EEVblog #473 - Microcontroller Voltage Doubler - Page 1
Forum section: electronics / general

author: Psi
author_rank: Super Contributor
timestamp: May 25, 2013, 12:20:11 pm
content: So.. what to do next.Multiplying the voltage up is cool and all but its not really a "doubler"So.. for no reason at all, presenting the unnecessarily complex 3 stage doubler.(yeah i know it's bad. I couldn't get LTSpice to do what i wanted so the push/pull driver sections are overly complex and require two i/o)

author: Alana
author_rank: Frequent Contributor
timestamp: May 25, 2013, 12:37:25 pm
content: Crazy thing - i'm working on a project and Dave is doing video that exacly matches what i need to learn to finish it.Is there a formula that allows to calculate/estimate how much current can i get from such voltage doubler depending on caps value and frequency?

author: NiHaoMike
author_rank: Super Contributor
timestamp: May 25, 2013, 01:49:08 pm
content: I have done a few designs that used the Lx node of an existing switching regulator to derive additional rails, like getting a -12V rail from a 12V to 5V buck regulator. I put a 10 ohm or so resistor in series from the Lx to the cap in order to avoid current spikes false tripping the protection.

author: c4757p
author_rank: Super Contributor
timestamp: May 25, 2013, 02:04:32 pm
content: Quote from: Alana on May 25, 2013, 12:37:25 pmEssentially the same as the ripple calculation for a transformer input rectifier cap, though you will have more losses so assume significantly less "ideal" situations. Best to just test it. It's rather easy to throw together a little boost converter instead, if you need current. The only disadvantage IMHO is approximately double the cost. You can even drive it from an MCU pin.The first capacitor (the one that takes the square wave input) is harder to calculate because its value depends almost entirely on the rise time of the square wave.

author: LaurenceW
author_rank: Frequent Contributor
timestamp: May 25, 2013, 02:08:05 pm
content: OK, so what happens if you connect the OUTPUT back to the DC input, in place of the 3V DC?? I might just have to try it...It might be worth mentioning that the price you pay for these doublers is a considerably higher input current, since the output POWER has to come from somewhere, of course. << Oh! there - I think I've just answered my first question - the voltage will just collapse under its own load...And I reckon you COULD get to the moon on 1mA (and about 5MV should do it...)

author: Rufus
author_rank: Super Contributor
timestamp: May 25, 2013, 03:41:12 pm
content: Quote from: Alana on May 25, 2013, 12:37:25 pmThe pump capacitor is charged (or 'loaded') with the supply voltage less a diode drop. That charge is pumped to the output leaving a diode drop voltage on the capacitor.Q = CV so the charge transferred to the load on each cycle is C * (Vsupply - 2 * Vdiode). Multiply that by the pumping frequency to get charges per second which is amps. These pumps are current limited.Circuit resistances mean the capacitor won't fully charge or discharge so reality will be somewhat less and reduces with increasing frequency.10kHz, 3.3v, 100n and Schottky diodes with say 0.2v drop works out to about 2.9mA.Edit: I should point out that is the current limit into heavy load. The lighter the load the higher the output voltage and so the more charge left on the capacitor and less charge transferred per cycle.

author: Jon Chandler
author_rank: Frequent Contributor
timestamp: May 25, 2013, 04:20:55 pm
content: Dave's example of using a doubler to operate a 5v LCD from 3 volts is handy, but there may be another method that requires less current.I've found that thelogicof many 5 volt character LCDs works fine on 3.3 volts; the problem is that the Ve (contrast) input requires a negative voltage when the supply voltage drops below a certain point. Rearranging the circuit slightly, it can be used to generate a negative voltage. Connecting the negative voltage to Ve through a pot may make those "mute" 5 volt LCDs happy on 3.3 volts or lower.

