18v from 9v works !!

[MartyMart]

Just built the 18v voltage doubler from GGG using a MAX1044 chip and it works!!
Powers my Mutron III but I think it's a bit under rated as that needs
to have about a 500mv PSU, I can't seem to find how much it draws
though ... ?
Its "power in" is a 300mv 9v regulated PSU.
Any options for a more "robust" 1044 type device to give me 18v ?
This one will work for smaller pedals with less current draw

Cheers,
Marty.

[Ben N]

For that big a power draw (and that big a pedal), wouldn't it make more sense to just get another transformer?

Ben

[waldo041]

hey marty i thought that the mutron 3 used a bipolor( +9v, 0, -9v )power supply. i did not know that it ran on 18 volts. how do i get mine to do that? and was there any whine from the max1044? thanks waldo

[PharaohAmps]

The MAX660 will do the job - it supplies (as I recall) 100mA instead of the ~15mA that the 1044 will do.  It is not pin compatible with the 1044, though.  You'll need a new PCB layout to make it go.  I think it does have boost like the 1044 so you should be able to bump its osc. freq. outside the audible range (no whine.)

Matt Farrow

[MartyMart]

waldo :
There's no whine at all, just an occasional "low battery" warning and a
little "distortion" from the mutron.
as it doesn't sound quite right, I assume its a bit starved !!

Matt:
Thanks for that, I'll see if I can find one and check its data sheet for
pinout etc

Marty.

[TheBigMan]

The MAX660 will do the job - it supplies (as I recall) 100mA instead of the ~15mA that the 1044 will do.  It is not pin compatible with the 1044, though.  You'll need a new PCB layout to make it go.  I think it does have boost like the 1044 so you should be able to bump its osc. freq. outside the audible range (no whine.)

Matt Farrow

How do you calculate mA output for a charge pump circuit?  I've been trying to find out what the ICL7660S will supply but with little success thus far from the datasheets.

[gez]

How do you calculate mA output for a charge pump circuit?  I've been trying to find out what the ICL7660S will supply but with little success thus far from the datasheets.

Try and get hold of a copy of Practical Electronic Design Data by Owen Bishop.  It's no longer in print but I've seen copies on amazon for around 50p.

According to this useful little book, for the 7660 (negative voltage conversion):

Vout = -Vs + (70XI)

Where Vs = supply voltage and I=current draw.

This will give you an idea of the loading effect the current draw has (calculation is approximate, but try a few figures with a calculator and it's not too impressive).

[Transmogrifox]

If you assume that the charge pump itself is ideal and doesn't consume any power of its own, then you can just think of it as:
power in equals power out.

Power, P = VI  (V=Voltage, I=current)

If the current input maximum is 300 mA at 9V, the power input in watts is
9*300mA = 2.7 W

Your output voltage is 18V, so remember P=VI, and Pin=Pout, so:
Pin = 2.7W => Pout=2.7W
Vout = 18V

Pout = Vout*Iout, and Iout is the number you want to know.

2.7W = 18V * Iout

=> Iout = 150mA

That is the absolute maximum current that could theoretically come out at 18V.

We know the charge pump dissipates power, so you can get a much more realistic estimate of current out by saying
Pout = Pin - Pd
Where Pd is the Power wasted withing the IC chip itself.

You can get a good estimate of Pd (usually) from a datasheet, then just plug it into the equation above and get your maximum output current.  You may also work it backward to find out what your input current must be to source an output pumped up to a higher voltage.

The charge pump can be loosely thought of as a DC step-up transformer.

[gez]

Vout = -Vs + (70XI)

Where Vs = supply voltage and I=current draw

The 70 in the formula above is the approximate output impedance of the device (available from data sheets), so it accounts for the 'internal' voltage drop and gives you an idea of available output voltage.  

Should have included that yesterday (makes it easier to understand).

[aron]

I used a charge pump in one of my Shakas but it started whining at certain settings. I decided to just use 2 batteries.

[MartyMart]

transmorgrifox, thanks for that great advice and i appologise if I mixed
up my mV and mA 's  !!
So there will still be a definate lack of available mA depending on chip/data sheet info ..........
But if I also use an "input" PSU rated higher, to power the charge pump, like 1200 mA 9v then I'm  helping with current draw "headroom" ?
Just as if I wanted to power 6 pedals.....
Do I understand this correctly ?
Thanks,
Marty.

[nelson]

How would I calculate the consumption of a particular circuit? to know whether the 1044 could provide adequate current?

[davebungo]

How would I calculate the consumption of a particular circuit? to know whether the 1044 could provide adequate current?

You'd be better off measuring it with a DMM to give you an idea and be sure to do it with controls in different positions.  I suppose depending on the complexity of the circuit you could probably work out the expected bias currents and take op-amp current from the relevant datasheet and then try to add them all up but I think it's a bit of a waste of time if the circuit is any more than one or two op-amps and a few trannies - strange things happen when devices start to saturate or clip which may increase the actual current drain beyond your prediction.

[nelson]

The circuit I have in mind is pretty complex, 2 opamps a clock and an MN3007 so I would prob just be better off measuring it. I wanted to incorporate the 1044 in the design......sigh.

[TheBigMan]

Thanks for the info all.  Where's my calculator gone?  

[davebungo]

Trying to calculate the current drain is a really useful exercise though because it gives you the experience to predict in the future what a circuit will use just by looking at it (and be correct to within 20%) so I wouldn't say "don't try", just do both and try to make your calculations match with what you measure.

[nelson]

Thanks, I will give it a try......I am off to look at datasheets.

[hardi_ami]

Hello people

Take a look at this : http://focus.ti.com/docs/prod/folders/print/pt5047.html
http://focus.ti.com/lit/ds/symlink/pt5047.pdf

"The PT5040 is a series of 3-pin boost-voltage Integrated Switching Regulators (ISRs). These ISRs are designed for use with +5V bus systems that require an additional regulated +8V to +20V with up to 1A of output current. These ISRs are packaged in the 3-pin, single in-line pin (SIP) package configuration."

This one gives +18V with a usual 9V input, with 85% efficiency (as they say ...)

I'm going to try it for my (being built) Shaka Express.

Bye
Michael

[ninoman123]

Cool chip, seems expensive...might be better off with a MAX in there. Plus that package seems big....just my opinion.

[hardi_ami]

Sorry I forgot ..

Expensive, yes. But TI sends free samples, so if you just need one, you can have it packaged and shipped to your home for free by UPS. (only available for the 20V version, not the 18V)

Anyway, I'll tell you how it works when I get mine.

See you
Michael