Voltage doublers, like those close to be one stage of Cockroft-walton multipliers, are well known. They allow to double the voltage of an AC input and can be useful at the output of a transformer for instance. Of course, the price is that the output current is half the input current. My question is: is there a circuit (hopefully analogical and passive), not a transformer, that performs the dual task: to input an AC current, and output an AC (or even DC) current with half the input voltage, and twice the input current.

INSIGHT: it would suffice to find a way to charge 2 capacitors in series and to discharge them in parallel.

  • 1
    \$\begingroup\$ What you are asking for is not the double of a cockcoft walton circuit. It (CW) takes an AC voltage and produces a DC voltage. \$\endgroup\$ – Andy aka Jul 26 '16 at 8:41
  • \$\begingroup\$ thx. I've edited the question. \$\endgroup\$ – MikeTeX Jul 26 '16 at 8:57
  • \$\begingroup\$ Gut feel is : not without active devices (transistors, FET switches, etc). Diodes turn ON above a certain voltage : you need something that turns OFF above a certain voltage. Non-existent, so take an active device and turn it on or off. Then it's some form of buck converter... \$\endgroup\$ – Brian Drummond Jul 26 '16 at 9:15
  • \$\begingroup\$ @BrianDrummond You're right, there is an active component needed. But it wouldn't make it a buck (there is no inductor), it would make it more look like a charge pump. \$\endgroup\$ – dim Jul 26 '16 at 9:56
  • \$\begingroup\$ It's possible to make an amplifier without using a semiconductor but not really desirable. What's the problem with a transformer? \$\endgroup\$ – Andy aka Jul 26 '16 at 10:43

You can build such a circuit, but it requires some active device. You can't do it with just diodes and capacitors. Here is one divide-by-eight that takes mains AC as input and outputs DC. It has about 85% efficiency at 4W. It could be improved in several ways, but as it is, it is quite simple:

enter image description here

R10 is the load. In this example, it draws about 4W with 220VAC input (The output voltage is about 32V). You can't draw much more without the efficiency dropping dramatically.

Here is how it works: when the input AC sine is positive, the PMOS is blocking and the eight capacitors in series are charged through the top diode D30 and all the shottkies (PMEG6030) in series (the other diodes are not conducting). So each capacitor ends up being charged at VIN/8. When the sine is negative, D30 stops conducting, but the PMOS conducts. This makes all MMDB3004 conducting and the eight capacitors becomes all paralleled. The charge is then transferred to the output capacitor C4.

This, in fact, works exactly like a charge pump. You can divide by what-you-want instead of eight, by adjusting the number of capacitors and diodes. Of course, efficiency will be affected if there are too many.

This circuit is working on half-wave (one half for charging, one half for discharging). It would be possible to make it work on full wave, but it would become much more complicated.

Also note that the choice of the components is critical. All the diodes, except the shottkies in series must withstand the mains voltage. The shottkies and the capacitors must withstand the maximum output voltage (input voltage divided by eight). The PMOS must withstand the mains voltage, and have relatively low RDSon, otherwise, efficiency drops a lot. R1 must be rated for mains voltage.

Finally, from a safety point of view, I wouldn't recommend this circuit, as there is no isolation. Also, the size of each component makes it not more compact than a small transformer. Probably not cheaper as well, given the number of components required (when dividing by a high ratio), and given the required mosfet (it would be possible to reverse the whole circuit and use a cheaper N-channel fet, though). All in all, this circuit certainly isn't the best choice, actually.

| improve this answer | |
  • \$\begingroup\$ Thank you for this detailed answer dim. Actually, a single stage of circuit may advantageously be used at the output of a transformer, say to abase 60V to 30V, and double the input current. Yet, I am still surprised that there is nothing like a dual of a CW multiplier, that charges the capacitors in series and discharges them in parallel without active devices. \$\endgroup\$ – MikeTeX Jul 26 '16 at 18:41
  • \$\begingroup\$ In fact, the dual of the multiplier is the circuit I shown. And it is just "by chance" that the multiplying version (compared to the dividing version) doesn't need active devices and can be made only with diodes and capacitors. The reason behind this has been given by Brian Drummod. Anyway, in the application you mention (60V to 30V), this circuit would indeed make more sense, provided you don't need much current on the output. \$\endgroup\$ – dim Jul 26 '16 at 19:08

Maybe using a dual of the capacitor? Something like this:

enter image description here

Conceptually, while the capacitor in the voltage doubler maintains a voltage on a node that never goes to zero, in this circuit the inductance stores energy and maintains a current in one branch that never goes to zero.

enter image description here

The average current on \${L_1}\$ in this picture is 1.5 time the \${I_{D1}}\$ (the average current on \$D_1\$ is 1,6A) but if we reduce the ripple the rectified current would be around 3.2A, thus \$ I_{L1} = 3.2{\text{A}} = 2* \overline{I_{D1}}\$.
The question is: how we can reduce the ripple?

Ngspice give me similar results.

| improve this answer | |
  • 1
    \$\begingroup\$ Where do you put the load? \$\endgroup\$ – MikeTeX Jul 26 '16 at 18:37
  • \$\begingroup\$ Good thinking, @Antonio. I was trying to come up with something similar while travelling today. I always think of Cs and Ls to be complimentary as are, in some ways their Vs and Is. The CW voltage doubler would tend to give a constant voltage output. Would this tend to give a constant current output? \$\endgroup\$ – Transistor Jul 26 '16 at 20:39
  • \$\begingroup\$ @MikeTeX, in serie with central L. \$\endgroup\$ – Antonio Jul 27 '16 at 13:08
  • 4
    \$\begingroup\$ Since we're swapping L & C and I and V we probably need to drive with an AC current source instead of a voltage source. \$\endgroup\$ – Transistor Jul 27 '16 at 23:03
  • 1
    \$\begingroup\$ Antonio, please put a load in your experiences otherwise you'll see nothing real. \$\endgroup\$ – MikeTeX Jul 28 '16 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.