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I replicated a voltage doubler circuit from this example I found online: enter image description here

... and as expected, I am getting 12V from a 6V input. I would like to increase the multiplier to more than 2x, but I am having trouble understanding how I can connect a second IC4049 to do so. My attempt was to hook up the second IC4049 like so:

  • pin 1: 6V
  • pin 8: GND
  • pins 3, 5, 7, 9: pin 11 of first IC4049
  • pins 2, 4, 6, 10: negative leg of C2

The output across C3 is still 12V, so it doesn't work. What is the correct way to connect the two ICs up?

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1 Answer 1

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If you connect a second one like this:

schematic

simulate this circuit – Schematic created using CircuitLab

(only one oscillator is required, so you can use all the inverters in the second chip in parallel if you want).

You'll get about -5V, so the difference between the two will be about 15V.. that will allow you to connect more LEDs in series.

You might want to put reverse biased diodes across each of the two output capacitors, your C3 and my C2.

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  • \$\begingroup\$ Hi Spehro, thanks for the reply. My skills in electrical is somewhat limited, so I may have misunderstood your circuit. With the second 4049, I connect 1 to +6V, 8 to GND, the inputs to the inverters to the oscillator, and the outputs to the positive leg of C1 in your circuit. Then between the positive leg of C3 of my circuit and the positive leg of C2 of your circuit I will get +12 -(-5) = 17V? \$\endgroup\$
    – user2218339
    Commented Jan 10, 2015 at 23:28
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    \$\begingroup\$ Perfect, except it's the negative leg of C2, but you meant to write that. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 11, 2015 at 2:10
  • \$\begingroup\$ Hi Spehro, I found a few 1N5817 schottkys lying around so I replaced all my diodes with those to see what would happen. I expected to gain 1 or 2 V at the output due to the lower voltage drop, but I'm actually seeing an output of 24V instead (about 17.8V previously). Does that look normal to you? \$\endgroup\$
    – user2218339
    Commented Jan 11, 2015 at 23:36
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    \$\begingroup\$ Seems suspiciously high.. is your supply voltage actually 6V? Does your meter have a fresh battery? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 11, 2015 at 23:47
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    \$\begingroup\$ Maybe a small load will fix that. Or maybe you invented an over-unity device. ;-) \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 12, 2015 at 2:26

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