Charge pump capacitor values

[Sage]

I'm building this charge pump circuit, but with an LT1054 instead of a MAX1044:



Notably, this schematic uses 10uF caps for both C1 and C2.  However, the example circuit for a similar voltage doubler in the LT1054 data sheet suggests a 100uF cap for C2.  I've set this up on a breadboard and both values seem to work.  What difference does it make?  From what I've read, it sounds like higher values reduce current ripple... is that the case here or are there other effects?

[italianguy63]

Look at the specific spec. sheets for the different charge pumps.  There are slight differences.  Each tend to have sample circuits on them with the values.  100uF may very well be OK.  Looks like it is a filter cap to me.

MC

[italianguy63]

Edit--

That is as long as it is on the END of the chain.  In my VERY LIMITED knowledge of these things, the pump acts like a "bucket brigade" passing bites (chomps) of power down the line.  If you have a cap that is too large in the chain, the charge of it may be too big for the next one in the line to overcome.. i.e. it will keep filling and not let the charge pass.  So, as long as the VERY LAST cap is large, you are OK.  But, the ones down that chain I would think should all be the same size (and similar to the one on the "bucket" side)-- 10uF.

That's what I think anyway.  If I am wrong, someone here with superior knowledge or experience will chime in!  I love this place--   

MC

[anotherjim]

Every cap after the diode pairs is the reservoir cap. The one between the diodes is the charge pump cap - that has to be sized to fully charge and discharge in the on and off times of the pump chips' output, but the reservoir needs to hold the charge from every pump cycle and deliver current to the next stage. The reservoir cap will have an optimum size -  it depends on the load current and switching speed of the pump.

[italianguy63]

What he said.

[Mark Hammer]

Just make sure the cap voltage rating is based on what is optimal for the highest voltage produced.  So, while a 16V rating may be fine for a 9V supply, if your goal is to end up with a tripled output  (somewhere in the vicinity of just over 25VDC), you probably want to be using 35VDC caps.

[Sage]

Just make sure the cap voltage rating is based on what is optimal for the highest voltage produced.  So, while a 16V rating may be fine for a 9V supply, if your goal is to end up with a tripled output  (somewhere in the vicinity of just over 25VDC), you probably want to be using 35VDC caps.

Yeah, I'm using 50V caps to be safe.  I'm actually taking a 12 VDC input (hence the LT1054 instead of the MAX1044) and tripling it to roughly 36V (34V when you factor in voltage loss from the diodes) and then using an LM317N to bring it down to a steady 30V.

Every cap after the diode pairs is the reservoir cap. The one between the diodes is the charge pump cap - that has to be sized to fully charge and discharge in the on and off times of the pump chips' output, but the reservoir needs to hold the charge from every pump cycle and deliver current to the next stage. The reservoir cap will have an optimum size -  it depends on the load current and switching speed of the pump.

Not a lot of load current; I'm driving a pair of buffers with a TL072, and that's it.  The internal oscillator runs at 25 kHz.  Like I said, the schematic in the datasheet suggested 10uF for the charge pump cap and 100uF for the reservoir, I was just wondering, is that a necessity or just a general purpose suggestion?  A 100uF capacitor is a bit bigger than a 10uF.  Does the reservoir just have to be greater or equal in value to the pump?

[armdnrdy]

Read the regulation and capacitor selection sections.

http://cds.linear.com/docs/en/datasheet/1054lfh.pdf

[anotherjim]

Reservoir definitely should be bigger than pump. It is the bucket in which all hope resides - you don't want it to get empty, nor be so full that it can't take any more. 10 x greater than pump cap seems a reasonable "rule of thumb".

[Transmogrifox]

Youch!  I looked at these things on DigiKey and depending on whether you get the LT part or the TI part ranges from about $3.00 to >$7.00 for these little beasts.

You can get a little 200 mA 100 uH inductor for about $1.25 and for a few pennies more with either a couple transistors and MOSFET, or a 555 timer you can make a simple inductor boost converter.  Probably the whole boost converter would cost you $3.00 including caps,diodes and active parts.

As for inductors, you can probably make one even cheaper with a little coil of wire and a nail.  Charge pumps aren't known for power conversion efficiency, so you might even be competitive using a nail and wire for conversion efficiency.

I know that's a little off-topic, but OTOH maybe for future projects it's something worthy of consideration.

[PRR]

> The reservoir cap will have an optimum size

The ivy-tower "optimum" is infinite for all these capacitors.