author: einstein
author_rank: Contributor
timestamp: May 25, 2013, 04:59:41 pm
content: figure 23 page 19:http://www.st.com/st-web-ui/static/active/en/resource/technical/document/datasheet/CD00002294.pdfI use the l6208 to drive a stepper motor.i have question about D1,D2, Rp; Cp, Cboot, is that also a dickson voltage doubler? Wat's the function of cboot and rp? This circuit gives me 18V when i feed 12V in to drive the mosfets in the upper half of the h-bridge? I think Dave gave me the answer for a question for my school project i have to do in the last year secondary eduction.

author: Bored@Work
author_rank: Super Contributor
timestamp: May 25, 2013, 05:17:39 pm
content: Quote from: einstein on May 25, 2013, 04:59:41 pmYes, it is a charge pump. Rp limits the current when charging/discharging Cp. Cboot is the same as Dave's second cap in his video. Only that it is connected to Vs, instead of ground. This reduces the voltage over it, while the output Vboot is still above Vs. I.e. less charge is needed to get it up to a voltage above Vs.

author: einstein
author_rank: Contributor
timestamp: May 25, 2013, 08:01:56 pm
content: Thank you for your answer, and Dave thank you for the explanation.

author: JoannaK
author_rank: Frequent Contributor
timestamp: May 25, 2013, 10:50:04 pm
content: Quote from: Jon Chandler on May 25, 2013, 04:20:55 pmNot a bad idea. I have not tested this, but it since since the digital cmos logic usually works with allmost any voltage the idea seems valid. The reaction speed of the logic may be a bit lower, but it's not much of an issue with an text mode lcd.

author: codeboy2k
author_rank: Super Contributor
timestamp: May 25, 2013, 11:15:53 pm
content: Quote from: LaurenceW on May 25, 2013, 02:08:05 pmYou get to start an indiegogo free energy campaign

author: andyturk
author_rank: Frequent Contributor
timestamp: May 25, 2013, 11:38:00 pm
content: Attached is a screen grab from David's video. It's a close-up of his scope showing the various voltages.What's going on with the "slice marks" in the green trace? It's as if the transitions on the yellow and blue traces are blacking out the green. Interestingly, the fuscia trace is solid all the way across even though it too must be crossed by the yellow and blue transitions.

author: Psi
author_rank: Super Contributor
timestamp: May 26, 2013, 01:46:32 am
content: Quote from: andyturk on May 25, 2013, 11:38:00 pmIt's probably the order that the scope draws them on screen.CH4 (fuscia) was drawn last so is 'on top'I'm not sure what governs the order as it doesn't seem to be 1 2 3 4.It's probably whatever order that the channels were enabled in.

author: Pat Pending
author_rank: Regular Contributor
timestamp: May 26, 2013, 06:14:52 am
content: In the closing edits what movie is that sound track from?

author: stryker
author_rank: Regular Contributor
timestamp: May 26, 2013, 06:17:52 am
content: HiI'm not sure I can put my question into words that will make sense, but here goes...Would it be possible to take the PWM signal through a parallel input to the first stage via a transistor not gate, and feed that inverted pulse to smooth out the output by shifting the 3V upward on both halves of the wave? ie, the red shifted pulse would become the blue line?There has to be a good reason why not I'm sure, and I thought I'd best ask before I try it and break somethingCheers,Geoff

author: nitro2k01
author_rank: Frequent Contributor
timestamp: May 26, 2013, 07:09:34 am
content: Quote from: Pat Pending on May 26, 2013, 06:14:52 amGoogle is your friend. WarGames (1983.)

author: smashedProton
author_rank: Frequent Contributor
timestamp: May 26, 2013, 07:17:02 am
content: I just watched the movie! Very good. What an aloha nerd

author: nitro2k01
author_rank: Frequent Contributor
timestamp: May 26, 2013, 07:17:33 am
content: Stryker, I'm not sure I'm understanding what you're saying, but if you mean having two voltage multiplier stages running in inverse phase, yes that would be possible. You could keep the first stage(s) separate and feed them an inverse phase square wave, and then in the final stage, the diode would be connected to the same capacitor. You could it even generalize it and run x stages at 360/x degree phase.Is it worth it, that's another question. If you can't get enough juice or a clean enough voltage out of this circuit, then you probably need a whole other voltage boosting solution anyway, like a proper DC/DC converter with an inductive element.

