Trying to apply capacitance multiplier.

Discussion in 'General Electronics Chat' started by hp1729, Oct 15, 2016.

  1. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    Capacitance multiplier.PNG Design 907 A.PNG Design 907 B.PNG
    Searching the Internet for a capacitance multiplier I came up with the simple circuit above. The objective was to apply it to the oscillator above and getting the result shown.
    Doesn't work. Are there any suggestions for a capacitance multiplier that will work in the application?

    VCC is 5 V. So V out high is about 3.3 V, about the same as the white LED on the output. No need for a resistor.
    I know, no capacitor on the control line. It works just fine without one.
     
    Last edited: Oct 15, 2016
  2. SLK001

    Well-Known Member

    Nov 29, 2011
    821
    229
    If you are expecting the capacitance to be C times some multiplier, you are out of luck. A capacitance multiplier is mostly used for smoothing a signal (like a full wave rectifier). Here, a smaller capacitor can be made to look like a much larger cap. A capacitance multiplier cannot be used for timing, like in your application. Here's a capacitance multiplier I have used in the past. The cap looks like C times the hfe of the transistor.
    Xer.png
     
  3. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    Thanks.
     
  4. DickCappels

    Moderator

    Aug 21, 2008
    2,662
    633
    BR-549 likes this.
  5. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
  6. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,805
    1,105
    LTspice says it works ok in conjunction with a 555. That means 555-based long-period timers can be achieved without the need for (inevitably leaky) electrolytic capacitors.
     
  7. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    Design 907 D.PNG


    Design 907 D.PNG
    That works. I changed the 5.1Ks to 10K and it did double the effect. Thanks.
    (edited to add ...)
    Replacing R10 with a wire still works.
    I'm not sure I could explain how it works. I took the op amp out so I effectively just added R3 to the circuit and it went back to flashing fast, So the op amp does do something. But the circuit looks like it is just a buffer for the voltage on the capacitor.

    What am I missing? How does it multiply the effect of the capacitor?
     
    Last edited: Oct 16, 2016
  8. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    That was the objective. Normally about 100 uF is the limit. So now we can go easily 10 times that. I haven't tried that yet.
    My Spice libraries are not to well developed, nor are my skills. I can usually build a small circuit quicker than I can put it up in Spice. Parts I got. :)
    Thanks.

    I haven't tried it yet but 1 M for timing resistors, 100 uF cap and a x 1,000 multiplier should give me about 2 1/2 days for a cycle. :)

    Problems with the 555 on these long periods ... the first timing cycle would be longer, On the first cycle the cap is completely discharged and won't reach Threshold (2/3 VCC) for a while. Following cycles the cap bounces between Trigger (1/3 VCC) and Threshold.
     
    Last edited: Oct 16, 2016
  9. OBW0549

    Well-Known Member

    Mar 2, 2015
    1,328
    890
    I've used this circuit, and my own experience makes me seriously doubt that you'll achieve anywhere near a 2.5 day cycle-- at least not with an LM358 op amp. Note the formula for effective leakage current in the diagram in post #4. That's one of the big weaknesses of this circuit: although you can achieve very high capacitance multiplication ratios, the price you pay is VERY high effective leakage current unless you use an op amp with extremely low Vos and Ios (a MAX44246, for example).
     
  10. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    For long delays a CD4060 based timer is more reliable, takes much smaller capacitors, and is easier to calibrate.
     
    DickCappels and OBW0549 like this.
  11. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,805
    1,105
    Trust Murphy to spoil things ;) .
     
    OBW0549 likes this.
  12. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    That makes good sense. Thanks.
     
  13. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    Also makes good sense.
    thanks.
     
  14. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    Design 907 x cap mult.PNG
    But I can't explain how it works. It does. I can change the value of R1 and get the predictable results. Taking the circuit apart, drawing A is the basic circuit. If I take the capacitor out we have circuit B. I put in 2 V and get 2 V out. The output follows the input just the same as drawing C. drawing D just shows the voltage on the capacitor. So where does the multiplying come from?
     
  15. OBW0549

    Well-Known Member

    Mar 2, 2015
    1,328
    890
    It's pretty simple, really. Current injected into (or drawn out of) the junction of R1 and R2 by whatever is trying to charge or discharge C1 gets split two ways; part goes through R1 and part goes through R2. But only the part going through R1 actually ends up charging or discharging the capacitor-- the part going through R2 just gets dumped into (or sucked out of) the op amp output.

    Since the voltages on the left ends of R1 and R2 are the same (because of the op amp operating as a voltage follower), analysis is easy and you end up with the effective capacitance multiplication factor being equal to the total current divided by the current through R1. Omitting some 7th-grade algebra, we get:

    Capacitance multiplication ratio = [1 + (R1/R2)].
     
  16. hp1729

    Thread Starter Well-Known Member

    Nov 23, 2015
    1,964
    219
    Thank you.
     
Loading...