Not just infinite uFd, but infinitely low internal series resistance and infinitely high shunt resistance.

Since Mouser does not stock infinite caps, and the price might be infinite, the "optimum" is purely commercial.

If the capacitance is low, the output voltage is low, and the ripple is high.

If the capcitance is high, bulk is large and budget is wrecked.

The reactance of 10uFd at 20KHz is under 10 Ohms; of 100uFd is under 1 Ohms.

The load can not be less than 240 Ohms (double-12V is 24V, 100mA max current is 24V/0.1A or 240 Ohms); you may not be using this much.

The pump-cap carries double the current and half the voltage, so should be much-much less than 240 Ohms. Probably under 6 Ohms.

So *IF* you expect high current, you should lean toward 100uFd.

If you are just powering an opamp or two, your current is more like 10mA, load is 2,400 Ohms, caps should be under 60 Ohm reactance, 10uFd may be ample to get your double-voltage.

Ripple is tougher to decide. An ideal opamp will reject power ripple, so as long as ripple is not huge, everybody is happy. However opamps with high rejection at low frequency may let-in a lot of ripple at 20KHz. I'd err on the high side (large output cap), but at low current this may still be 10uFd. And if 10uFd doesn't do it, 100uFd is probably not utter happiness neither. And aside from a grossly under-sized last cap, supersonic troubles are more likely due to long leads common between pump and output.

[italianguy63]

Youch!  I looked at these things on DigiKey and depending on whether you get the LT part or the TI part ranges from about $3.00 to >$7.00 for these little beasts.

The LT1054 is a nice piece.  The MAX1044 has overvoltage issues (fragile), and I avoid it.  For bang for the buck, I recommend the TC1044SEPA from Mouser-- right at $1 a copy.

And just to confuse things.. DON'T go on FleaBay and do a search on LM2596 or LM2577 Buck Converters...  I still don't know how the Chinese can make these so cheaply.

MC

[Sage]

The load can not be less than 240 Ohms (double-12V is 24V, 100mA max current is 24V/0.1A or 240 Ohms); you may not be using this much.

That's disconcerting.  How do I measure, and is there any way around that if I'm not?

The pump-cap carries double the current and half the voltage, so should be much-much less than 240 Ohms. Probably under 6 Ohms.

So *IF* you expect high current, you should lean toward 100uFd.

If you are just powering an opamp or two, your current is more like 10mA, load is 2,400 Ohms, caps should be under 60 Ohm reactance, 10uFd may be ample to get your double-voltage.

Just that single TL072 that we were discussing in the split rail thread.  While bulk is not something I want to add too much of, a pair of 100uF caps isn't a dealbreaker for me, and the cost difference is negligible in my case since this is a one-off.  Unless there are any other advantages to dropping the reservoir caps to 10uF, it sounds like this is an exercise in "what can I get away with" rather than "what's my ideal solution," so I'll stick with 100uF unless you know of a good reason not to.

Ripple is tougher to decide. An ideal opamp will reject power ripple, so as long as ripple is not huge, everybody is happy. However opamps with high rejection at low frequency may let-in a lot of ripple at 20KHz. I'd err on the high side (large output cap), but at low current this may still be 10uFd. And if 10uFd doesn't do it, 100uFd is probably not utter happiness neither. And aside from a grossly under-sized last cap, supersonic troubles are more likely due to long leads common between pump and output.

Assuming I use the LM317N to stabilize the voltage at 30V, wouldn't that filter the ripple for me?

[MrStab]

Just to offer a real-world example that should hopefully illustrate the theory:

With 10uF caps, i achieve around 27V. with 22uF caps, it's 31V. Rated at 63V. Not too friendly on real-estate, but does the job.
That's using an ICL7660S, with the input voltage regulated via. a zener at 9.1V.

[Mark Hammer]

If I may be so bold as to use a seasonally-adjusted analogy: bigger baseball glove, you catch more balls...up to a point.

The caps are essentially "catching" a series of pulses, hanging on to them, and summing them.

[PRR]

> That's disconcerting.

What? That the size of your power-contraption is proportional to the power you need? Or that you do not know what power you need?

> that single TL072 that we were.......

Sorry, had my head in my sewer-pipe for two days and can't remember nothing.

One TL072 is (you could look it up) around 5mA. Far-far less than the 100mA maximum of this doubler. You probably do not need the biggest caps.

> bigger baseball glove, you catch more balls

Well, or bigger balls.

One TL072 is more like a grape.

