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Small-signal equivalent circuit models - How?

  1. Dec 11, 2007 #1
    [SOLVED] Small-signal equivalent circuit models - How?

    Basically I've got an exam in circuitry this thursday and I've noticed that every single exam has one of these amplifier circuit excercises in them where the first part is usually finding the small-signal equivalent of said model and the other parts are based on this.
    I am however finding it rather hard seeing how I'm supposed to go from circuit model to small-circuit equivalent, as the book never mentions it thoroughly, it basically just says; "hey here's a circuit, here's the equivalent, have fun".

    So anyway, I'll illustrate an example and say what I think I've understood of it.
    Here's an amplifier circuit with a MOSFET transistor:
    [​IMG]

    And then I'm supposed to create the small-signal equivalent. The first thing I think I know is that when constructing the small-signal equivalent is that we're now only interested in the small signal, and for the small signal we can view any capacitors as short-circuited.
    Alright. At the same time, I think we can short-circuit all voltage supplies and remove all current supplies, and then basically evaluate where the current from the small-signal Vsig will travel to see how the equivalent small-signal circuit will be. The transistor should also be replaced with a voltage-controlled current source.

    Am I missing anything here now?

    Basically following what I've said, Vsig should be in series with Rsig, and the voltage-controlled current source should be paralell with RD and RL.

    I know this is supposed to be how it looks:
    [​IMG]

    The thing is, even with what I know, I could never ever have eventually ended up at this solution by myself. Do you think anyone here can give me a nudge in the right direction on how to design these small-signal equivalent circuits or direct me to a webpage which could describe it for me?

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 12, 2007 #2
    Okay I've been trying really hard to figure this out and I think I understand now - There's just one question remaining really;
    When designing the small-signal equivalent circuit - How do I know which of the ports Drain, Source and Gate to chose as ground?

    It's pretty obvious from the upper example that I chose gate as ground as, well, obviously it's connected straight to ground, however, for other excercises? Anyone?
     
  4. Dec 12, 2007 #3

    CEL

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    You have your input signal in one terminal (in your example the Source) and your output signal in other (the Drain). The third is the common terminal (the Gate).
    The common terminal is not always connected to ground. It may exist a resistor between the terminal and ground that provides negative feedback to the amplifier.
     
  5. Dec 12, 2007 #4

    chroot

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    When you make a small-signal model, you are implicitly linearizing your circuit around its dc operating point. That means, for the sake of analysis, all of the independent current sources go to zero. (Current sources now drive zero current and thus are open circuits; voltage sources drive zero voltage and are thus short circuits.) You do leave your independent stimulus source, though -- Vsig in this case. Otherwise you'd have nothing to analyze.

    You cannot necessarily consider capacitors to be shorts unless you are told that the signal frequency is much lower than the bandwidth of the RC's.

    Finally, you substitute the transconductor model of the MOSFET device. Sometimes you need to provide its ro, sometimes you do not -- it depends on how small ro is compared to other resistances at that node.

    The source of a MOSFET is the terminal which provides its majority carriers. An NMOS device uses electrons as majority carriers, so its source is the terminal with the lowest voltage, nearest to ground. For a PMOS, the source is the terminal with the highest voltage.

    - Warren
     
    Last edited: Dec 12, 2007
  6. Dec 12, 2007 #5
    Wow, that's some really helpful information, and I learnt some extra's!
    Thanks to both of you!

    I think I understand how I'm supposed to design the small-signal equivalent circuit now.
    I can draw up the terminals source - drain and gate, fill in the voltage controlled current source between drain and source, and after that see where the "cords" go from gate, drain and source. Uh, man it's kind of hard to explain but you get the picture.

    I think I'm on to it, anyhow. My only problem is knowing wether "ground" should be at Gate, Drain or Source in my equivalent circuit model. I've seen a few excercises where the ground symbol have been straight beneath source, yet on the circuit equivalent they chose to put it at drain, making the equivalent circuit somewhat different then the one I made.
     
  7. Dec 12, 2007 #6

    chroot

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    Well, remember that when a voltage source (like the supply) is "zeroed," it becomes a short. Since the supply voltage source is between ground and VDD, when you zero it, you short VDD and ground. This means that, for the purposes of small-signal analysis, VDD and ground are now the same node. This makes sense, if you think about it -- small-signal analysis deals only with small ac signals, and the supply does not respond at all to ac signals. It's an "ac ground," in the sense that it is irrelevant to ac signals.

    Depending upon how the transistor is connected, you may end up with ground at any terminal in your small-signal model. If a node is connected directly to VDD in the original circuit, it will be grounded in the small-signal model.

    - Warren
     
    Last edited: Dec 12, 2007
  8. Dec 12, 2007 #7
    Hmm, alright then. So I just need to make sure all nodes connected to one terminal AND ground is also connected to the same terminal and ground in the small-signal equivalent? I'm fairly sure I got it now, your help is extremely appreciated at this moment.
     
  9. Dec 12, 2007 #8

    chroot

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    I don't know what you mean. All I'm trying to say is that you zero the supply, so that anything that was connected to VDD is now connected to ground.

    - Warren
     
  10. Dec 12, 2007 #9
    Ah yes, I do understand that. I'm just asking, really; When creating the small-signal equivalent model - Every node connected to ground in the original circuit needs to be connected to ground in the small-signal model, aswell as to the same resistances/voltage supplies they were in the original circuit, correct? In which case I'm fairly sure I know what to do when designing these circuits. I was just working under an incorrect assumption when asking some of my previous questions which you somehow cleared up for me.

    Thanks again!
     
  11. Dec 12, 2007 #10

    chroot

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    Yes, ground remains ground. Other nodes, however, like VDD, become ground, too.

    - Warren
     
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