. The INDUCED voltage in inverted and opposes the supply. You have to reverse voltmeter leads to measure induced (voltage drop) voltage.

    Being able to control the voltage drop across the resistor, independently of the power supply, is extremely handy.

    Let's say we need a certain voltage somewhere on a pin or terminal in the circuit.

    By limiting the current thru the resistor, we can set and apply that voltage (which is voltage drop) at that point.

    So, a secondary benefit of limiting current.........CAN be setting a voltage.

    Edit...You can think of a resistor as an in-phase voltage inverter. The amplitude of inversion is controlled by current. A current device.
     
    Last edited: May 7, 2016
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  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Yes.

    Per Ohm's Law, if there is a current through a resistor, then a voltage difference develops across it. But also, if a circuit impresses a voltage across a resistance, it *pushes* a current through the resistor based on Ohm's Law (as long as it can supply the required current). Your circuit has several relatively fixed voltages that restrict the voltage that it impresses across R3. For your circuit, it goes like this:

    First, eliminate R1 unless there is a specific reason to keep the 12 V away from the Q1 collector. It makes the calculations much more messy, and does nothing that R3 can't do. Connect Q1's collector directly to Vcc.

    Choose an LED current. Let's start with 10 mA.

    The LED is a diode and has a forward voltage drop, Vf. For a standard little red or green LED, assume 2.0 V. Reds run around 1.8-1.9 V, and greens are around 2.1 V.

    The base current will be about 1% of the emitter current, so that's 100 uA. Across 10 K ohms, that's 1 V.

    The base-emitter junction is about 0.7 V, so the total voltage drop from Vcc to the emitter is approx 1.7 V.

    The voltage across the resistor is 12 - 1.7 - 2.0 = 8.3 V.

    Since you want 10 mA through the LED, that sets R3 at 830 ohms. 820 is the closest 5% value.

    The way this works is that there are three relatively constant voltage drops that do not change much even with wide changes in the output current. Yes, the voltage across R2 is directly proportional to the output current, but it is less than 50% of the total of all of the "fixed" voltage drops. So if you rerun the calculations for 5 mA or 20 mA (a 4-to-1 range), the calculated R3 resistor value range will not be that large.

    I said "push" above because voltage is the quantity of Electromotive Force, or EMF (in the same way that amperage is the quantity of current). In the classic analogy, voltage equates to the force of water pressure, and pushes electrons around.

    ak
     
    Last edited: May 7, 2016
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  11. Dr.killjoy

    Thread Starter Well-Known Member

    Apr 28, 2013
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    I wanted to say Thank you very much for helping out and I finally understood what is going on and how it works ..


    Thanks Again
    Jay
     
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