Connect with us

LM13700 voltage controlled LP filter derivation

Discussion in 'General Electronics Discussion' started by george2525, Sep 7, 2016.

Scroll to continue with content
  1. george2525

    george2525

    170
    0
    Jan 30, 2015
    Hello

    I would very much like to understand how the following is derived if anyone has an idea. A full derivation would be great but some method pointers and/or references would be equally appreciated. I have chosen the simplest circuit from the datasheet (the LP filter) in the hope that if I can understand this I can move onto the more complex ones. I understand how the LM13700 works in terms of its internal circuit but things like this leave me confused as ive only worked with discrete (and simpler) circuits in the past

    the official formula is here also

    Thanks to anyone who knows how to approach this

    lm13700 LP.PNG
     
  2. george2525

    george2525

    170
    0
    Jan 30, 2015
    OK I have managed to derive this but only if I ignore the darlington pair and pretend that Vo = Vcap

    if I do it like this the answer is the same as the datasheet.

    But is this correct? surely the darlington pair affects the labelled Vo and it cant be the same as Vc right?

    anyone?
     
  3. Alec_t

    Alec_t

    2,980
    806
    Jul 7, 2015
    Without analysing the circuit in detail I would expect the Darlington to give a DC offset of Vo relative to the cap voltage but not to affect the filter's AC frequency response. By Vc I assume you mean the cap voltage, not the control voltage?
     
  4. george2525

    george2525

    170
    0
    Jan 30, 2015
    Yes by Vc I meant Vcap. If you do "Vo/Vin" = Vcap/Vin then solve for fc it gives the quoted value.

    I understand that the frequency response will be the same in terms of cutoff but in my analysis I use V- = [RA/(RA+R)]*Vout which is fine but the I also use Vout = Iout*Zc which doesnt make sense, yet I get the correct result.

    The darlingtons and the negative voltage are very confusing to me
     
  5. OBW0549

    OBW0549

    157
    117
    Jul 5, 2016
    Just think of the Darlington pair, the 10K resistor connected to the Darlington's emitter, and the negative supply as a unity-gain buffer amplifier: high input impedance, low output impedance, and a gain of 1.

    It's only there to isolate the capacitor C from whatever load is connected to the output, so the frequency response will be independent of loading effects.
     
  6. Ron Glum

    Ron Glum

    1
    0
    Sep 11, 2016
    The Darlington is simply a high input impedance buffer. It will have an offset voltage but because of the feedback this will appear at the cap rather than the filter output. Sounds to me like you approached it correctly. The output voltage is the amplifier output current multiplied by the capacitor impedance. The amplifier output current in turn depends upon the gm setting and the voltage difference between its input terms. I've attached a short derivation where I used 1/SC as the capacitor impedance but 1/(jwC) would be fine. The cutoff frequency is where the real and imaginary terms become equal in the denominator.
     

    Attached Files:

    • x-1.png
      x-1.png
      File size:
      96 KB
      Views:
      106
  7. george2525

    george2525

    170
    0
    Jan 30, 2015
    Thanks. I can do that derivation and I have the same answer as you for Vo/Vin

    what would be realy helpful is an explanation or a link that explains why the gain of the darlington is 1

    also does the negative -15V have any effect on the feedback (seems not)? if not then what is its purpose. Im inclined to think it has something to do with the +1 gain but im not sure how to approach this. I do understand why the darlington has large input R and small output R but its gain troubles me here.
     
  8. Alec_t

    Alec_t

    2,980
    806
    Jul 7, 2015
    The voltage on the emitter of the Darlington is always 2 diode drops below the base voltage. Therefore any input voltage change at the base is matched by the same output voltage change at the emitter. Hence the gain is 1.
     
  9. george2525

    george2525

    170
    0
    Jan 30, 2015
    But if the gain is 1 then wouldnt Vout = Vc ?

    I did a small signal analysis on the darlington and 10k resistor alone alone (with Vcc and the 10k connected to ground and no R and RA) and I can get a gain of around 1 if re1 and re2 are small (<<10k) but Im not sure this is the correct approach

    I got Vo/Vin (for darlington with 10k alone) as 10k/(re2 + 10k + re1/(B+1)) which looks ok

    typing in "gain of darlington pair" into google gives nothing
     
  10. Alec_t

    Alec_t

    2,980
    806
    Jul 7, 2015
    I was considering voltage change; not absolute voltage. dv(out) = dv(in).
     
  11. OBW0549

    OBW0549

    157
    117
    Jul 5, 2016
    The gain is unity because those two transistors are in the common-collector configuration, a.k.a. an emitter follower.

    Enter "darlington emitter follower" into Google and you'll get beaucoup stuff.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-