Charge Pumps; What's the catch?

[seadi123]

I mean you can't just get 18v out of 9v without compromising something? What is it, more current draw? More heat?

[Mark Hammer]

Essentially 3 catches, in my view:

1) There are limits to how much current a single one can supply.
2) Inserts a HF clock into the mix.
3) Takes up more space and parts.

None of those are dealbreakers, as long as the circuit is compatible with those conditions.  So, if the enclosure can accommodate more components, here are no other clocks to conflict with in the circuit and no bandwidth requirements that might make the clock audible,  and the circuit has no heavy current requirements, go for it.

[Groovenut]

Most chips are ~99% efficient, so you lose 1% in their process. You also lose a diode drop for each diode in the process. You never actually get 18v from 9v. You can minimize those loses with good design, but you will still be limited by the things Mark mentioned above.

[R.G.]

Mark is correct on all of that.

Looked at another way, you don't really get 18V from 9V. You get something that is twice the 9V minus at least one diode drop, and also minus however much the "18V" sags under load. A simple  charge pump IC often sags as though it has a 40 to 80 ohm resistor inside. So you lose another 0.8V for every 10ma of current you draw. Turning on an LED can sag your power voltage significantly.

Then there are the laws of thermodynamics. The first law says "you can't win". The second law says "you can't break even". In this context, you can't get more power out of the 18V than you supply through your 9V supply, so at the very best case, you will need twice the current from the 9V supply that you take out of the 18V supply. Anything else violates the first law, making energy out of nothing. The second law tells you that there are always losses; in this case, the diode loss, the heating of conductors along the way, the ripple voltage from the switchng conversion, all contribute to the useful DC you get out being less than what you put in.

So yes, there are compromises; current is limited, there is a high-power, high frequency clock running around loose on your power supply wires, it takes more parts and space. It delivers less than twice the voltage you put into it, and eats more than twice the current you get out, and generates some heat along the way.

There are always compromises. Formal training in the sciences teaches you that all known processes start with the three laws of thermodynamics, and these set some boundaries for the evaluatiuon of new situations.

[Mark Hammer]

But, all of those party-pooper caveats aside, it beats having to use two 9v batteries, and also permits use of a 9V wallwart and standard  plug.  Although, having a power brick that can provide supply voltages higher than 9VDC is probably a better solution than use of a charge pump.  Chances are pretty good that unit will be able to provide the current you want/need, and even if it is a switching supply (with its own HF clocks), it's nice to have thse clocks in a separate enclosure; hopefully prevented from crawling their way along the power lines to your pedal.

[electrosonic]

More parts can mean more modes of failure. Charge pumps can be particularly unforgiving when fed too high an input voltage.  The convenience of the charge pump (ie plugging into your 9vdc daisy chain vs having a dedicated wall wart for that pedal) is offset somewhat by the added complexity and decrease in robustness of the pedal.

Andrew.

[amptramp]

Charge pumps are not all that efficient.  You only see them used for small current drains.  What a charge pump does is switch voltage onto a pair of capacitors than activate different switches to put them in series.  This charges an outboard capacitor that retains most of the higher voltage.  As the switch goes through infinite ohms to zero ohms, there is a dissipation that occurs due to the charging current through resistance of the transistors.  Although the parts count is small, the losses are high, which is why they are restricted to low power.

The next step up the efficiency ladder is the flyback power supply.  It switches an inductor on then off and the voltage spike is rectified and used as the output.  Televisions using CRT's used this system for about sixty years to generate the high voltage from the horizontal sweep circuit.  Electrically, it looks similar to an ignition system.  The efficiency advantage comes from the fact that when you switch on the inductor current, the switch goes to low resistance before the inductor current has a chance to rise, so turn-on efficiency is high.  There are a number of other power supply topologies such as the forward converter, the push-pull converter and the resonant converter that all offer advantages.

You can design whatever you need to accomplish the task when certain criteria are set.  Want a converter with low levels of interference?  The resonant converter does this and has the advantage of high efficiency because switching devices are turned on at zero voltage and off at zero current, meaning the losses are minimal.  Want small magnetics?  The push-pull converter does not need a gapped core, so it can be the better choice above a certain power output.  There is even an inductively driven push-pull that has poor electrical characteristics but has one saving grace: it can tolerate all semiconductor devices being turned on briefly at the same time, so it is used in weapons systems that have to withstand the gamma photocurrents arising from a nuclear blast.

