[HN Gopher] Pump up the charge: Getting -5V from a 5V power supply
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Pump up the charge: Getting -5V from a 5V power supply
Author : zdw
Score : 13 points
Date : 2021-10-10 22:13 UTC (47 minutes ago)
(HTM) web link (nicole.express)
(TXT) w3m dump (nicole.express)
| tuatoru wrote:
| Obtaining dual supplies from a single supply:-
|
| https://sound-au.com/project192.htm
| peterburkimsher wrote:
| Thank you! The Bose Sounddock for iPod that I found in a
| recycling bin uses a dual +/- 18V power supply, and I was
| wondering how to power it from batteries. That link taught me
| that the extra voltage headroom may not be required, it might
| run on 12V, and it would in theory be possible to build a +/-
| dual voltage circuit. Now I just have to decide whether it's
| worth building that and learning more about power electronics,
| or buying a new speaker and having a somewhat safer device.
| vintagedave wrote:
| I'm struggling to understand one part of the article. Could
| anyone expert in electronics help explain this, please?
|
| > So what if we immediately disconnected the capacitor from the
| rest of the universe, including its power supply and ground?
| Well, if we did that after the capacitor was charged, we'd find a
| differential voltage on the power supply. What if we hooked up
| ground to the side that had -5V before?
|
| Is this a typo - should this say "to the side that had +5V
| before?" (ie change minus to plus)?
|
| I think I have a handle on the design, which is that it's
| charging a capacitor with +5V, changing the charging side to
| ground, discharging the capacitor through what used to be the
| input (you can tell I'm not an electrical engineer, I suspect
| these terms are very faulty) and causing a -5V flow that way. If
| that's true, I think it's a typo (?); if not, and much more
| likely, my understanding is flawed. Insight here would be much
| appreciated.
| mastax wrote:
| > Is this a typo - should this say "to the side that had +5V
| before?" (ie change minus to plus)?
|
| Yes.
|
| > I'm struggling to understand the design, but I think it's
| charging a capacitor with +5V, changing the charging side to
| ground, discharging the capacitor through what used to be the
| input (you can tell I'm not an electrical engineer, I suspect
| these terms are very faulty) and causing a -5V flow that way.
|
| That's correct enough. It's all a matter of perspective. I'd
| probably say you discharge it through what used to be _ground_
| , though as the article notes technically ground is the source
| of electrons - but we usually talk about holes and positive
| (conventional) current.
|
| Imagine measuring the voltage on a AA battery: 1.5V. Then you
| flip it around and measure again: -1.5V. Voltage is all
| relative to where you measure, usually ground. It's a tricky
| concept.
| morcheeba wrote:
| The theory on this is correct, but the ICL7660 is not designed to
| provide the kind of current you'd need to run a 12W audio
| amplifier. It maxes out around 50mA - see figure 7 of the
| datasheet[1]. 50mA*17v = only 0.85W.
|
| Because I can't see the datasheet of the audio amplifier, it may
| be wired in a way that it doesn't use significant current from
| this supply (for example, in the pre-amp part of the circuit).
| But then we wouldn't be getting the advantage of 17v supply to
| make it louder than other games.
|
| When overloaded, simple switching power supplies like this lose
| their regulation. In this case, the switcher operates at 10kHZ,
| so it might introduce a high-pitched sound to the output.
|
| You can probably get away with connecting the -5v input to GND to
| make the circuit run at 12v instead of 17v. Or, look for a more
| capable negative power supply.
|
| [1] https://www.renesas.com/us/en/document/dst/icl7660-datasheet
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(page generated 2021-10-10 23:00 UTC)