Electronics BJT using Re moddeling

Discussion in 'Homework Help' started by skusku, Jul 28, 2012.

  1. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Hi guys

    This is my 1st post here
    I would like to ask some help with electronics homework.

    This is the circuit
    http://postimage.org/image/9k24jiu8b/
    I forgot to add the transistor in the pic - It is a Q2N2222

    Design a voltage-divider bias circuit to operate from a 12V supply. The bias conditions are to be Vce=3V, Ve=5V, Ic=1mA. Take Vbe=0.7V
    Use E-12 values for resistors. Use Approximate circuit analysis. As a rule of thumb use voltage-divider current approximately equal to one tenth of the transistor collector current.

    ----------------------------------
    Basically I want to work out the 4 Resistors.

    The lecturer said was that approximate method means Ic=Ie.
    The thing im not sure about is the part about the "voltage-divider current being one tenth of transistor circuit.

    This is what ive done so far:

    R4=Ve/Ie = 5/1mA = 5K ohm (E-12=4.7K)
    R3=Vrc/Ic= (Vcc-Vce-Ve)/Ic=(12-3-5)/1mA= 4K Ohm (E-12=3.9K Ohm)

    Vb=Vbe+Ve = 0.7 + 5 = 5.7V
    R2< (1/10).Beta.R4 = (1/10).(50).4700=23500 (E-12=22k)
    [Ive taken Beta from the datasheet as 50 where Ic=1mA and Vce=10V]

    Vb=(R2.Vcc)/(R1+R2) Rearrange equation where R1 is subject gives:
    R1=((Vcc.R2)/Vb) - R2 = 24315 Ohm (E-12=22K)

    Can you verify if this is correct?
    There is a simulation for Part 2 as well, to determine Beta and re using pspice, but lets get to that part once Resistor values are verified.

    Hope you can help me :confused: :D
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Why don't you fallow this rule
    voltage-divider current approximately equal to one tenth of the transistor collector current

    Ic = 1mA so Voltage divider current is equal Ic/10 = 100uA.

    So we have

    R2 = 5.7V/100uA = 57K = 56K
    R1 = (12V - 5.7V)/100uA = 63K = 68K

    But you resistor value are also looks good, so don't worry too much.
     
    skusku likes this.
  3. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Jony I think you are correct with the "voltage divider current approximation part". I was not so sure by what they meant and where that current flows (thats why i colored in the parts to better distinguish the currents). But it makes so much sense now! Thanks alot.

    The last part of Beta and re from the simulation: I will change the values in pspice and rerun the app tomorrow(its already midnight now and pspice is not installed on my pc Im at now). I will post my findings(questions) here then. Thanks so far!! I appreciate it alot.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Note that it is one-tenth "of the transistor-collector current", which is a bit more specific and meaningful than just "of transistor circuit"

    What might still be unclear is why this rule of thumb is what it is.

    In a voltage divider, if the current in the two resistors isn't the same, the voltage at the junction will be different than what you calculated by ignoring the base current being drawn out of the junction. How much of an affect this has is dependent on how large the difference is as a fraction of the nominal (unloaded) current. So we want the current flowing through the voltage divider resistors to be much larger than the base current. On the other hand, we want to use the smallest voltage divider current we can get away with for power dissipation and other reasons. So we might say that we want the voltage divider current to be 10x to 20x the base current and not a lot larger than that. Okay, but we don't know the base current and we are actually assuming it is zero. Well, small signal transistors generally have betas that are in the 100 to 200 range, perhaps a bit higher. So the base current will be something around Ic/100 and we want the voltage divider current to be 10x that, or Ic/10.

    Will this put you exactly where you would like to be? No. But it will put you in the ballpark and for many elements of practical designs, 'in the ballpark' is good enough, and 'good enough' is also known as 'job complete, start production and pay me'.
     
  5. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Last part says Determine Beta and re of the circuit using pspice, take Vt=26mV

    What ive done is:
    Beta=Ic/Ib=968.2uA/6.341uA=152 Ohm
    re=Vt/Ie=26mV/974.6uA=26 Ohm

    In our book where we should write down the answers for re it looks like this:

    re(from simulation) = ..................k Ohm

    The answer I get for re is just the 26 Ohm. Is it possible for re to be in the kilo ohm range anyways?
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    How can a current gain β = Ic/Ibcan have Ohm dimension ?
    Check the units again.

    As for re - a small signal emitter internal resistance seen from emitter terminal when we looking into emitter with base shorted to ground.

    re = Vt/Ie ≈ Vt/Ic = 26mV/1mA = 26Ω = 0.026KΩ
    But I'm 100% one hundred percent sure that simulation will give you different answer.
     
  7. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Sorry for β, that was just a stupid mistake where i noted the ohm along.

    With re I simply typed in 26mV/Ie in the simulation part(as on a calculator). Why will the simulation give me a different answer than that one?
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    OK I understand you simply use a calculator. But simulation program give as a opportunity to measure re in simulation. And simulation will give the different answer because Vt is temperature dependent. And most of the simulation program perform there simulation for 27°C so Vt will not be equal 26mV.
    Also exact formula for re is equal re = Vt/Ie
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    If I were told to measure re from the simulation, I would need to ask whether we were expected to measure Ie from the sim and then plug it into the formula (which seems implied since you are told what value for Vt to use), or should I use the differential definition, namely re = d(vbe)/d(ie). I think the latter is far more educational.
     
  10. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1

    Well as I understand it we should measure using whatever value pspice gives us for Ie, and then use 26mv for Vt. The lecturer says pspice sees the transistor as 1 whole unit with standard values. Now he hinted that we should change some properties of the transistor in order to get the right values. Im assuming its got something to do with Beta.
     
  11. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    I have NO idea what your lecturer is going on about.

    If you are going to change some properties to "get the right value", where are those "right values" coming from? If you have some actual data from a real circuit, or have nominal data for a particular transistor, then this may be reasonable to do.

    Depending on the model level being used, changing the properties of the transistor (other than intended user-defined parameters for that model, such as various physical dimensions or temperature) is not too easy.
     
Loading...