PICAXE Charge Pump

hippy

hippy

Ex-Staff (retired)
Anyone with experience of charge pumps want to cast an eye over this ...
Code: 
    -.- +5V
     |
     `---|>|----.----|>|----.----|>|----.----> +15V (ish)
              __|__ +     __|__ +     __|__ +
           C1 ==.==    C2 ==.==    C3 ==.==
                |           |          _|_ 0V
     O1 >-------'           |
     O2 >-------------------'
Ignoring diode drops throughout, with O1 Low, C1 charges up to +5V. With O1 then taken high and O2 low, the voltage at C1+ is 5V plus the 5V C1- is raised to, so it dumps +10V charge into C2. With O2 taken high, it then dumps 15V into C3.

As best I can see O2 can be driven as the inverse of O1. This could therefore be done with any PICAXE with PWMOUT and an inverter, or a 14M in full-bridge mode ( which is even better because there's a programmable delay between O1 going low before O2 goes high etc ).

Am I missing anything obvious ? How would I even go about choosing an ideal PWM frequency ?

My actual target voltage is 10.5V-11.5V, so guess I would drop the final voltage through however many diodes it would need.

Edited by - hippy on 23/05/2007 03:46:16
 
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boriz

Senior Member
Looks ok. But you might need to limit the current with resistors to protect the PIC.

As for frequency. It depends on the CR constant and the load current. I’d just guess then experiment.

Edited by - boriz on 23/05/2007 05:41:14
 
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moxhamj

New Member
Neat circuit. Maybe try 1Khz and 10uF but it isn't all that critical. To get timing constants I tend to breadboard the circuit with a bit of a load and alter the frequency till get the maximum volts.
 
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Jeremy Leach

Senior Member
Definitely need current limiting resistors, but Hippy, I'm pretty sure it won't do what you're saying.

<i> with O1 Low, C1 charges up to +5V. </i>

Yes

<i> With O1 then taken high and O2 low, the voltage at C1+ is 5V plus the 5V C1- is raised to, so it dumps +10V charge into C2. </i>

No. Assuming the capacitors are all the same, then if C2 is 'empty' of charge and C1 'full' then you'll get a resulting voltage of 2.5V at the junction of C1 and C2 (ie the at the diode). Plus you'll be reversing the voltage on your electrolytic cap C1 !

If both are full with 5V charge you'd get 7.5V (I think).

(Think of it as if you have two buckets, one full of water, the other empty .If you connect a tube between bucket 1 and 2, you won't get all the water flowing into bucket 2 !)

Edited by - jeremy leach on 23/05/2007 08:07:15
 
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Michael 2727

Senior Member
What sort of current do you need from the output ?
Prolly won't get more than a few mA Max.
Resistors Yes. Try 220R @ 50/50 duty.
 
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moxhamj

New Member
Re Jeremy Leach, I haven't got my trusty breadboard with me right now but I am pretty sure it will work as Hippy describes. It is similar but not quite the same as the first few stages of a Mosmarx or Cockcroft/Walton voltage multiplier <A href='http://www.patentmonkey.com/PM/patentid/5313384.aspx' Target=_Blank>External Web Link</a> I have built these and they do work. 4001 diodes may not be quite fast enough, but given the 20mA output of a picaxe, 914 diodes will probably be the diode chosen (10mA). The mosmarx circuit does not seem to be out there on the internet but from memory it swapped the diodes for transistors with active switching and thus greatly decreased the voltage losses.

I'm not sure current limiting resistors are needed - under no load there is negligable current draw (cap leakage) and as long as the output is &lt;4-5mA the input will be less than 20mA. But maybe put them in for testing purposes.

