Voltage Doubler

JamesDougherty

I have been trying to make a simple voltage doubler for a·little side·project I am working on, but I can't get it to work right.

Here are two of them that I have tried with no success.
http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/UsefulCircuits/VoltageDoubler.htm
http://www.creative-science.org.uk/multipliers.html

It looks simple enough, but it just does not like me. Where the the voltage is supposed to be double it was reading almost half the supply voltage. Any ideas would be greatly appreciated.

Thanks


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James Dougherty

Ariel Productions
http://www.arielproductions.com/

agfa

is your supply voltage AC?

JamesDougherty

That has to be the problem now that you mention that... It is DC that I was supplying.

By the way, thank you for your help.



[noparse][[/noparse]EDIT]

I know they make the rectifiers to go from AC to DC, but what do you use to go from AC to DC? Is there any easier way to amplify (double) the voltage?

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James Dougherty

Ariel Productions
http://www.arielproductions.com/

Post Edited (JamesDougherty) : 1/15/2009 2:17:47 PM GMT

sailman58

I know that there are DC circuits that will multiply voltage because my PI robot has a circuit that maintains a constant voltage to the motors. With fresh batteries it decreases the voltage and when the batteries have run down it boosts the voltage. So far the folks at Pololu are not talking about the circuit, but they do plan to sell that circuit separate from the PI robot in the future. I am sure that somebody on this board could come up wit a circuit that does the same thing. Unfortunately, I don't have the knowledge myself.

Ron

Bean

Here is the typical voltage doubler for a microcontroller.

The input to C1 must be a 50% square wave.

The output is (Vdd*2)-(2*diode drop), so for 5V you will probably get about 8.5V or so.



Bean.




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·The next time you need a hero don't look up in the sky...Look in the mirror.

JamesDougherty

Wow, thank you Ben! Will that work with DC?

Bean (Hitt Consulting) said...

Here is the typical voltage doubler for a microcontroller.

The input to C1 must be a 50% square wave.

The output is (Vdd*2)-(2*diode drop), so for 5V you will probably get about 8.5V or so.



Bean.

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James Dougherty

Ariel Productions
http://www.arielproductions.com/

Bean

You only need a 50% square wave output from the microcontroller (+5 to Gnd) to drive it.

Bean.

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·The next time you need a hero don't look up in the sky...Look in the mirror.


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JamesDougherty

Too awesome, I think it will work perfectly! I will try it tonight and let you know how it goes.

Thank you so much for your help!

Bean (Hitt Consulting) said...
You only need a 50% square wave output from the microcontroller (+5 to Gnd) to drive it.

Bean.

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James Dougherty

Ariel Productions
http://www.arielproductions.com/

SteveW

Don't forget that there's no free lunch - double the voltage means (less than) half the current.
If your pin is only good for 20mA, then trying to suck more than 10mA out of this will lead to disappointment, and maybe smoke. This is actually quite an unpleasant thing to do to an output pin. It'll be seeing damn close to a dead short at the start of each cycle. Some careful messing with the square wave frequency and capacitor sizes may be called for. I'd suggest that the capacitor at the left be 1/10 the size of the one to the right, as a first guess.
Also, schottky diodes are good in this application - lower forward voltage drop, lower capacitance, better efficiency all round. BAT54S if you're in surface mount territory, gets you the 2 diodes in series that you want.

Steve

RickB

Use schottkey diodes for lower loss and pay attention to the frequency vs capacitor value. Look up Cockroft-Walton multipliers.

Rick

JamesDougherty

Thank you both for the information; I appreciate it. I looked into the Cockroft-Walton multipliers and it appears that you need AC current. I'm using DC so I don't think it will work like the one Bean had posted. However, I did save the link in case I ever need an AC version.

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James Dougherty

Ariel Productions
http://www.arielproductions.com/

sailman58

Hi Bean,

Funny you should show up now, I just put one of micro boards in my bag so I have something to play with at Cabin Fever this weekend. Have you tried it with the Linux IDE?

Ron

SteveW

JamesDougherty said...
Thank you both for the information; I appreciate it. I looked into the Cockroft-Walton multipliers and it appears that you need AC current. I'm using DC so I don't think it will work like the one Bean had posted. However, I did save the link in case I ever need an AC version.

They _all_ need AC. That's the way they work. If you just want something that turns 5V into 10V, you need something like http://search.ebay.co.uk/360121702406
5V in, 10V out. They cost rather more than a few Cs and diodes, though. (There's AC stuff going on inside, but it's hidden from you, and you pay for that privilege)

If you can generate a square wave, then that _is_ your AC, and you can feed that into your doubler. When you stop generating your square wave, you'll stop getting twice the voltage. Since you can easily generate a square wave under interrupts, there's no real need to stop...

What do you need this 10V for, anyway? How much current? How stable? How cheap? More information in the question will get you better answers.

Steve

Post Edited (SteveW) : 1/15/2009 6:04:53 PM GMT

Phil Pilgrim (PhiPi)

There's a difference between AC and pulsed DC. Bean's circuit requires the latter. But the input square wave does not have to go below ground, as AC would. The long and short of it is this: if you want to build a voltage doubler from the ground up, you will need one or the other. There's simply no way to go from V to 2V without modulating V somehow. Commercial DC-DC converters that appear to do this just hide the details internally.

