Imagine a micro-controller's pulse output is coupled to a voltage multiplier with many stages such as in the case of a Dickson voltage multiplier. We then can increase the DC "voltage" level of the pulse this way..

But since we still use the micro-controller as supply, unlike in amplifier case we are limited with the micro-controller's pin's max current output right??

So we have a situation where we can increase the voltage but we are still limited with the max current can be driven by the load.

If I understood it right in my above paragraph, what can this be used for? Any useful applications?


The output of the voltage multiplier will deliver less current than the circuit driving it. If your multiplier can draw 30mA on the input side then it will never deliver 30mA on the output.

Voltage multipliers are used to generate high voltages from lower voltages - duh! You use that kind of thing when you need high DC voltages. The ouput of a voltage multiplier is DC even though it starts with AC or pulsating DC.

Wikipedia gives an explanation of how they work (with diagrams) and also names some uses - extremely high voltages for physics experiments being one use.

Another common use is (was) to generate the high voltage for cathode ray tubes - that's the old fashioned TV tubes in monitors and TVs. They needed several thousand volts DC to operate.

A good reason to use a multiplier is that it can be difficult to rectify high AC voltages. That is to say, you could use a transformer to get the needed high voltage, but then you'd have to convert the high AC voltage to DC. High voltage parts are more expensive and/or harder to get. Using a multiplier lets you use parts rated for lower voltage to generate the high voltage. Your parts only have to be rated for the difference in voltage between each stage of the multiplier rather than for the whole voltage range from low to high.

In answer to the questions in your comment:

  1. It is a bad idea to drive anything directly with the output of the processor's output. Use the output from the processor to control a transistor that drives the multiplier. This protects the processor and allows you to have a higher output current.
  2. Frequency limits wil be given by the diodes and the capacitors you use. Diodes all have an upper limit to how fast they can switch. The capacitors can only charge as fast as the available voltage can make them, so you are limited there as well.

Do read the Wikipedia article and the information in the external links.

  • \$\begingroup\$ how bout using from TTL level to RS232? \$\endgroup\$ – user16307 Jan 28 '16 at 11:43
  • \$\begingroup\$ ok so the current supplied will even be less than uC's max pin. we need a very high load impedance right? another thing: is there a limitation for the input freq. here? \$\endgroup\$ – user16307 Jan 28 '16 at 11:46

A multiplier doesn't store current, is stores charge. As such, the available current depends on how much charge is available and how quickly it must/can be discharged. If you need more continuous current then you can have the MCU drive a H bridge connected directly to the supply.

  • \$\begingroup\$ im not trying to drive more current or build anything. im asking just to understand. so what i described was wrong? ok i will put this way: lets say uC is connected to a multi stage voltage multiplier and its DC level increased a lot. and lets say the max current can be driven from the uC's pin was 30mA. will this limitation remain at the multiplier's output or not? and where are these multipliers used in practice? that was my question. \$\endgroup\$ – user16307 Jan 28 '16 at 10:28

Fundamentally, energy will be conserved. For the case of a DC input, DC output voltage multiplier, this usually means:

$$ I_{in} \cdot V_{in} > I_{out} \cdot V_{out}$$

In other words, the input power must be greater than the output power. This is not strictly true; you could have a circuit that takes a small amount of input power for a long time, and provides a very brief high power output. The more general case would have the time-integral on both sides of the inequality (since energy is the integral of power). Also, there will always be some power loss (inefficiency), so you will need to account for that.

I highly recommend you simulate your circuit in LTspice. It's free, and I've actually used it for the same type of circuit. Don't forget to include the expected load current in your simulation. It will help you size your capacitors and will help you avoid many other mistakes.


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