Edited by - Dr_Acula on 23/05/2007 08:32:40
 
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Jeremy Leach

Senior Member
I must admit I'm still in thought over this ... need to dust down some old books, so not saying I'm definitely correct <img src="wink.gif" width=15 height=15 align=middle>
 
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moxhamj

New Member
Jeremy, I think your argument is correct and the volts end up being shared equally, but the numbers you gave are just for the first pulse. If the first pulse starts with 10V on C1 and 5V on C2 there will be 7.5V on C1 and C2 at the end of the pulse as you say. But then the negative of C1 goes to 0V, the positive charges to 5V and when that output goes high C1 again goes to 10V. But C2 already has 7.5V so the charge on C1 and C2 end up equal at 8.75V. Gradually with each pulse the charge on C2 goes closer and closer to 10V.


 
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Jeremy Leach

Senior Member
I think you're right Drac, the voltage will gradually increase, but I'm not convinced I got the numbers quite right. It's been a long time since I messed with capacitor equations. Breadboarding would be quicker !

I would still think resistors are needed, but again needs experiment.

 
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moxhamj

New Member
I think your numbers are right. Take two ideal caps of the same value, one with 10V on it and one with 0V on it, join them together and they will both end up with 5V. In the real world both have ESR values so the first cap is discharging via its internal resistor into the second cap which also has a resistor. So a little bit of energy gets lost in these resistances and maybe it is 4.95V.

Re the current protection resistors, there have been posts recently about driving leds from picaxes without resistors. I think this is because the picaxe ends up current limiting, but I'm not a huge fan of this as it exploits a peculiar feature of picaxes and may encourage some of our school age readers to leave out resistors for leds in other circuits. The proper answer for this problem is thus 5V-0.6V diode drop = 4.4V at max 20mA = 220 ohms at O1 and O2 (which is exactly the value Michael2727 suggested).

C3 is presumably a much bigger value and acts to both smooth ripple and power whatever device needs 11V?
 
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Jeremy Leach

Senior Member
My thinking for the resitors is that for a capacitor i=C*dv/dt, so any step change in voltage applied to a capacitor could in theory generate huge current. This will be limited by internal resistance etc, but surely could still be too high?
 
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moxhamj

New Member
Absolutely - current could be very high. Short out a big cap and see the spark - ESR is probably not much more than the resistance of the wires coming out of the cap.

Some real world applications mean one might get away with leaving resistors out. Picaxes current limit. Ditto 555s. Transistors would be current limited dictated by the base resistor. But switch the caps using a relay and there would be sparks for big caps. And even small caps like 10uF might still put a nasty dip on the power supply which might reset the picaxe. So yes, best to include the resistors.
 
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Jeremy Leach

Senior Member
Another interesting thought here - going off on a tangent ...

Let's say the picaxe itself is powered from the output of the charge-pump, via a regulator, and the input to the charge-pump is from a battery. If you can initialise the output voltage somehow to get the picaxe running then you could keep the thing running until the output voltage isn't sufficient to drive the picaxe. If the pump multiplied sufficiently then this could run from low voltage batteries - say a single AA cell.

This is the sort of thing I mean...

Code: 
                                                 D2
           .-------o----------------------------->|o-----o----.
           |       |                               |     |    |
           |       |                               |     |   .-.
           |       |                               |     |   | |
        D1 -      ---   PICAXE                     |     |   | |R1
           ^    C1---  .-------.                   |     |   '-'
           |       |   |       |             .------.    |    |
           |       '---|OUT  V+|-------------|      |   ---   o-----.
           |           |       |             |      |   ---   |     |
    1.5V  ---      .---|ADC    |             |      |  C2|   .-.    |
  Battery  -       |   '-------'             '------'    |   | |    |
           |       |       |                LDO  |       |   | |R2  |
           |       |       |           Regulator |       |   '-'    |
           |       |       |                     |       |    |     |
   GND o---o-------)-------o---------------------o-------o----o--o  |
                   |                                                |
                   |                                                |
                   '------------------------------------------------'
Assuming the Picaxe is already running (somehow!), then..

