NPN exercise - math is not matching the LTSpice simulation

Discussion in 'Homework Help' started by PsySc0rpi0n, May 27, 2014.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hi once more...

    I was trying to calculate Ic, Ib, Ie and Vce to the attached circuit, but my math is not matching the LTSpice simulation.

    I have tried to simplify the circuit eliminating the Base Voltage Source and Base Resistor and replacing the Zener by a Voltage Source with the same value but looks like it's not the same thing because LTSpice is not showing the same results...

    Can anyone help me finding what am I doing wrong here?

    PS: I'm writing now my calculations!

    I'll put them here in a bit!

    By the mesh's laws:

    Vcc = RcIc + D1 + Vce + ReIe
    Zener - Vbe -ReIe = 0
    Ic = βIb

    Note: Ie = (β+1)Ic/β and i have assumed that the voltage drop at the LED is 0.7.

    12 = 100Ic + 0.7 + Vce + 220(257/256)Ic
    5.6 - 0.7 - Re 220(257/256)Ic = 0
    Ic = 256Ib

    Ic = 0.02218A
    Vce = 4.181V
    Ib = 0.0000867A
     
    Last edited: May 27, 2014
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Try a simpler circuit. Use two voltage sources, transistor, resistor at the base, led between voltage source and collector.

    I can give you an example from my textbook (but it uses motor instead of led):
    Rb=240 Ohm
    Vcc=12 V
    Vin=12 V (this is voltage that connects to Rb so you have Vin to Rb to Base of 2N2222)
    Vbe(on)=0.7 V
    Vce(sat)=0.1 V
    Beta=75
    Rc(motor)= assume 5 Ohm; Stick the led there. I think we will have to ignore the Rc. Does LTSpice have volt and current meters that you can stick into the circuit?
     
    Last edited: May 27, 2014
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Thanks for your reply...
    For now I really need to understand what's wrong with my math in my circuit because I'm going to have exam tomorrow and I need to sort that circuit out.

    After that, I can try your example. So, please, don't take me wrong, but I really need help to figure out my mistake in my circuit for now!

    Thanks
    Psy
     
  4. jjw

    Member

    Dec 24, 2013
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    You have the zener voltage 5.6V in your equation, but because it is higher than 5V the zener is not conducting and can be omitted from the circuit.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Why would you assume that the voltage across an LED is only 0.7V? Typical values are in the 1.8V to 3V range. In the lack of specific information, 2V is commonly assumed.
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    So I have to give a value to the Voltage source that is connected to the base in such way that I still have over 5.6V to correctly polarize the Zener, right?

    I can try it with 2V... Let me just repeat the calculations!

    I still don't know If I can eliminate the Voltage source and the R and replace the Zener with a 5.6V voltage source as I have done in the circuit at the right in my 1st post!
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    As has already been pointed out, if you have a 5V source, then the Zener diode is in cutoff and has no effect on the circuit. Remember, a Zener diode does not produce energy, so how can you apply 5V to a resistor/diode combination and all of a sudden have the Zener diode be 5.6V? That would mean that the resistor would have 0.6V across it in such a way that current was flowing from the Zener diode and into the voltage source which would mean that the voltage source would be absorbing power. Does that make any sense?
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I got that mate... I know that that voltage source must be higher in order to have a voltage higher than the Vzener so that it can be correctly polarized.

    What I was looking is to know how to calculate the Voltage source value so that it could still be possible to correctly polarize the Zener! And also to know if the other circuit is equivalent to the original one!
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    If, by correctly "polarize the Zener" you mean make it conduct a reverse current, then the voltage source has to be greater than the Zener voltage of the diode (as you stated in the previous paragraph).

    You have one circuit that has a 5V source and a Zener diode that is cutoff. You have another circuit that has a 5.6V source. You say that you understand that you need a voltage source higher than 5V in order to turn on the Zener. Yet you are still wondering if the two circuits are equivalent?
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I think we are misunderstanding each other...

    I got that I need to have an higher voltage source to make the zener to conduct a reverse current. I got that those 2 circuits are not equivalent because of the 5V voltage source is too low.

    What I want is to know How to calculate the Voltage Source value so that the left version of the circuit is equivalent to the right version, knowing that the right version is a correct simplified circuit of the original one (left version without any value for that Voltage Source).
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Okay. I think I understand what you want. The answer is that you can't.

    All you know is that the voltage source is enough greater than 5.6V so as to get the transistor base current (and minimum Zener current) to flow through the 1kΩ resistor. But it can be anything above that voltage -- more current will flow in the 1kΩ resistor, but all of the additional current will simply flow down through the Zener.

    Now, in the real world there would be a limit on how much current could flow before you exceeded the maximum current rating of the Zener. In addition, the voltage of the Zener would change slightly as more current flows in it.
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, but can I simplify the original circuit (the one on the left) and work with the simplified one (the one on the right), I mean, is the simplification correct?
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, you can but only if you increase V2 value on the original circuit.
    Because now Zener diode don't work as it should. And this is why Vg1 is not equal 5.6V
     
  14. PsySc0rpi0n

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    Mar 4, 2014
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    I'm confused..

    Is the circuit on the right equivalent to the one on the right assuming V1 has enough voltage to correctly polarize the zener?
     
  15. Jony130

    AAC Fanatic!

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    For DC calculation you can this equivalent circuit.
     
  16. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, using the circuit on the right, I have made the following math:

    12=2+100Ic+Vce+220((257/256)Ic)

    0.7 + 220((257/256)Ic) - 5.6 = 0

    Ic = 256Ib

    and I got the following results:

    Ic = 22.17mA
    Vce=2.88655V
    Ib=86.7uA

    But LTSpice doesn't say the same!
     
  17. Jony130

    AAC Fanatic!

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    What values LTspice gives you ?
    What is the difference in percentage?

    For 5.6V we have Ie = Ic = (5.6V - 0.7V)/220Ω = 22.28mA;
    VRc = 100Ω*22.28mA = 2.228V. And Vc = 12V - 2V - 2.228V = 7.772V
    And Vce = 2.8V
     
  18. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    LTSpice reports the following values:

    Ic = 22.146mA
    Ib=90.8512μA
    Vce = 1.20519V

    And the Vc you calculate, LTSpice shows 6.097V

    I'm measuring Vce probing the circuit between Vclo2 and Vem2 and not Vcol2 to GND...
     
  19. Jony130

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    I still don't see any problem. The answer is simple. We assume VLED = 2V but simulation show 3.68V. And this is why we are slightly off for Vce and Vc. But we are ok with the current. And for the diode the current is important not the voltage. Also R4 is not necessary, serve no function.
     
  20. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    How have you measured the VLED???
     
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