We are accustomed to using charge pumps to produce a higher supply voltage than that provided by a 9vdc supply. The basics of the chip are that it uses an oscillator circuit to produce pulses that are effectively "summed" by a network of diodes and caps, to produce a higher voltage.
But keep in mind that you're starting out with DC (from a battery or adaptor), and then essentially converting it to AC, before reconverting it back to a different DC. What if you start out with AC?
I was trying to organize the files in my "Temp" directory, and re-stumbled onto a scanned page from an old 70's circuits book that provides simple diode/cap circuits for doubling and tripling the supply, while at the same time rectifying AC to DC. In theory, one could take a 9VAC supply (something which I imagine is readily available in the wallwart discard pile at thrift stores, or even available from more and more of the pedalboard power brick units), and double or triple it (minus whatever is lost to the diodes). So if you have a 9VAC source, you could use it to provide a 26VDC supply with just a few parts. And the best part is that you aren't limited with respect to current, as you would be with a charge pump chip.
I have the vague feeling I may have posted this before....but having just stumbled onto it, I'm posting again!
Maybe it's been posted before, but it's good to remember it again.
People wanting the general case and more info can google "voltage multiplier" and "%^&*croft-Walton".
Is it just me or is it becoming harder to find AC wallwarts? Something about all those eco-directives to use a little standby power as possible and making everything a switching supply in the process?
You can find wads of them in thrift shops and pawnshops. There's an always-increasing bin of wallwarts that somebody couldn't remember "what it went with".
A question, however.....
If I have a power amp chip that will "work" off a rectified 9VAC (12.7VDC), but provide more power at higher supply voltages, do I gain advantages by doubling or even tripling, or are the current limitations of the wall-wart going to over-ride all of that? Assume here that I have a 9vAC unit rated at 1.2A.
Quote from: R.G. on February 21, 2012, 01:29:10 PM
People wanting the general case and more info can google "voltage multiplier" and "%^&*croft-Walton".
What-croft?
Rhymes with "rock". Starts with "C". ;)
Quote from: frequencycentral on February 21, 2012, 03:58:20 PM
Quote from: R.G. on February 21, 2012, 01:29:10 PM
People wanting the general case and more info can google "voltage multiplier" and "%^&*croft-Walton".
What-croft?
c0ckcroft replace the 0 with a 'O'
> What if you start out with AC?
AC is going out of style. Not just Vampire Load; it's got the the point that a 37KHz switcher is cheaper than 50/60Hz 120V/230V iron.
Note also that we have a lot of DC supplies; adding an AC wart to the mix increases confusion and possible smoke.
> thrift shops and pawnshops
Stock up.
> if you have a 9VAC source, you could use it to provide a 26VDC supply
No. Each stage charges to the Peak of the sine AC wave. "9VAC" makes around 12V DC, or 24 or 36 V DC.
The "Delon" plan makes a lot of sense for many purposes:
(http://upload.wikimedia.org/wikipedia/en/thumb/2/23/Bridge_voltage_doubler.svg/200px-Bridge_voltage_doubler.svg.png)
Ground the bottom, call it +24V. Ground the middle, call it +/-12V. The losses are small. When used single-ended (ground and +24V) the ripple is 120hz and small. (Worked bipolar, the ripple is two out-of-phase half-waves.)
If one end of the source must be common, or for more-than-double, we go to the Greinacher schemes. These tend to sag more, more-and-more as the leverage rises.
> a power amp chip that will "work" off a rectified 9VAC (12.7VDC), but provide more power at higher supply voltages, do I gain advantages by doubling or even tripling, or are the current limitations of the wall-wart going to over-ride all of that? Assume here that I have a 9vAC unit rated at 1.2A.
Assuming simple (not bridged) amps: The full-roar amp presents a load to the DC supply roughly six times higher than the load impedance. 8 ohm load reflects as 50 ohms to the supply. 4 ohm load reflects as 25 ohms to the supply.
Note also that a simple (not multiplier) DC load must be multiplied by about 1.6 to find the load on the AC dupply.
So 12V DC to an amp with 4 ohm load draws 12V/25= 0.5 Amps.
The AC supply must deliver 0.5A*1.6= 0.8 Amps.
Try doubler. 25V out, 25 ohm loading, 1A DC, 1.6A load on the AC supply.
Your 1.2A wart will sag or smoke at full roar.
Here is a link for a tube microphone power supply. This could be adjusted for other tube circuits. It is a tripler.
http://recordinghacks.com/articles/one-tube-microphone-from-berlin/
I have posted about this in the past at this forum.
Usually if you double the voltage , you have to half the current .
And if you attempt to draw more then about half or a quarter current the voltage will sag .