C2 provides the power to the Picaxe via a low dropout regulator.
R1 and R2 present a voltage to an ADC input pin of the Picaxe. This voltage is
monitored, and if it drops below a certain threshold the Picaxe 'tops up' C2 by:

1. Taking the output low, giving time for C1 to charge via D1 (and current limter resistor I've not
shown)
2. Raising the output to V+ which dumps charge into C2.

This might be a simple trigger circuit shown below.
It's a bit difficult to draw, but it works via a
Double-Pole Double-Throw switch.

I'm thinking here of a scenario where you use 2*AA batteries instead of 4, to
save space.

Switch is in position1 when 'off'. In this position C1 is fully charged via R1 because
it's directly across the battery. When the switch is thrown to 'On' the top of C1 will
go to 2*3V = 6V which is a trigger voltage applied via a diode into the main C2 capacitor of the other diagram. This will be enough to activate the Picaxe and start
the cycle up, so once powered it should
stay powered.

Basically this circuit will do a one-off doubling of the battery voltage, in order
to activate the picaxe. So unfortunately
it wouldn't provide enough umph on just
a single 1.5V cell.

Another idea relating to the other diagram,
is that it might be possible to use the ADC
input as an interrupt instead, and set the
resistors to trigger an interrupt when
'top-up' is required.

Code: 
                          1
              .----o-----o
              |    |     __--o-.
      -|<-----)----)-----o     |
              |    |+     2    |
 Trigger      |   ---          |
              |    -          ---
              |    |+ +3V     --- C1
              |   ---(2*AA)    |
              |    -          .-.
              |    |          | | R1
              |    |      1   | |
              |    .-----o    '-'
              |    |     __--o-o------o + V intocircuit
              '----)-----o
                   |      2
                   |
                   '------------------o GND


Edited by - jeremy leach on 23/05/2007 21:53:23
 
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c6jones720

Member
You see circuit diagrams of chips that do similar things to that all the time in the back of magazines like &quot;new electronics&quot;. Im sure there are quite a few charge pump chips.

Here: you can have 5V for the price of 3.3v
http://cn.maxim-ic.com/appnotes.cfm/appnote_number/3202
 
hippy

hippy

Ex-Staff (retired)
Thanks for all the comments and they've been very useful.

I was wondering about currents through the PICAXE. Given I/O is rated at 25mA source or sink, at 50:50 duty that could be considered 50mA if only passing current 50% of the time, bit if it ends up switching between sourcing and sinking that would keep the 25mA limit.

What's intriguing me is - and I may be in fantasy or ignorance land here - is there anything other than minuscule current flow at all through the I/O pins ?

Parcels of charge are pulled in from +5V to C1 where they accumulate so, like filling a bucket, nothing actually flows out the bottom, so no current sunk by O1.

When O1 is put high, the parcels of charge in C1 are then tipped into C2, but no current goes through 02 as above, and no current is pulled out of O1.

There will be inrush current when first powered up and C1-C3 settle at nominally 5V, but that should all be current from +5V. Once the system is free-wheeling, there should be very little current flow either if C3 doesn't discharge that much.

The 11V probably only needs to deliver sub 1mA; enough to put a PICmicro into programming mode and that voltage is not used to do the internal programming. I don't think that some ripple would even be too much of a problem. I can diode feed C3 into further caps while dropping the voltage and smoothing further, and I'm sure one or more 100nF will help in there.

I appreciate this won't be 100% efficient and there will be losses and I think Jeremy and Dr Acula are right about the levelling and bumping up of capacitor voltages. In my case, because I want nearer 10V than 15V losses would work to my advantage but Sod's Law dictates it will be better than hoped for :)

Feeling confident, I'll build one without R's using a sacrificial PICAXE-08, 10uF's, 1N4148's, bit-bang PWM at up to ~2kHz and see what happens. I know there are chips which can do this better, but I'm again on a 'use what I've got' project. I'll let you know if it explodes or not :)
 
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Jeremy Leach

Senior Member
Er, maybe I shouldn't have mentioned buckets ;-)

Whenever the charge changes on the cap there will be current flow ( i = C*dv/dt etc), and what flows into one end will flow out of the other. So the output pin will source/sink the current to charge/discharge the cap.
 
