Q About How MAX1044-type IC Chips Might Malfunction - RG?

[Paul Marossy]

What happens when a switched capacitor voltage converter chip or it's external circuitry is faulty? Can it draw an abnormally large amount of current? There's a guy with a Parker Fly that has his 9V battery draining down to 3.5V in roughly 30 minutes. I estimate that the circuitry draws around 5mA or so when operating normally. I have a hunch that it may have something to do with the onboard LT1044 chip, but I'm not really sure - I don't know enough about those chips. Anyone? 

Here is the schematic for the Parker Fly: http://www.parkerguitars.com/media/pdfs/original_wire_diagram.pdf

[Paul Marossy]

OK, what if one of the external caps was bad? I am pretty sure that the circuit makes use of the boost function to keep the oscillator out of the audio range. And power consumption goes up with frequency in these sorts of chips, so what if somehow it was oscillating at say 500kHz instead of the more usual 50-100kHz - wouldn't the current used go up dramatically? Or how about if it used an external oscillator source and that got whacked somehow?

Anyone? 

[Sir H C]

Does it otherwise work?  Those chips should be pretty bulletproof, but if one of the transistors shorted you would not get the voltage kicking up as it should.  I would have to go with the shorting capacitor or other short in the system.  The chip might well be getting hot from the excessive current, but that doesn't mean that it is bad, just that it is being overworked (but it could be bad too). 

You can not really kick the clock speed up by a factor of 10 with these things and have them work well.  Your losses in switching start to kill you.  Also you have to have some deadtime between the switching of the devices and if you kick the frequency up by 10, the deadtime remains the same and is more loss.

[Paul Marossy]

Yeah, that's the funny thing - it works normally at waht I assume is the right voltage, but it drains the battery in 30 minutes or so.  It looks to me like they use it to double the 9V to 18V. This is all very perplexing... 

[Sir H C]

Probably doing the higher voltage for the piezo.  I would have to guess something else is eating up that current and it should be getting warm. 

[Paul Marossy]

Yeah, but what?! I think there's a short in his power supply or something.

[Sir H C]

Finding what is the tricky part.  Can you ring out the circuit and measure the resistance on the output?  How much stuff is there?  Other method is to feel what is "hot" and that is the culprit.

[Paul Marossy]

Well, I can't do that because I don't have the guitar in my possession - it's half way across the U.S. from where I live. 

[Sir H C]

Ah, that is a problem.  I don't think it is the chip though.  Those things are tough, so they shouldn't die.  I would have the owner look for chafed wires and capacitors that look like they have blown.  Those are the most likely culprits.

[Paul Marossy]

Yeah, I think I would have to conclude that it's a short in the power supply somewhere.

[Sir H C]

Oh, the output voltage from the charge pump should be twice the battery.  The droop is directly related to the current that the output is using.  So, by measuring that output voltage, you can determine if the shorting is on the output or input side of the circuit.

[Paul Marossy]

Oh, the output voltage from the charge pump should be twice the battery.

Yeah, minus 0.5V 

[Sir H C]

Oh, the output voltage from the charge pump should be twice the battery.

Yeah, minus 0.5V 

Well it all depends on the loading.  The FET switches can be seen as resistors in series with the charge capacitors and so if you don't have enough time to charge the capacitors you won't see full voltage, but if you do, you should be able to hit 2x input voltage.

Yeah, I have had to design several of these charge pump ICs over the years...