-Phil

Post Edited (Phil Pilgrim (PhiPi)) : 1/15/2009 6:51:57 PM GMT

SteveW

Phil Pilgrim (PhiPi) said...
There's a difference between AC and pulsed DC.
-Phil

Well, sort of. But not when the first component in the circuit (all of them) is a coupling capacitor [noparse]:)[/noparse]

I still think we need more information about what this 10(ish) volts is intended for - I smell a potential calamity.

Edit: Not really calamity. Much more likely is a drooping doubler, delivering V-2diodes, rather than V*2. Forcing non-trivial power through doublers (and above) takes beefier drivers than an SX pin... (I usually hang them off the driven node of switchers, which also causes trouble if the switch mode controller chip is a bit fussy about overcurrent)

Steve

Post Edited (SteveW) : 1/15/2009 6:55:29 PM GMT

SteveW

Phil Pilgrim (PhiPi) said...
The long and short of it is this: if you want to build a voltage doubler from the ground up, you will need one or the other. There's simply no way to go from V to 2V without modulating V somehow.

I like a challenge...

Best I can come up with, though, is a LED/photocell (or photodiode, if you need _pitiful_ currents.)
No moving parts or nodes, though, and galvanic isolation for free... Bill of materials isn't dreadful, but it's going to take some space, and efficiency will be woeful.

Steve

Phil Pilgrim (PhiPi)

Ah, but you're still modulating the DC — at about 400THz for a red LED!

-Phil

SteveW

Phil Pilgrim (PhiPi) said...
Ah, but you're still modulating the DC — at about 400THz for a red LED!

I'd probably be using a UV LED for efficiency, so it's even worse than that [noparse]:)[/noparse]
Still, my 'scope (1GHz) and spectrum analyser (21GHz) are both convinced that 400THz is DC, and that's good enough for me...

Edit: Also, remember when 400THz was an unthinkably high frequency? Just seems like a matter of time before we're blurring the lines.

Steve

Post Edited (SteveW) : 1/15/2009 11:05:27 PM GMT

Dennis Ferron

No I'm pretty sure the electricity coming out wouldn't modulate at 400 THz; I think the way the physics works is that a light photon knocks exactly one electron loose as it strikes. That electron, once knocked loose, goes on its merry way (as a DC current). For the electricity to modulate at 400 THz, the electron would have to travel first one direction, then get sucked back and go the other direction, and sucked back and go the first direction again. It wouldn't get very far a that frequency. But the photon that struck it is already vanished, being absorbed to create the higher energy state of the electron.

Remember, just because the color of the light creates a wavelength equivalent to 400 THz, doesn't mean all the waves are hitting the "high" and "low" ("on" and "off") at the same time. It's a bunch of incoherent waves, not something in unison like a radio signal.

Maybe a ruby laser striking a stream of electrons in a cathode ray tube could induce that kind of 400 THz oscillation in the electrons it strikes (i.e. if you created a free-electron laser, itself pumped with another, conventional laser), but then I'm pretty sure it would all be re-emitted as red light before the electrons left the tube.

Edit: Although I have often wondered, if you could make such a setup, and get rid of enough parasitic capacitance, whether the wire itself would throw off red light like a radio antenna for light, if only for a fraction of an inch as the electrons enter the wire. Alas, at such high frequencies, any capacitance at all would cancel out the effect.

Post Edited (Dennis Ferron) : 1/15/2009 11:30:43 PM GMT

Philldapill

Phil, I think there is some confusion about your prior statement that Bean's circuit uses Pulsed DC. That would yield nothing... You MUST have a negative voltage applied to the inputs for you to double the voltage. i.e. some sort of AC signal. A 50% square wave without any negative voltage won't do anything.

Phil Pilgrim (PhiPi)

Phill,

That's really not true. When the input to C1 is grounded, it charges (via D1) to 5V - V_fwd(diode). When the input is subsequently raised to 5V, it's still carrying that charge, so its high side has a voltage relative to Vss of 10V - V_fwd, charging C2 (via D2) with 10V - 2V_fwd. When the input to C1 returns to ground, it again charges to 5V - V_fwd, and the process continues.

If a negative voltage were required for switched capacitor voltage converters, chips like the MAX232 or ICL7660, for example, would never be able to operate without a negative supply voltage.

-Phil

Post Edited (Phil Pilgrim (PhiPi)) : 1/16/2009 8:45:10 AM GMT

SteveW

Philldapill said...
Phil, I think there is some confusion about your prior statement that Bean's circuit uses Pulsed DC. That would yield nothing... You MUST have a negative voltage applied to the inputs for you to double the voltage. i.e. some sort of AC signal. A 50% square wave without any negative voltage won't do anything.

I believe you are wrong, and that a square wave between 0V and 5V will drive that circuit perfectly.
Look, there's a capacitor in the front of the circuit. Feed it anything you like (within reason), the capacitor will remove the DC, and pass the AC, it's what they do. 0V,5V, or 20V, 25V, or -2.5V,+2.5V, all fine.
Also, for what its worth, it works in practice, too. I have millions of these things out in the field, running as doublers, triplers and inverters. It's a standard way of generating those irritating low current feeds that you find yourself needing, and comes very nearly for free when you've got a handy switch mode PSU generating a high current rail.

Steve

Philldapill

Arg. You guy's are right. I misunderstood what you had said. Of course it works... [noparse]:)[/noparse]