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Michael 2727

Senior Member
PWMOUT 2, 249, 500 = 4KHz @ 50/50 Duty
5V - 1 doide drop 4.4V
1&#181;F
68 Ohms
<A href='http://ourworld.compuserve.com/homepages/Bill_Bowden/rc.htm' Target=_Blank>External Web Link</a>
http://ourworld.compuserve.com/homepages/Bill_Bowden/rc.htm

Note: when using the calculator the instantaneous
voltage can never be as high as the input voltage
as theoretically this can never happen due
to ESR, leakage etc.
You must enter a voltage slightly lower or
you will only get &quot;infinity&quot; as an answer.

A good example of this -
Try charging a Super Cap and watch the voltage
on a multimeter.
A 1Farad Super Cap will keep you waiting 4 to 5
mins to get up to 4.95V from a 5V supply.



Edited by - Michael 2727 on 23/05/2007 15:33:07
 
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boriz

Senior Member
If the CAP is empty/full, when the output pin toggles, it looks like a SHORT CIRCUIT to the PIC. For charging AND discharging. This condition remains until the Cap is charged/discharged. But then you toggle the output again&#8230;

It&#8217;s a bit like wiring the output directly to ground OR +5v then driving it with a square wave. Protection circuitry in the PIC may keep it from dieing for a while. But eventually it&#8217;s going to break. ESR is usually so small, it&#8217;s not really a factor. I&#8217;d definitely use a resistor.
 
hippy

hippy

Ex-Staff (retired)
Jeremy : <i>Er, maybe I shouldn't have mentioned buckets ;-) </i>

Maybe not :) but it set me thinking, 'exactly what is a capacitor, and how does it work', a &quot;bucket for charge&quot; is how I recall it from happy school days Physics ...

<i>Whenever the charge changes on the cap there will be current flow ( i = C*dv/dt etc), and what flows into one end will flow out of the other. So the output pin will source/sink the current to charge/discharge the cap. </i>

Agreed, current flows into the capacitor, but does it really pass through, or does it stay there ? My problem is I cannot, these days, comprehend how it all ties together; electrons, current, charge. And let's not mention which way electrons actually flow in a circuit !

<b>Anyway ... Good News - It Lives ! </b>

A quick re-cap to save going back a page -
Code: 
    -.- +5V
     |
     `---|>|----.----|>|----.----|>|----.----> +15V (ish)
              __|__ +     __|__ +     __|__ +
           C1 ==.==    C2 ==.==    C3 ==.==
                |           |          _|_ 0V
     O1 >-------'           |
     O2 >-------------------'
All driven from an 08 running -

Code: 
 Do
    Low 1 : High 2
    Low 2 : High 1
Loop
3 x 1N4148 diodes, C1 and C2 are 10uF/16V, C3 is 10uF/25V.

With ( an arbitrary ) 2K2 across C3/0V, +V=5.00V, C1=6.35V, C2=9.61V and C3=12.76V. With C2 removed and the C2-C3 diode shorted, C1=6.40V, C3=8.73V - DMM readings so there's probably some ripple being averaged out.

I don't fully understand that over 5V on C1. It's the same with download cable connected or not. I'll <i>assume </i> it's 'free energy', patent the idea, and save the planet. Coupled with Jeremy's self-powering scheme we will start Perpetually Powered PICAXE Panopoly Inc <img src="smile.gif" width=15 height=15 align=middle>

The Lab Supply shows 5mA, 3mA with +V to the circuit disconnected, and there's ~2mA through the 2K2. It shows 3mA with the 2k2 removed, which fits with my 'zero current' theory. The PSU does 'max out' at turn-on with inrush current because I've limited current to 25mA.

