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circuit explanation

Discussion in 'General Electronics Discussion' started by lotus, Dec 6, 2010.

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  1. lotus

    lotus

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    Dec 4, 2010
    I have found this circuit in Keith Billings book. Can anybody please explain the working and logic behind this circuit
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    If you posted a link, you might have to try doing it again.
     
  3. lotus

    lotus

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    0
    Dec 4, 2010
    I have attached the circuit now
     

    Attached Files:

  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Weird.

    Pass.
     
  5. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Definitely a weird circuit by itself, but it needs to be seen in context with both the power supply driving it and the type of load. I don't know what use it could have though.
    It's a kind of switched capacitor voltage multiplier. When the voltage is above the treshold, A1 output is high and Q1 does not conduct. R1 charges C1 and R2 charges C2.
    When the voltage goes below the treshold, A1 output goes low and Q1 conducts, thereby series-connecting the capacitors, doubling the power supply voltage (for a while).
    The process will repeat if the power supply is below the treshold voltage (determined by R3/R4 & Ref V).
    There will be a substantial ripple voltage generated, so both the supply and the load will need to be able to deal with this.
     
  6. lotus

    lotus

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    Dec 4, 2010
    thanks a lot
     
  7. selva

    selva

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    Apr 26, 2010
    Hello Resqueline,

    I hope u made a mistake in that highlighted text. I think Q1 conducts when A1 output is high. I hope this circuit is to prevent the load from sharp edges, more or less like a basic caps operation, but i don't why they made it clumsy with a op-amp.
    Sorry if i have made any mistake.

    Warm Regards,
    Selva
     
  8. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    I reviewed the circuit again and I find that I mistook Q1 for a PNP instead of an NPN.. (I would have drawn an NPN the other way around.)
    That inverts the function and it could look like it becomes a ripple filter, but..
    However; when the voltage is below the treshold, A1 output is low and Q1 does not conduct, R1 charges C1 and R2 charges C2 (to full supply voltage).
    When the voltage goes above the treshold, A1 output goes high and Q1 conducts, thereby series-connecting the capacitors, increasing the power supply voltage by 50%.
    The process seems not to repeat by itself.
    I still don't quite see the point of the circuit, actually even less.. It would be interesting to see some of the text and the rest of the circuits associated with it.
    And why don't someone run this in a simulation?
    There's no problem in placing questions about anything, even the best can make mistakes, I just rewrote this reply completely more than 3 times! ;)
     
    Last edited: Dec 15, 2010
  9. selva

    selva

    25
    0
    Apr 26, 2010
    Hello Resqueline,

    I just made that question to make sure, that operation which i understood is correct :D. I would like to try it in simulation, but i get stuck with pspice simulator, i repetitively get a same error (i forgot that error).

    I'll try again with pspice and if i face any error again, i'll let u know. By the mean time can u suggest me any other simulator for windows as well as ubuntu.

    Warm regards,
    Selvam Anand:)
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    Yeah, I was going to try to simulate it, but I have not had the time.

    I would recommend that you have a non-zero impedance of your source voltage. This may be modelled in the simulation, or you can put a resistor in series with the power supply.

    Start with 1 ohm.

    You will also probably want a load on the output. Start with 100 ohms.

    Maybe add some ripple to your input voltage too.
     
  11. lotus

    lotus

    14
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    Dec 4, 2010
    when i use square wave(8-14V) as source then it dont remove ripple
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    What is the impedance of your source?
     
  13. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Note that we didn't say it would remove ripple, only it might be interesting to see what might happen to the ripple (on top of a varying voltage).
    Also, as the circuit most likely outputs squares/spikes it would be more advantageous to test it with a sine or a triangle wave.
    And lastly, what did you use for component values? Did you make certain that the divider voltage exceeded the reference voltage?
     
  14. lotus

    lotus

    14
    0
    Dec 4, 2010
    I hav used these values
    R1 = R2 = 100 ohms
    C1 = C2 = 10uF
    R3 = 50k ohms
    R4 = 106.4k ohms
    Vref = 6.8V
    Load = 100 ohms
    Vinput = 8-14 V (pulse with Tr = 1ms)
     
  15. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Ok, so R1*C1 charge (& discharge) time constant is 1ms, and the switchover point for the comparator is 10.0V. I'd say the load is a little on the heavy side, I'd try 1k.
    Also you didn't reply to steve's question about the source impedance. If that is zero then no response will be had no matter what. Put a diode in series with Vinput.
    Vinput should also not be a square (or any pulse form) in order to see what might happen. Make it a relatively slow sine or triangle varying around 10V (+/- 2-4V).
     
  16. lotus

    lotus

    14
    0
    Dec 4, 2010
    I hav attached the simulation results
     

    Attached Files:

    • sim.zip
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  17. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    How come you have put a capacitor across the load, being 10 times larger than C1 & C2? It was not there in the original diagram..
    The red & blue curves shows the same thing of course, but I don't understand why the two yellow curves follow each other perfectly..
    And then there's the source impedance, being extremely low. How about using a diode there also?
     
  18. lotus

    lotus

    14
    0
    Dec 4, 2010
    that capacitor across load is the filter capacitor. there is only one yellow curve which is Vgs
     
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