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If I have a 5V input source and I wanted an output voltage of 500V, can I use the output of one DC/DC converter to serve as the input voltage to a second DC/DC converter?

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    \$\begingroup\$ Usually you would use a multiplier for this, not a converter. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 5 '15 at 4:41
  • \$\begingroup\$ and remember that the current available at the 500V source will be reduced by more than 100 times that of the 5V source (power out = power in x efficiency) so you could end up demanding a big current from the 5V source. \$\endgroup\$ – JIm Dearden Jun 5 '15 at 6:05
  • \$\begingroup\$ As long as each dcdc is isolated for at least double thar \$\endgroup\$ – JonRB Jun 5 '15 at 6:18
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    \$\begingroup\$ See Voltage multiplier. \$\endgroup\$ – CL. Jun 5 '15 at 7:32
  • \$\begingroup\$ Or just get a dc-dc converter that outputs 500V. \$\endgroup\$ – Andy aka Jun 5 '15 at 8:42
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As someone mentioned in the comments, you want to use a voltage multiplier. However most voltage multipliers require AC as their input since the AC causes the circuit to switch back and forth. So you would need to create an AC voltage (perhaps using a 555), step it up with a transformer (optional) and then feed it into a conventional voltage multiplier.

But there is a version of a voltage multiplier that converts DC to DC, it is called a Dickson charge pump.

enter image description here

The multiplier for the circuit is equal to the number of diodes; so the above circuit multiplies the input by five times.

The gotcha with this circuit, is you need to provide an external clock as shown in the diagram. This could either be provided by a microcontroller (if you already have one on board) or a 555 circuit.

Using several of these in series, you could easily bump 5v up to 480v, using multiples of 32x and 3x. (For the 32x, rather than have 32 diodes in series, you could split that up into five doublers of two diodes each. Or something in-between.

I don't know how exact your requirement of 500v is. If you need exactly 500v, I'm sure that can be achieved.

Note you will lose a diode drop for each diode, so as a minimum, you would lose 13 * 0.2v = 2.6v, bringing the total down to 477v. You can avoid the diode drops by using MOSFETs instead.

Don't expect to get much current out of this, maybe a few mA.

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  • \$\begingroup\$ Just make sure to use capacitors and diodes which are rated for the high voltages. In the case of the diodes especially, they need to be rated for potentially the full 500V in reverse depending on how the startup and shutdown waveforms look. \$\endgroup\$ – Tom Carpenter Jun 9 '15 at 23:26
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You can do it, but you'd want to connect all of the inputs, in parallel, to the 5V source, and all of the outputs, in series aiding, to the load.

One requirement for that to work is that the inputs and outputs of the converters must be isolated from each other, and a caveat is that if your converters are of the Power-Out-Equals-Power-In persuasion, the power the load consumes must be the power input to the converters.

That means that if your load takes 1 milliampere from the 500 volt side, the converters must take 100 milliamperes from the 5 volt source, plus whatever it takes to feed the converters.

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    \$\begingroup\$ The boost converters I've used all had a common ground shared between the input and output. Therefore you couldn't parallel the inputs and put the outputs in serial at the same time. \$\endgroup\$ – tcrosley Jun 10 '15 at 1:00
  • \$\begingroup\$ Of course, and I'm sure, however, that there are isolated boost converters. Good point, though, and I've edited my answer to remove any ambiguity, thanks. \$\endgroup\$ – EM Fields Jun 10 '15 at 1:54

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