I've no scope so cannot say what the cycle frequency is, how nice or nasty the signal on C3 is, or how full of EMI and switching spikes, but if someone wants to try it out, my PICAXE-08 didn't release any magic smoke.

For completeness, at +V=3.00V, C1=3.58V, C2=5.28V, C3=6.91V
 
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Jeremy Leach

Senior Member
If you're measuring DC voltage across top of C1 and GND then you would expect a voltage between 5V and 10V.

When O1 is low, C1 will charge (if given enough time) to +V minus the forward drop across the diode (call it 0.5V) = 4.5V.

When O1 is high the voltage will rise to the high voltage of the output pin (call it 4.7V) plus 4.5V = 9.2V. It will discharge slightly also, but ignoring that, we've got 4.5V and 9.2V each for half a cycle, which is an average of 6.85V. You get 6.35, so that drop must be due to C1 charging C2.

(PS:I had initial nagging doubts about my circuit being another perpetual motion farce, but there is an energy source so I still 'think it could still work!').


Edited by - jeremy leach on 23/05/2007 19:25:19
 
hippy

hippy

Ex-Staff (retired)
boriz : <i>If the CAP is empty/full, when the output pin toggles, it looks like a SHORT CIRCUIT to the PIC. For charging AND discharging. </i>

Food for thought there. I'll certainly agree that when C1 is discharged and O1 low it's effectively a short to 5V-vDiode, and C2 discharged with O2 low is a short as well, but I'm not so sure about when O1 or O2 are high. In that case it looks to me like a 5V-to-5V short which should be okay, but I haven't thought this through entirely.

The bottom line - whether its imminent demise is premature or not - probably rests upon what currents are drawn in any shorted states and for how long. The pins can handle 25mA continuous and more for shorter periods. It's beyond my knowledge or measuring skills to investigate that.

I have also found this <A href='http://www.voti.nl/wisp628/index_1.html' Target=_Blank>External Web Link</a> which is identical ( for an identical application ) but using smaller (C1,C2=1uF) caps which would limit the time of any inrush or short-circuit currents as they charge. Full credit therefore to the renowned Wouter van Ooijen. Not sure why he didn't just take a +12V feed off the MAX232.
 
hippy

hippy

Ex-Staff (retired)
Jeremy : <i>If you're measuring DC voltage across top of C1 and GND then you would expect a voltage between 5V and 10V. </i>

Of course ! It's not a static voltage. Thanks; have an extra Panopoly Share.

You also set me thinking about how PWM etc could alter the voltage for tuning the cycle frequency, but then I thought; measure the output voltage and adjust Pause's or vary real PWMOUT duty to get the actual output voltage I want.

Who dares say PICAXE's aren't fun :)

Edited by - hippy on 23/05/2007 19:42:59
 
premelec

premelec

Senior Member
OT - Did I hear 'buckets' ? There is an interesting IC called a 'bucket brigade' where hundreds of stages pass charge forward on each pulse [like an analog shift register]- no fire is put out but interesting audio effects can be had...
 
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skyhawk

New Member
<BLOCKQUOTE><font size=1 face=arial>quote:<hr height=1 noshade>The Lab Supply shows 5mA, 3mA with +V to the circuit disconnected, and there's ~2mA through the 2K2. It shows 3mA with the 2k2 removed, which fits with my 'zero current' theory. The PSU does 'max out' at turn-on with inrush current because I've limited current to 25mA.[/Quote]

I must be misunderstanding something. Shouldn't 12.76 volts across 2K2 produce a current of 5.8 ma?

<hr height=1 noshade></BLOCKQUOTE></font><font face='Verdana, Arial, Helvetica' size=2>
 
hippy

hippy

Ex-Staff (retired)
skyhawk : <i>I must be misunderstanding something. Shouldn't 12.76 volts across 2K2 produce a current of 5.8 ma? </i>

Interesting and intriguing observation. This is where I'm out of my depth. My guess is that once the 12V is charged up full, it just needs a small charge to keep it topped up, and that's the only charge which then comes through the 5V line. The Lab PSU isn't calibrated and is probably also averaging or missing peaks.