> ...illustrate the theory: With 10uF caps, i achieve around 27V. with 22uF caps, it's 31V.

Yes. Above some bare-minimum, we expect large change of cap (2.2X) to give small change of voltage (1.15X).

[Sage]

> That's disconcerting.

What? That the size of your power-contraption is proportional to the power you need? Or that you do not know what power you need?

> that single TL072 that we were.......

Sorry, had my head in my sewer-pipe for two days and can't remember nothing.

One TL072 is (you could look it up) around 5mA. Far-far less than the 100mA maximum of this doubler. You probably do not need the biggest caps.

Ah, I think I got mixed up.  You said:

The load can not be less than 240 Ohms (double-12V is 24V, 100mA max current is 24V/0.1A or 240 Ohms); you may not be using this much.

I thought you meant I wasn't drawing enough current.  I'm guessing you meant it can't be more than that?  I'm definitely below that threshold.

[tubegeek]

bigger baseball glove

Stick to hockey, you Canuck!

The load can not be less than 240 Ohms (double-12V is 24V, 100mA max current is 24V/0.1A or 240 Ohms); you may not be using this much.

That's disconcerting.  How do I measure, and is there any way around that if I'm not?

I think, from your reaction, you maybe have misunderstood what PRR meant.

What he's saying is: if you do an Ohm's law calculation, using the maximum specs of the chip (24V and .1A, I'm assuming) the worst-case scenario for your load (the single op-amp, which is all that you are powering with that power supply) is that your circuit must look like an impedance no lower than 240 ohms from the supply's point of view.

BUT: you are well within that limitation because your circuit is drawing far less current (1/20th of it) so it looks to the supply like an impedance 20 times larger, so let's say 4800 ohms. (You could round to 5K if you like that better to calculate with.)

Then what comes next is a voltage division. PRR is comparing the reactance of the cap to the equivalent resistance of the load to check what the ratio is. The good news: as the load's equivalent resistance rises, the smaller the cap can be while maintaining at least the same ratio with the true, larger resistance. (Smaller cap = larger reactance.)

The ratio between the cap and the load forms a voltage divider that establishes how much loss of charge the capacitor will suffer when the load is drawing off of it, and if I followed the description correctly, we're shooting for something like at least a 40:1 ratio (OR LARGER - larger is better here but we can't go too big with real components and still fit them into a walnut shell) between the resistance-equivalent of the load and the reactance of the cap.

Now: knowing PRR's old-school methods as I do, I'd bet the next thing he did was pull out a reactance nomograph and see what the reactance of various capacitors would look like at whatever frequency the boost converter is running at. This is the way it was done when you had to wash your hands to pull out your slide rule to figure stuff out unless you had that nomograph laminated to the back of your clipboard. Then you could leave the slide rule in the desk and keep it clean if you were in the middle of a stinky sewer pipe problem.

(He uses fewer words than I do but sometimes I like to translate for him. This way he can get back to fixing his sewer. And what I get out of it is an insight into how his mind works, which is humbling. PRR's methods cut VERY quickly to the chase, I find.)

One more thing:

PRR: my condolences. There is no such thing as a lovely sewer pipe issue. Hope you are done soon.

EDIT:

Here is PRR's secret weapon:

http://www.rfcafe.com/references/electrical/frequency-reactance-nomograph.htm

[PRR]

> wasn't drawing enough current.

Taking rough numbers: 24V at 100mA is 24V/0.1A which is like 240 Ohms.

(I'm not sure this multiplier will OUTput 100mA; that may be an input limit?)

Using TL072 current consumption (which hardly varies with voltage 8V to 36V), and Grant's 27V-31V, we get more like 5,000 Ohms. Which is less current. It also points to smaller power-parts, particularly caps, rather than the show-off circuit in the multiplier datasheet.

[PRR]

> pull out your slide rule

No. The sliderule gives you the significant digits but NOT the decimal point.

And in this case we don't need *any* precision beyond the rough order-of-magnitude (the decimal point).

You memorize several land-marks to get your place.

I remember that 0.01uFd against 1K Ohms is the top of the audio band. If you have (or can fake) a 600 or 1K node (such as an interconnection), 0.01u across it will shave the top of the audio band (and cut-out a little hiss and some supersonics).

These converters run near the top of the audio band.

So if 0.01uFd is 1K, then 10uFd must be about 1 Ohms at the top of the audio band. The sliderule says there is a number 796, the calculator says the answer is 0.7957747.. Ohms, neither any more useful than "about 1 Ohm".