[anotherjim]

This thing is cheap...
https://www.bitsbox.co.uk/index.php?main_page=product_info&cPath=347&products_id=3269&zenid=5a1dc1tj3v6n8net5joqdq19i5
Can that really put out 5A at 30v? That's 150W  If it is 96% efficient, then that's just 6W dissipated in the controller.
Or even cheaper...
https://www.bitsbox.co.uk/index.php?main_page=product_info&cPath=347&products_id=3202&zenid=5a1dc1tj3v6n8net5joqdq19i5

As Mark said...

2) Inserts a HF clock into the mix.

Guitar bandwidth falls off over 5KHz. Why would HF (at least 30kHz and many closer to 200kHz) be a problem?

Because it can cause aliasing distortion in a digital audio input that has a low sampling rate and/or insufficient filtering.

Because it can cause audible heterodyne whine with any other RF source or sampling rate. This audible whine is the difference in frequency between two RF sources. Other RF source could be another charge pumped pedal or an unsuitable SMPS power brick.
A classic is the original Electric Mistress which design had no input RF filtering proceeded by a charge pumped boost pedal. The flangers BBD clock intermodulates with the charge pump frequency. Because of a common ground connection, this could still happen with the boost pedal bypassed.

For most of us, sensible layout, screening, good supply filtering and upper bandwidth filtering of the audio path is usually sufficient.
So when you see a 10pF cap in the circuit, don't leave it out even though that circuit generates no RF and you worked out it can't have any effect at audio frequency. It's there to save you from a problem you won't know you had so long as it's fitted.

[amptramp]

This thing is cheap...
https://www.bitsbox.co.uk/index.php?main_page=product_info&cPath=347&products_id=3269&zenid=5a1dc1tj3v6n8net5joqdq19i5
Can that really put out 5A at 30v? That's 150W  If it is 96% efficient, then that's just 6W dissipated in the controller.
Or even cheaper...
https://www.bitsbox.co.uk/index.php?main_page=product_info&cPath=347&products_id=3202&zenid=5a1dc1tj3v6n8net5joqdq19i5

These are not charge pumps, they are buck regulators that provide a lower voltage at the output than the input.  Even so, 96% efficiency is remarkably high.

[anotherjim]

Ron, I thought they can go either way. Note the spec..
"Input 4V - 32Vdc.
Output 1.3V-30Vdc"
Suggests with 4v in, it can output 30v. No?

[amptramp]

Ron, I thought they can go either way. Note the spec..
"Input 4V - 32Vdc.
Output 1.3V-30Vdc"
Suggests with 4v in, it can output 30v. No?

In the title, it says "Step Down" meaning the output is lower than the input.  There is a class of regulators called "buck-boost" that can output voltages above or below the input but this is not one of them.

[printer2]

Ron, I thought they can go either way. Note the spec..
"Input 4V - 32Vdc.
Output 1.3V-30Vdc"
Suggests with 4v in, it can output 30v. No?

In the title, it says "Step Down" meaning the output is lower than the input.  There is a class of regulators called "buck-boost" that can output voltages above or below the input but this is not one of them.

The giveaway is the voltage ranges,

"Input 4V - 32Vdc.
Output 1.3V-30Vdc"

So at 4V input it can put out 1.3V. At 32V it can put out 30V. So you will need to supply at least 2V over the voltage you want to put out, at low voltages at least 3V and at some point the overhead starts to drop until it needs only 2V overhead.

[Brainojack]

Can that really put out 5A at 30v? That's 150W  If it is 96% efficient, then that's just 6W dissipated in the controller.
Or even cheaper...

Careful when reading advertised 'maximums'.   The maximum output and efficiency are individually adjustable up to 30 V, 5 A, and 96% but that's not to say that they will be obtainable at the same time.  Most datasheets give the maximums for each to showcase the instrument but will offer graphs later on in a datasheet that show operations as particular parameters are altered (Vout vs Iout)