1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Op-amp voltage regulator

  1. Jun 24, 2016 #1
    1. The problem statement, all variables and given/known data
    FIGURE 3 (on page 7) shows a PIR (passive infra-red) detector and its
    associated amplifier, as used in burglar alarm systems1.

    (a) The detector is powered from a 12 V unregulated supply that needs
    to be stepped down to 5 V. Design a suitable 12-to-5 V voltage
    converter using an op-amp and a diode that has the forward
    characteristics given in FIGURE 3(b).

    (b) Estimate the quiescent voltages at the inputs and outputs of the two
    op-amps and the overall voltage gain (in decibels) of the circuit at the
    frequency of operation. State any assumptions made.
    2. Relevant equations
    [itex]V_-=V_o\frac{R_2}{R_1+R_2}[/itex]
    Vo = GV (V+ – V–)
    Vd = V+

    3. The attempt at a solution
    I think pic 2 is the circuit they are asking for.
    as for the rest.


    Any help would be appreciated.
    Thanks
     

    Attached Files:

  2. jcsd
  3. Jun 28, 2016 #2

    donpacino

    User Avatar
    Gold Member

    What is your actual question?

    Why don't you give the problem your best shot...
     
  4. Jun 28, 2016 #3
    (a)
    If we work with a forward voltage VD of 320mV this means the diode would require a forward current If of 1mA. The resister R3 would be requred to drop the remanig voltage and limit the current through the diode
    [tex]R_3 = \frac{5-0.32}{0.001}=4.68K[/tex]
    The voltage at V_ and V+ must be equal when the system is balance at the required output voltage.
    [tex]V_D = V_+[/tex]
    The inverting input is worked out as R1 and R2 form a potential divider.
    [tex]V_o=V_D\frac{R_1+R_2}{R_2}=5[/tex]
    [tex]\frac{V_D}{V_O}=\frac{R_2}{R_1+R_2}=\frac{0.32}{5}=0.064[/tex]
    So if we give R2 a value of “1” we can find the ratio of R2 to R1.
    [tex]\frac{1}{0.064}-1=14.625[/tex]
    So if we give R2 3.3k then R1 will be 48k

    As for part (b) I don't know where to star
     
  5. Jun 30, 2016 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    All very well done for part (a).
    As for part (b) you haven't given us the frequency of operation so can't do part (b).
    Can you flip the image 90 deg for us?
    Since they're asking for voltage gain you also have to define the input point. The overall circuit gain is volts/watt or volts/lumen or similar with infrared input quantity.
     
  6. Jun 30, 2016 #5

    donpacino

    User Avatar
    Gold Member

    assuming they don't give you a frequency, you might need to make it generic (the answer will simply have f as a parameter).
     
  7. Jun 30, 2016 #6
    No frequencies are given.
    That all the info I have been given.
    Although it says state any assumptions.
    I assume the RC feedback is some kind of high or low pass filter. I really don't know where to start.
    Thanks for you help.
     
  8. Jun 30, 2016 #7

    donpacino

    User Avatar
    Gold Member

    I would first calculate the operating points at DC.

    Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
    Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
     
  9. Jun 30, 2016 #8

    donpacino

    User Avatar
    Gold Member

    if you really dont know where to start, maybe do nodal anlysis or some other system at DC (at dc capacitors are open circuit). It should be easy to find the voltages at the IOs of the op amps.
     
  10. Jul 5, 2016 #9
    I really don't have a clue.

    so R1and R2 form a potential divider giving 2V at the non-inverting inputs of X3 and X1

    but I am not sure about the inverting side this looks like it is held low at 0V
     
  11. Jul 6, 2016 #10

    donpacino

    User Avatar
    Gold Member

    there is a property of op-amps that states with negative feedback the inverting and non inverting terminals will be equal
     
  12. Jul 7, 2016 #11

    LvW

    User Avatar

    I think, the whole circuit resembles a kind of bandpass amplifier. Why not using the mid frequency Fo as operating frequency?
    A circuit simulation can give the value of Fo.
     
  13. Jul 7, 2016 #12

    donpacino

    User Avatar
    Gold Member

    that could work (assuming the circuit is properly designed/tuned). I assume op is supposed to do this analytically.
     
  14. Jul 7, 2016 #13

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    There is no requirement to use that specific circuit is there ?

    There are other ways of doing the same thing and some of these other ways will certainly be easier to analyse .
     
  15. Jul 7, 2016 #14

    donpacino

    User Avatar
    Gold Member

    This looks like a homework problem. I doubt a "solution" would be resdesign the problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Op-amp voltage regulator
Loading...