differential amplifier project

Discussion in 'The Projects Forum' started by boby4u, Jun 6, 2007.

  1. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    Hi
    I have a question about differential amplifier.
    I`ve designed a differential ampli with two pnp transistors
    those are bc807-40 .I had never worked with a pnp tran before.
    the question is getting the most possibe gain from the circuit.
    as you see in te pictures I have connected a current mirror as
    a current source under the tho emitters.
    I have two Rc with the value of 21 k.and because my current mirror is giving me
    1.221 mili Am those Rc were the highest I could put there not allowing
    the transistors to go to saturation.
    Was my design correct or you see problems in that?
    do you have any Idea to make the gain higher?
    the schematic of my circuit is below:
    I look forward to your guidance.
    thankyouthankyouthankyouthank.

    [​IMG]

    [​IMG]
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I would have thought you needed to use a PNP constant current source and reference its base bias to the ground rail and its emitter via the emitter resistor to the ground power rail rather than using an NPN as you schematic indicates.

    You may want to try reconfiguring the circuit with this in mind and see what gain you can obtain from your simulation.

    hgmjr
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    This is what I had in mind.

    The 15V power source is a negative voltage. The plus sign is an artifact of my image capture process. Sorry if it is a slight bit confusing.

    hgmjr
     
  4. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    thanks for your help hgjmr.
    But I wanted the current mirror only for giving
    current to my transistors.
    the current mirror is giving me 1 mA and is constant.althought both of the ways are true.
    now I have a problem in calculating maximum swing of output voltage.
    imagine that we did not have any value for Rc. I want to understand how to find Rc
    according to the max swing of Vo ???
    Ido`nt khow how to calculate max swing of Vo??
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    By the phrase "calculate max swing", are you interested with finding the maximum useable headroom for the amplifier or are you interested in computing the gain of the amplifier (Vout/Vin)?

    hgmjr
     
  6. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    the question is finding the max swing of Vo.
    then by means of that finding Rc
    after that finding gain(Vo/Vin)
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I think hgmgr's point was that you can't use an NPN current source. You have created a monster whose differential pair is accidentally biased in the active region. See the annotated schematic below.
    You need a PNP current source, as hgmjr pointed out.
     
  8. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    thanks for your help hgjmr & RON H.
    YEAH AS YOU SAID MY FIRST circuit was really accidentally giving me
    the current needed. my npn was in saturation that is not accepted
    in a current source.
    I changed it to a pnp one but the design that hgjmr gave me for current
    source was giving me only pico ampers. so I changed it like this.
    I put the 15v dc source behind the emitter to bias my current source.
    Now I am giving 2 mA whicht I needed .thanks to both of you.
    here is the new scheme . the two upper transistors does not seem to be on
    saturation but my Vout is cutted (up & down). whatever I PUT in Rc place
    did not effect what do you advise me to do right now.

    [​IMG]


    [​IMG]
     
  9. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    If I bring the V(in) lower to 0.1v then Vo is not cut.
    but I don`t really know wether there is a spechial rule about putting Rc
    here is what my friend told me about it:
    [Vcc+V(sat)-V(BEon)]/2=MAX SWING OF vO
    THEN MAX SWING=Rc*Ic
    what do you think about that?
     
  10. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    By the term "cutted" I assume you mean "clipped" at the top and bottom of the waveform.

    That is most often a sign that you have more gain than you need for the signal level being applied at the input.

    I also notice that you are now using a dual supply scheme with +15V and -15V. That should give you the potential for a greater output voltage swing. Your very first schematic only had a single 15V supply I believe.

    Since your input signal is DC coupled and referenced to ground, the signal swing on the collectors is going to be limited in how far it can go in the negative direction to around +0.7 volts due to saturation. I think this is what you have already observed.

    If you can ac couple your input signal then you could bias you differential transistor pair to a more positive voltage so that you could take greater advantage of the available power supply range.

    hgmjr
     
  11. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    Thanks alot for your guidance.
    :cool:
     
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    You don't need 40mA through your current source bias diodes. That's a huge waste of current.
     
  13. boby4u

    Thread Starter Member

    Jun 1, 2007
    26
    0
    what should I to instead of that .
    I think putting another transistor beside Q4 will work.
    when the base & collector are connected to each other.
    I will go through that.
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    There are many ways of making current sources. Below are three possibilities.
     
  15. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Ron H,

    Would you deem it appropriate to include a fourth current source possibility; that of the classic "current mirror"?

    hgmjr
     
  16. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Sure, that is another possibility. A current mirror using discrete, unmatched transistors requires "ballast" resistors in order to get predictable results, and if you use a 1:1 mirror, the bias current has to equal the output current (of course, that is nearly the case in the 2 diode bias network I posted, although it was not necessary). In ICs, ratioed mirrors are possible by scaling transistor sizes. In discrete circuits, it gets a little messy.
    I like current sources with operational feedback, but it's probably overkill in this situation. The current is more predictable, but the diff amp gain will still change with temperature.
     
  17. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Actually what I had in mind was the use of one those dual-transistor IC's containing matched transistors. Transistor matching is probably not such a problem in simulation since the use of the same model for both devices provides a matched scenerio. That is one of the pitfalls of simulation. Real world circuits tend to bring their own set of variables that are often missing from the simulated version.

    Upon reflection the use of an dual-transistor IC may violate the premise of keeping it simple by basing the design on discrete transistors.

    Oh well, just a thought.

    hgmjr
     
  18. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I had temporarly forgotten about matched-transistor ICs. What you say about simulation is very true. It is possible to see the effects of mismatching by adding the parameter m to a semiconductor. In LTSpice, you do it by editing the transistor's part number. For example, you would right-click on the part number, and change it from "2N3906" to "2N3906 m=1.1", or whatever. M is the size-multiplication factor (in this case, the area of the B-E junction is what matters most).
     
Loading...