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Help analyzing active full-wave rectifier.

Discussion in 'Electronic Design' started by MRW, Oct 17, 2006.

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  1. MRW

    MRW Guest

    Hello everyone,

    Good evening!

    Active full-wave rectifier circuit:
    http://img273.imageshack.us/img273/6397/fullwaveactivedm9.jpg

    Section of active full-wave rectifier circuit
    http://img117.imageshack.us/img117/2217/fullwavepartza5.jpg


    I need some help analyzing the active full-wave rectifier circuit in
    the link above. I am wondering if someone can check if my work is
    correct.

    So far, I have the following:

    VA1 = Vin * (R2)/(R1+R2)

    VA1 = VA2 = VOA = VB1 = VB2

    Then I decided to analyze part of the circuit as shown in the second
    link:

    -I3 + I4 + I5 = 0

    [(Vin - VB2) / R3] + [VB2 / R4] + [(VB2-VOB) / R5] = 0

    VB2 * [ (1 / R4) + (1 /R5) - (1 / R3)] = [VOB / R5] - [Vin / R3]

    VB2 * [ (R5*R3 + R4*R3 - R4*R5) / (R4*R5*R3) ] = [VOB / R5] - [1 /
    R3]*[VA1*(R1+R2)/R2]


    To keep things simple, I've set aside the following:

    TMP1 = (R5*R3 + R4*R3 - R4*R5) / (R4*R5*R3)
    TMP2 = (R1+R2)/(R2*R3)

    So, I have:

    VB2 * TMP1 = [VOB / R5] - VA1 * TMP2

    Since VA1 = VB2, I have:

    VA1 * TMP1 - VA1 * TMP2 = VOB / R5

    VA1 * (TMP1 - TMP2) = VOB / R5

    Substituting Vin with VA1:

    (Vin * [R2 / (R1 + R2) ]) * (TMP1 - TMP2) = VOB / R5

    Letting TMP3 = [R2 * R5 / (R1 + R2)]:


    Vin * TMP3 * (TMP1 - TMP2) = VOB


    Vin = VOB * [1 / (TMP3 * [TMP1 - TMP2])]


    Have I done this analysis the right way? If not, where did I go wrong?


    Thanks!
     
  2. Ban

    Ban Guest

    You first have to look at your circuit and understand it. You know already
    it is a rectifier. This is not a linear function, where does it happen?
    There is no diode. Now you have a look at your opamps. Normally Vcc is on
    pin7 and GND/Vee on pin4. If you connect it the wrong way, you would see the
    blue smoke, what opamp is this supposed to be?
    Now to your homework.
    Analyze the circuit with two different cases: Vin>0 and Vin<=0. Why is Vee
    of the opamp tied to gnd?
    There is also a capacitor on the input, it will act like a highpass, can you
    simply ignore it?
     
  3. Eeyore

    Eeyore Guest

  4. Tom Bruhns

    Tom Bruhns Guest

    Of course the pin connections for a quad op amp are different than
    those that Ban gave, typical for a single op amp in an 8-pin
    package...but why would one not use two sections in the same quad
    package, instead of apparently two separate packages?

    Answers to Ban's questions should lead you in the right direction:
    there is a very good reason that Vee is indeed tied to ground.

    Yet another question: Does the circuit present the same resistance
    load to the input capacitor through both the positive and negative
    halves of an input waveform cycle? If it does not, what happens to the
    DC level on the right side of the capacitor when you put in a signal?
    Part of the answer to these questions has to do with what protection
    there is on the op amp input pins: what does the input pin current do
    if you try to drive the input negative?

    Cheers,
    Tom
     
  5. MRW

    MRW Guest

    Thanks, Ban!

    Actually, this is not a homework problem. I got this design from
    EDN.com (search for active full-wave rectifier). The SPICE simulation
    works, so I just wanted to see if I can understand its operations more
    by writing the transfer function.

    So far, my initial calculations indicate that VOB (output) will always
    be positive because it relies on the relationship between the resistors
    used in the circuit. But then again I'm not positively sure if I did
    the analysis the right way. I just used the two golden rules of opamps
    and also some nodal analysis.

    I'll try to plugin some values in my equation once I can get access to
    the university computer labs for Matlab.

    Thanks!
     
  6. MRW

    MRW Guest

    Thanks, Tom!

    I've only got the macromodel of one opamp from National Semiconductors.
    That's all they have. I really don't know much about SPICE modeling. I
    haven't delve much into it.
    Not really sure, I had the impression that the opamp inputs are
    high-impedance. I don't know how to go about answering this question.
    Any tips?

    I'm kinda new to electronic design, so please excuse my lack of
    knowledge.

    Thanks!
     
  7. Eeyore

    Eeyore Guest

    It has some diodes in it though !

    Graham
     
  8. Tom Bruhns

    Tom Bruhns Guest

    Well, most ICs have either intentional or parasitic diodes from inputs
    to at least the negative supply, so if you try to drag the input more
    negative than ground in this case, they start to conduct, and will only
    go "one diode drop" below ground at modest current.

    Presumably those op amps are ones that work when the inputs are at the
    same potential as the negative supply. The data sheet might say
    something like "input common mode range includes the negative supply
    voltage." And presumably the outputs will behave reasonably all the
    way to the negative supply voltage. Typically, they can't actively
    pull down quite all the way, but if you have a resistive load to the
    negative supply, the output just stops sourcing any current, and the
    resistive load drags the voltage down to zero. It's the inability of
    the op amp output to go negative that gives you the "diode" effect that
    makes the circuit a full wave rectifier. As Ban suggested, break the
    analysis into two pieces, one when the input is positive and one when
    it's negative. For a first pass, ignore any effects like current in op
    amp inputs when they are dragged negative, but if you build the
    circuit, such an effect would explain why you got a DC offset...

    Imagine, for example, +1V at the junction of R1 and R3 (the input,
    ignoring the capacitor). By inspection, what's the voltage at VA1, and
    therefore at VB1? (Recognize R1 and R2 as a 2:1 voltage divider.) If
    feedback is working around the second op amp, VB2 must equal (very
    nearly) VB1, and you can calculate the currents in R3 and R4, knowing
    the voltage at each end of each of them. The current in R5 must make
    the net current in the VB2 node zero. That tells you the voltage drop
    across R5. Be sure to get the direction right, and that drop added to
    the voltage at VB2 tells you the output voltage.

    The analysis for -1V at the input is different, because the output of
    the first op amp cannot go negative...it will just go to zero. And it
    should be clear how to do the rest of the analysis for that case.

    You should be able to find LOTS and LOTS of op amp macromodels. If you
    download LTSpice for free from Linear Technology, you'll get a whole
    bunch of macromodels included with it. And you should be able to find
    Spice macromodels from the TI, Analog Devices, and National
    Semiconductor web sites, to name just three of many, for a great many
    parts. You DO have to be careful about WHICH version of Spice the
    model was written for. Most seem to be Spice2, and often use a POLY
    keyword in some of the dependent sources. That's bad news for use in
    Spice3, and you need to translate them, if you use a Spice3-based
    simulator such as LTSpice. Well, as far as I know, LTSpice doesn't
    automatically do the translation for you, but it's surprised me with
    its capability before, so maybe it does. Anyway, a time-domain
    simulation over one cycle should show you just what's going on in the
    circuit, and make it clear how the "rectification" happens.

    Cheers,
    Tom
     
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