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Topic: Adding 3 volts to a small load (Read 1 time) previous topic - next topic

Unknownsymbols

There's this schematic here, it's supposed to be right. It should add 3 V to 12 V by means of 5 PWM V. Could someone explain, what must be the capacitors value, and what must be the PWM percent?
From bits around the internet there is indication that both 10 microF, but other places have it one can be smaller than the other. I don't quite understand the principles here. This is the charge pump, i believe which can operate only on alternating voltage or PWM. So here it's PWM. And i think the frequency can influence the capacitors - the frequenter the smaller. Usually PWM is 50 %, to make a fixed doubler or tripler, but by changing the duty, it can add more or less V. So here it's needed 3 with something. I believe diodes should be of smallest drop, 0,3 V. The frequency is Arduino's, ≼ 65 khz. The load is by the way very small (1 microA) so that's why it is used.

Grumpy_Mike

It is a charge pump, the clue is in the name. Its job is to charge up a capacitor and pump that charge on top of another charge, the output capacitor.

It is not a voltage pump, you can not control the voltage by a different frequency or duty cycle, changing those only change the charge transferred. Of course if that charge is being used to drive a load then as you take more charge out of the capacitor with the load the voltage will go up and down, but this is not the same as controlling the voltage.

Unknownsymbols

#2
May 30, 2017, 02:12 pm Last Edit: May 30, 2017, 02:26 pm by Unknownsymbols
But, will this somehow be able to make a steady 15 volt voltage? Because that's what is needed by the chip that's attached to it. Steady not as like regulated for different load, but as like not oscillating.

Grumpy_Mike

Quote
It should add 3 V to 12 V by means of 5 PWM V.
Assuming that last bit is "means of 5V PWM", no you can only add 5V to it without changing your circuit.

You have to change the PWM signal into a 3V PWM signal in order to add 3V to the output. You could do this with a simple transistor by useing the 3V3 voltage output of the Arduino.

MorganS

To make a steady 15V voltage, first make an unsteady voltage higher than this. A charge pump or voltage doubler can boost 12V up to about 24V very easily.

Then put a voltage regulator on that to make it "steady" and reduce it down to the voltage you actually need. a 7815 regulator will do nicely, so long as you aren't expecting to get more than a few hundred milliamps.

I recently needed a -3.3V power supply for a sensor amplifier, for less than a milliamp. Using a MAX232 chip, I had an unsteady -6V source, so a -3.3V regulator could easily create clean power from that source.
"The problem is in the code you didn't post."

Unknownsymbols

#5
May 31, 2017, 05:20 am Last Edit: May 31, 2017, 05:41 am by Unknownsymbols
Well, the charge pump voltage, should it be called unsteady? To decrease, yes, but I thought voltage regulator wasn't necessary in terms of steadness here, the load is very small, 0,001 mA, and it won't change.

3,3 volts would sink below 3 on 2 diodes. To increase 12 to 24, will it need a transistor again, because PWM is 5 V? So better, to compactize, maybe 5 V PWM should be added to 12, and those 16 reduced to 15 by regulator?

Can it be reduced by voltage divider or just placing more diodes along?


Also, can such a scheme in the right
, found there
be used the same to quatruplize 5 to 18, and then regulator / divider?

In any of those ways, the question would still be, what should be the capacitors, or what formules, that is in what points, calculate it?....

Wawa

Build it up with LTSpice, and you can see for yourself.
Leo..

Unknownsymbols

Oh, hee, how didn't i think to model it? Deamn.

Wawa

I think 100n caps and small schottky diodes will do the trick.
Leo..

Unknownsymbols

#9
May 31, 2017, 05:30 pm Last Edit: Jun 01, 2017, 05:24 pm by Unknownsymbols
Made it with LTSPICE. Yet can't figure it out. In that multiplier with diagonals, why the quintupler barely triples the voltage? And quadrupler just only doubles? On every next halfsection the voltage is 5 V oscillating, then 5 V, then 10 V oscillating, 10 V, then so on. it doesn't help to put 5 V as reference ground. Diagram seems coherent with that link.

