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So I came across this article on CircuitDigest where they explained the functionality of a voltage doubler designed with capacitors. The circuit shown is the following:

enter image description here

At first capacitor C3 is charged to 5v (source voltage) when Q (the output of the 555 timer in astable mode) is low. Afterwards, when Q is set to high the diode D1 avoids C3 from being discharged (since the plates now have a direct path to connect to each other) and instead it gets discharged via ground passing through capacitor C4 thus adding the charge to C4 and therefore doubling the voltage (on C4) at that instance.

What seems odd to me is how come the capacitor C3 was able to discharge? Don’t the plates of the capacitor need to be directly connected to each other? In this case it looks like C3 is being discharged through ground and not through the other plate of the capacitor. How is this possible?

If anyone cloud clear this out for me I’ll greatly appreciate it.

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  • \$\begingroup\$ A capacitor will charge/discharge if a current passes through it. In your opinion, will a current pass through the capacitor? \$\endgroup\$ – user253751 Mar 24 at 22:25
  • \$\begingroup\$ @immibis At first I would have said no, as in no current passes through the capacitor C3 since there is no direct connection between the plates. This however doesn’t seem to be the case... I thought that a capacitor could only be discharged through its plates. After all, the potential difference is defined between the plates not between one plate and ground, right? \$\endgroup\$ – rr1303 Mar 24 at 22:36
  • \$\begingroup\$ If the plates have to be directly connected in order for a current to flow, then how would it charge? And yes, that's why it's a potential difference and not just a potential. \$\endgroup\$ – user253751 Mar 24 at 22:44
  • \$\begingroup\$ Current merely appears to flow through a capacitor (if you treat it as a black box you can't see into.) But avoid that thought. Suppose pin 3 is at ground and \$C_3\$ has no stored charge (\$0\:\text{V}\$ across it.) Then \$D_1\$ will charge it up (which requires current and time.) Once charged (to about \$4.5\:\text{V}\$, let's say), pin 3 is driven upward to \$5\:\text{V}\$. Since the negative end of \$C_3\$ is attached to pin 3, this means the positive end of \$C_3\$ must be at about \$9.5\:\text{V}\$ now (add capacitor voltage to pin voltage.) \$D_2\$ charges \$C_4\$ to \$9\:\text{V}\$. \$\endgroup\$ – jonk Mar 24 at 22:45
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    \$\begingroup\$ Hey @jonk, it looks to me you have answered the question. Answering in comments is discouraged on this board, care to just turn that comment in an answer? \$\endgroup\$ – Vladimir Cravero Mar 24 at 22:52
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C3 discharges via the power supply. When the output of the 555 is high, you have basically the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Q28 is conducting, so current can flow clockwise around this loop. The voltage on C3 is in series with the power supply, so the combined voltage is enough to forward bias D2 and push charge onto C4.

C3 discharges by virtue of the fact that current flows out of its positive terminal, through D2, C4, the power supply and Q28, back to its negative terminal. Note that the same current is flowing in to the positive terminal of C4, charging it.

Internals of the 555 are from the datasheet.

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  • \$\begingroup\$ Perfectly explained, thanks. Just one thing, how is it that the current enters the negative side of the power supply and exits through the positive side? Aren't they suppose to be isolated since that would otherwise create a short? Thanks \$\endgroup\$ – rr1303 Mar 24 at 23:20
  • \$\begingroup\$ No. That's exactly what the power supply does all the time! \$\endgroup\$ – Dave Tweed Mar 24 at 23:45
  • \$\begingroup\$ You’re right! Everything makes sense now. Thanks \$\endgroup\$ – rr1303 Mar 25 at 0:01

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