author: stryker
author_rank: Regular Contributor
timestamp: May 26, 2013, 07:26:22 am
content: Hi,Quote from: nitro2k01 on May 26, 2013, 07:17:33 amYes, that would have been my question converted to englishYou're right it doesn't sound worth it, but it might be less of a cost than adding more stages to raise the voltage further and then a regulator to smooth things out that Dave mentioned in the vid.Thanks for validating my thoughts,Geoff

author: EEVblog
author_rank: Unknown
timestamp: May 26, 2013, 07:39:13 am
content: Quote from: smashedProton on May 26, 2013, 07:17:02 amThere is also a little known sequel to Wargames!Not that great, but worth watching for fans of the original.

author: trevwhite
author_rank: Frequent Contributor
timestamp: May 26, 2013, 07:51:22 am
content: Hi.I was wondering about the 'processor pin' Dave connected his circuit from. How do you ensure the voltage doubler doesnt exceed the output current of the processor pin. Dont want to be blowing the pin up by accident or straining it and shortening the pins' life.Trev

author: tom66
author_rank: Super Contributor
timestamp: May 26, 2013, 12:55:25 pm
content: I wonder if you could add an extension to this: Perhaps a way to regulate the output voltage using the internal comparator of most micros? That way, you'd only need a couple of resistors instead of an LDO. That would require a little microcontroller software and control loop analysis.And of course, you've also got silly circuits like the microcontroller boost converter I've seen before. It's the same as an ordinary boost converter, but with a diode in series with the switch pin, to protect the IO pin from excessive voltage. It reminds me of my attempt to make a boost converter with an LM555. I think I fried about three or four 555's that day.

author: Pat Pending
author_rank: Regular Contributor
timestamp: May 27, 2013, 03:58:59 am
content: Quote from: nitro2k01 on May 26, 2013, 07:09:34 amOh yeah! That's going back a while, thanks for the recap.While on the topic some might like "Colossus - The Forbin Project",Just discovered that is was based on a novel by "D Jones" - that's D as in Dennis

author: smashedProton
author_rank: Frequent Contributor
timestamp: May 28, 2013, 01:08:52 am
content: Well now that this thread is derailed, "real genius" is another great movie you all should like.

author: jancumps
author_rank: Unknown
timestamp: July 04, 2013, 07:37:53 pm
content: spotted one:

author: NivagSwerdna
author_rank: Super Contributor
timestamp: January 21, 2016, 01:54:29 pm
content: I've just watched this video and think it has a technique that might be useful for a project... I have a clock that needs a 100mS pulse at 24v, the clock has a 10k resistance (although I guess it's really a solenoid so probably behaves a bit more like an inductor than a resistor)... Anyway if I can store enough energy to move the solenoid in a capacitor and then discharge it into the solenoid then the hands will move on one minute.So... if I could charge a 470uF capacitor to 24v and then let it discharge through the clock and then spend the next 59.9 seconds recharging the capacitor it would be good.Given a supply of 9v I need to triple the supply...During the 100mS of discharge the current due to the clock needs to go through all the capacitors... does that imply that it is also going through the PWM pin and therefore I shouldn't actually drive the capacitors directly from my Microcontroller? It's around 2.5mA?Any pointers appreciated... I've never tried anything like this before.

author: MarkR42
author_rank: Regular Contributor
timestamp: November 19, 2018, 10:13:07 am
content: Hi,Could this technique be used, theoretically, to drive a much larger current, if instead of taking the pin directly from a micro. it was from a suitably rated H-bridge (or half H-bridge) chip? Assuming all caps and diodes are suitably rated?Those chips also usually have a current limit built-in, so they won't get fried so easily.I know this was originally intended for driving tiny loads, but what about, say, a motor which draws 1-2A?