Without a scope I cannot even tell if the cap is discharging near completely, and if it were discharged for half the time that would drop the average current down by half too.
 
hippy

hippy

Ex-Staff (retired)
Some more experimental results. Replaced C1 and C2 with 1uF/50V, and output went up to 14.25V across 2K2 and there's no obvious 'max out' current limiting shown on the PSU, meaning it's gone or is much shorter. the total current draw from the 5V PSU has now dropped to 3mA. The DMM initially reads 14.7V when connected to the 2k2 loaded output, then drops to 14.25V so it's not perfect, and there was a much smaller drop seen with the 10uF capacitors.

Undoubtedly the increased output is because the smaller caps charge and discharge quicker, and can therefore pass charge along much quicker. It's obvious that it's a complex field and a careful balance between cycling frequency and capacitor values, to maximise efficiency and voltage while reducing ripple and noise. I did find a PDF detailing the maths, and when I've got my PhD in <i>Euclidean Alegbra in the Context of Time-Space Inversion Scenarios </i> I'll understand it. In the meantime, a meter on the output suggests it's working. How well, is a differnet matter.
 
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Jeremy Leach

Senior Member
You'd only go off and be a rocket -scientist, we need you here, so stick to the multimeter <img src="wink.gif" width=15 height=15 align=middle>
 
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skyhawk

New Member
Since as a physicist I'm a believer in the conservation of energy, a power analysis is even more intriguing. The power supplied to the load is 74 mW. This power plus that dissipated in the diodes plus the power to run the PICAXE must come from the 5 V supply. A 5 V supply must supply 14.8 ma to deliver 74 mW.

In order to understand where that current is going, I looked at the charge transfer that occurs in the charge pump once it reaches it's steady-state cycle. The average current used by the charge pump is three times the average current supplied to the load. The source of the current is equally shared by the three inputs to the charge pump, the two PICAXE pins and the 5 V supply. Thus, with 5.8 ma through a 2K2 load the charge pump should draw 17.4 ma. The difference between 17.4 ma and the previously calculated 14.8 ma repressents the power loss in the diodes.
 
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inglewoodpete

Senior Member
Jeremy, On the topic of buckets and capacitors...

I use the analogy of a barrel with a diaphram across the centre. Each end of the barrel has a pipe, of course, equivalent to the two leads of the capacitor. I'll leave you to imagine the flow of DC and AC water 'currents' and 'voltages' (pressures). The model even supports what happens when too much voltage (water pressure) is applied: the diaphram (dialectric) ruptures.

The inductor on the other hand can be likened to a water turbine. When the water flow stops, the turbine will tend to continue to turn, tending to perpetuate the pressure and flow. This has similarities to the colapsing magnetic field generating a voltage of a the same polarity.

Just my Friday thought-of-the-day <img src="wink.gif" width=15 height=15 align=middle> Now we can all get back to rocket science...
 
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moxhamj

New Member
That is a great analogy. I have a vibration damper on a piston pump which is a diaphragm and has compressed air on the other side. It smooths out the vibrations, in exactly the same way as a capacitor on a supply rail. The boost inductor circuit and the waterhammer ram pump have much in common. In the water analogy resistors are narrow pipes. Transistors I suppose are like the hydraulic servo on power brakes or power steering.

One could design a whole electronic circuit out of hydraulics.
 
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Jeremy Leach

Senior Member
Yes, that is a great analogy Pete, much better than buckets - which I realised can be misleading, although good enough for bucket brigade delay lines.

Great one for inductors too. I get a lot from this forum !
 
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Jeremy Leach

Senior Member
The capacitor voltages in this have been bugging me, so here are some calculations I've done, going back to some basics. I may have messed up, but hopefully not (If I have I'd like to know how for my own understanding!)...