Otherwise it operates on all values from about 100 pF to 100 mcF on silicon diodes. The shape is different, but on smallish scale. LTSPICE also revealed diodes by which the drop is so little, won't have to account for it: 1N5817. It's not near 0,3 V - it's some 0,01 V at minimum. At those diodes capacitors have to be larger - it comes to those 10 mcF.

MorganS

... can boost 12V up to about 24V very easily.
Well, the charge pump voltage, should it be called unsteady? To decrease, yes, but I thought voltage regulator wasn't necessary in terms of steadness here, the load is very small, 0,001 mA, and it won't change.

3,3 volts would sink below 3 on 2 diodes. To increase 12 to 24, will it need a transistor again, because PWM is 5 V? So better, to compactize, maybe 5 V PWM should be added to 12, and those 16 reduced to 15 by regulator?

That's why I said "about". You do lose a bit in diode drops and resistive losses. It's still called a doubler, even if it's not quite double.

The voltage doubler output is not regulated. It will drop under load. It will have voltage ripple due to the fact that it's only fed incoming power 50% of the time. The ripple will get worse with more load.

You have 12V DC right? You want 15V? Then the simplest and most reliable solution is to double it and add a regulator. "Adding voltages" is a very bad way to do things.

Another way to do it is use a boost converter. This has a chip in it which monitors the output and regulates to a precise output voltage. Something like this, from Pololu. That's an example of overkill for this application but if $12 is cheaper for you than doing the engineering yourself, it's a good option. The specs page says it can output up to double the input, so it's obviously a voltage doubler inside it, just with more control and higher efficiency at higher loads.
"The problem is in the code you didn't post."

Unknownsymbols

12 is available, yes, but it would need another cable to tug in, so to use only 5 and PWM if possible it's kind of better.
The adapters-converters, i've thought about, they are even cheaper somewhere, but soldering the contacts is even more of enginning level than calculating diodes. Also it would be sort of bulky in that space. Anyway i reckon that as option, but can some1 please write anything about that schematic of tripling the 5 volts?
The initial scheme from the first post operated as expected in SPICE, it adds 5 to 12. But the more general multiplier gets wacky for some reason.

MorganS

So 12V isn't available? Then tripling or quadrupling 5V seems like a good way to go. Depending on how accurately you need 15V (you haven't said yet) then I would probably go for 4x plus a regulator.

Well, there's not enough detail in the first diagram to say whether you've simulated it correctly. Did you pick a real diode or ideal diode? What's the ESR of the capacitors? What's the frequency of the PWM? What load did you put on it?

Then the same applies to your second question.
"The problem is in the code you didn't post."

Unknownsymbols

#13
Jun 01, 2017, 08:08 pm Last Edit: Jun 05, 2017, 07:12 pm by Unknownsymbols
12 V is just a bit unpreferable.

About the model, it's the next drawing, in the post 5. the status is in post 9. I couldn't include the painting inline.
See, in the article it classically is 3 half sections, to triple the voltage. But in the running model, it needs 6.

The exactitude is not important, just the principle, why more stages. Load always 1 microA (maybe less sometimes). Needs to be 14,7 .. 15,3 volts,
As in the picture - i picked the 1N5817 diodes, capacitors were 10 or 9 mcF. Some real capacitors from the top of the list. The results are the same.

I didn't check it with PWM, only usual square pulse on frequency is 500 hz, or 100000, but it doesn't affect it too.

Unknownsymbols

#14
Jun 05, 2017, 07:15 pm Last Edit: Jun 06, 2017, 09:27 pm by Unknownsymbols
Oh, yes, i got it; i guess. It's erric, that example. Actually it's needed 2 full sections, thus 4 half, to add 2 times. Putting 5 V as ground, and then 4 cnapacitors and 4 diodes are needed.

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