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Consider this voltage doubler:

enter image description here.

I simulated the circuit using NON-polarized capacitors and WITHOUT the transformer as in this schematic:

enter image description here

and the output voltage (at C1) was still doubled. I have a couple of questions about this:

  1. Why would people want to use polarized capacitors?

  2. What is the purpose of the transformer?

Finally, after playing around with the circuit,

  1. What happens if I reverse the orientation of both diodes? It seems that it inverts the sign of the output voltage, but this is only possible using non-polarized capacitors, which brings me back to my first question.
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1) Why would people want to use polarized capacitors?

Because, usually, they have more capacitance per volume and are cheaper. Search for values like 1000uF, 2200uF, 4700uF or even bigger of ceramic ones and will find the answer. If your value is low, you don't need polarized caps.

2) What is the purpose of the transformer?

If the purpose of circuit is only a voltage doubler, its for isolation. But you may achieve bigger voltages ratio with the transformer. An isolation transformer is a safety device. Since the right part of circuit is floating, if you touch only at one point of the circuit you will not be electrocuted. In spite off, for example, if you touch the positive side of the last capacitor it will be grounded and the negative will be -V.

3) What happens if I reverse the orientation of both diodes? It seems that it inverts the sign of the output voltage, but this is only possible using non-polarized capacitors, which brings me back to my first question.

If you reverse the diodes you must reverse also the capacitors.


Also, generally, non polarized capacitors have lower inductance and resistances. So they act faster and are better suited for high frequencies circuits.

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    \$\begingroup\$ +1. Note also the polarity of the first capacitor. When the transformer secondary goes negative the diode will clamp the + side of the cap at -0.7 V. From then on, in normal operation it will always be positive with respect to the left-hand side. Obviously, if you reverse the diodes you reverse the caps too. \$\endgroup\$ – Transistor Feb 10 '16 at 10:33
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Try loading that circuit with a resistor across C1.

You will find that, when delivering current to a load, its output voltage falls far below the open circuit voltage - especially if you add further voltage multiplication stages.

The problem is that the impedance of those capacitors quickly adds up, and is effectively doubled by each multiplication stage, so it's a poor voltage source into anything other than a high impedance load.

I've seen questions about the effective source impedance of a voltage multiplier - and never seen a simple satisfactory calculation of it - including my own attempts.

However the fix is to reduce that impedance - increase the capacitance - and high value capacitors are often polarised. Simple as that.

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