I need to power a device (it's an analog bucket brigade delay) from a 9v power supply or battery. The device takes 15v. Is there anyway? I have a few 7660s IC chips, if that helps. I've made charge pumps before but they were integrated into another circuit and I'm not totally confident how they work.

The schematic:

enter image description here

  • \$\begingroup\$ How much current does the bucket brigade pull? \$\endgroup\$ – ThreePhaseEel Nov 13 '16 at 22:43
  • 1
    \$\begingroup\$ The unit for voltage is a capital V, not a small v. \$\endgroup\$ – Marcus Müller Nov 13 '16 at 22:44

Datasheet: http://www.experimentalistsanonymous.com/diy/Datasheets/MN3011.pdf

If you look at the "Application Circuit" Reverberation Effect Generation Circuit at the top of page 53 (the last page) you will find that it is almost identical to the circuit you show. It also shows that the power supply voltage is rated for "+6V ~ +18V" To my view, that means that it would be ideal to power from one or two 9V batteries in series. No need to fool around with voltage conversion (UP or DOWN).

It should make no significant difference whether you use the original AN6551 op amp chips or the more modern TL071 and TL072 chips.

  • \$\begingroup\$ I notice in the instruction booklet for this kit it says a minimum of 14V and a max of 16V. Are they just playing it safe? \$\endgroup\$ – TCassa Nov 17 '16 at 8:59

There are many boost converters available in the form of modern integrated circuits. And there are many very small complete converter boards made from these chips available on Ebay. It should be quite easy to find a "boost DC converter" on Ebay which can handle 9V input and 15V output. I found several for less than US$1 and with free shipping:

2A Max DC-DC Boost Step Up Conversion Module 2V-24V to 5V-28V Output L3

As @ThreePhaseEel mentioned, you really must know how much current your circuit will require at 15V so you know which solution to select. Note also that boost converters are not terribly efficient so you can plan on losing a significant amount of battery capacity because of the conversion. Note also that the rectangular 9V "transistor radio batteries" have the lowest mAH capacity of any generally-available battery.

In fact, it may well be better to use TWO 9V batteries, or to experiment to see how your 7660 chips operate on 9V?

  • \$\begingroup\$ When you say try the 7660 chips - is that something I can do without buying anything extra? I found a schematic to boost one to 18V or +/-9V but nothing to bring it specifically down to 15V. \$\endgroup\$ – TCassa Nov 14 '16 at 10:02
  • \$\begingroup\$ You said: "the device takes 15V". I assumed you meant the 7660 chips? Perhaps you should reveal the whole picture so we know what you are talking about here?????? \$\endgroup\$ – Richard Crowley Nov 14 '16 at 12:40
  • \$\begingroup\$ Sorry I meant the actual bucket brigade delay takes 15V. I want to power it with a 9V battery. Is it, whether using the 7660 or not, possible? \$\endgroup\$ – TCassa Nov 14 '16 at 21:07
  • \$\begingroup\$ Since you never identified your BBD, it is not possible to discuss whether it might be operable from 9V or 18V. You could probably operate your 7660 at either 9V or 18V, but since we don't know what your circuit looks like, that would be guessing as well. \$\endgroup\$ – Richard Crowley Nov 14 '16 at 21:31
  • \$\begingroup\$ Sorry, I have now located the schematic. Added to OP. \$\endgroup\$ – TCassa Nov 14 '16 at 22:37

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