Code: 
                    o
                    |  +
                   --- |
                 C1--- 4.8V
  Picaxe            |  |        o  ^
  Output  + o-------o  |        |  |
          |                    --- Vi
       4.7V                  C2--- |
          |                     |  |
          | o-------------------o-------o GND
This diagram shows the state the instant the Picaxe output goes high (The diode between the top of C1 and top of C2 omitted because it has only just started conducting).

C1 has been charged to 4.8V ready to dump some of this charge into C2.
Assume the output voltage of the Picaxe is 4.7V.
Assume C2 is already charged to an initial voltage Vi.

For capacitors Q = C*V. Plus let's make it simpler and say C1 = C2. Therefore:
The initial charge on C1 = Qi1 = C * 4.8
The initial charge on C2 = Qi2 = C * Vi


Now let's show the circuit with the diode, and for the sake of simplicity assume it's effectively a wire...


Code: 
                    .-----------.
                    | ^         | (D1)
                   ---|         |
                 C1---V1        |
  Picaxe            | |         o  ^
  Output  + o-------o |         |  |
          |                    --- V2
       4.7V                  C2--- |
          |                     |  |
          | o-------------------o---------o GND
The capacitors, now connected, will swap charge and settle to voltages V1 and V2.

The final charge on C1 = Qf1 = C * V1
The final charge on C2 = Qf2 = C * V2

But we've just connected the 'plates' of the two capacitors together, so the total (positive) charge on the joined plates is the sum of the initial charges = Si = (C * 4.8) + (c * Vi), and this charge can't change.

The negative charges have to balance the positive, which means:

Qf1 + Qf2 = Si ...Eqn1

The voltages must also add up in the circuit, ie: 4.7 + V1 = V2
But V1 = Qf1/C, and V2 = Qf2/C, therefore:

4.7 + (Qf1/C) = Qf2/C
or (4.7*C) + Qf1 = Qf2 ...Eqn2

Substituting Qf1 from Equ1 and expanding Si gives:
(4.7*C) + (C * 4.8) + (c * Vi) - Qf2 = Qf2
Therefore :
Qf2 = [(4.7*C) + (C * 4.8) + (c * Vi)]/2

Therefore:
V2 = [4.7 + 4.8 + Vi]/2

<b> V2 = (9.5 + Vi)/2 </b>

This makes sense because if C2 is fully charged initially (ie Vi = 9.5) then V2 remains at 9.5. However, what it does show is that if Vi = 0, then V2 = 4.75 so it doesn't look like there's a worry with C1 being electrolytic, because it won't get a reverse polarity applied to it.

This result ignores the diode drop so isn't very accurate. If there's no load then starting from Vi = 0, repeated charge cycles would show V2 going up something like this...

4.75V
7.13V
8.31V
8.90V
9.20V
9.30V
etc


Edited by - jeremy leach on 25/05/2007 15:20:40
 
Last edited by a moderator:
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piclt

Member
This seems an interesting topic ! !
Hippy's charge pump would seem to be similar to the Dickson charge where it can have many stages driven by 2 anti phase clock pulses. The odd stages driven by one pulse and the even stages driven by the anti phase pulse. The web link gives a good writeup of this also improvements using static switches instead of diodes
http://www.eecg.utoronto.ca/~kphang/ece1371/chargepumps.pdf

We should maybe start another discussion about turning the diodes and capacitors round the other way and generate a -ve voltage from the positive pulses, Thas might confuse readers ??

 
I

inglewoodpete

Senior Member
Thanks for that link, piclt. I had been searching for charge pump discussion papers but could only find AC-driven ones. The pdf has a good discussion of the theory, which I have filed away for future reference.
 
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DavidGw

New Member
Hi Hippy,

I ran across this document on microchip's web site. Go to their site http://www.microchip.com and do a document search for &quot;DS40040C&quot;. Download the PDF and open it and look for TIP #10 and TIP #11.

Dave


Edited by - noesisdg on 28/05/2007 01:24:26
 
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