Can you increase the current of a negative voltage?

[Dan123456]

Hey all 

I have recently become quite intrigued with negative voltages and have been playing around with different ways of creating them 

One way I have been doing this is with a charge pump. The issue with this route is that they don’t produce much current.

I am wondering, is it possible to somehow increase the current with an higher current positive input?

I am probably doing a terrible job of explaining this in words so please see the diagram below  The exact numbers are arbitrary and am more just wondering does the “Current amplifier thingy” exist?

If feel like the answer is going to be really obvious and either going to be “of course you can you idiot, just use an op amp / whatever” or “No you dingus! That would go against the laws of physics because X” but I can’t work out which one it is 

Sorry, this is probably a really dumb question! Negative DC voltages are really quite new to me so I am probably missing the basics (even with stuff like when to use negative numbers into the data sheet formulas is tripping me up at this point  ) 

Thanks in advance 

[Kim Christensen]

For higher currents, it's better to use an inductor based DC-DC converter instead of a capacitive charge pump.

EDIT: Check out this simple circuit that you can experiment with. You can also get dedicated chips for this sort of thing.

[Dan123456]

For higher currents, it's better to use an inductor based DC-DC converter instead of a capacitive charge pump.

EDIT: Check out this simple circuit that you can experiment with. You can also get dedicated chips for this sort of thing.

Indeed! Thank you so much 

One of the other ways I have been creating negative voltages is with a LM2596 and that provides significantly more current 

I should have said in my original post that I realise this probably wouldn’t be the best way to go about things! I am more just hoping to understand if this is theoretically possible, even if not practical, just to further my understanding 

[IanB]

If you are starting with batteries as in your picture, you could create a split supply from the batteries themselves. For example, using two 9 V batteries will give you a split +9 V/0/−9 V supply with balanced voltages and current. Or, if using AA cells, you could just use a few more of them to get the negative rail, e.g. use 6 cells to get +4.5 V and −4.5 V.

[Andy Chee]

Another simple technique to generate +ve and -ve rails referenced to ground, is with an AC adapter with, say, 12VAC output and build a full wave voltage doubler:



The trick with this circuit is to use the centre connection point between the capacitors as the GND point, thus creating +ve and -ve rails!

[Siwastaja]

Take a look at standard buck converter - any buck IC does. You can wire it such that output is connected to ground and the "ground" pin then becomes your negative voltage. When wired like this, the topology becomes so-called "inverting buck-boost". Much more efficient than a charge pump, and able to deliver a lot of current.

[Kim Christensen]

I am more just hoping to understand if this is theoretically possible, even if not practical, just to further my understanding 

It is theoretically possible to get more current, but not very efficient or practical.

This is because of the way the energy is transferred in a capacitive charge pump circuit. Every time the capacitor is charged and discharged, power is dissipated in the circuit resistance. This would include the ESR of the capacitor, internal resistance of the battery/supply, and the resistance of the switching device and wiring. Current will spike at the start of every switching cycle, resulting in high I2R losses.
An inductor is the opposite. Current is zero when voltage is first applied across it and builds until it saturates. So there's less R losses as long as the switch turns off before saturation.

[Dan123456]

I am more just hoping to understand if this is theoretically possible, even if not practical, just to further my understanding 

It is theoretically possible to get more current, but not very efficient or practical.

This is because of the way the energy is transferred in a capacitive charge pump circuit. Every time the capacitor is charged and discharged, power is dissipated in the circuit resistance. This would include the ESR of the capacitor, internal resistance of the battery/supply, and the resistance of the switching device and wiring. Current will spike at the start of every switching cycle, resulting in high I2R losses.
An inductor is the opposite. Current is zero when voltage is first applied across it and builds until it saturates. So there's less R losses as long as the switch turns off before saturation.

Ah perfect! Thank you so much!

I wasn’t sure if it would be theoretically impossible due to the electrons moving in opposite directions and “crashing” into each other causing the flow to go to 0 (or just resulting in a smaller current from the “stronger” source - 1A-100mA = 900mA) or what not 

Sorry again if this train of thought makes absolutely no sense! This negative stuff is still brand new to me and quora seems to be the only place that has been giving me some answers that half make sense… and we all know how accurate that place can be 

[IanB]

Sorry again if this train of thought makes absolutely no sense! This negative stuff is still brand new to me and quora seems to be the only place that has been giving me some answers that half make sense… and we all know how accurate that place can be 

It might be worth remembering that there is no such thing as a "negative voltage". If I take an AA cell and stand it on its base, we can say it has a voltage of +1.5 V. If we turn it upside down, we can say it has a voltage of −1.5 V. But it is exactly the same battery, with exactly the same voltage. So what has changed? In fact, nothing. It is neither +1.5 V, nor is it −1.5 V. It is merely 1.5 V.

In reality, the concept of "negative voltage" only exists if we compare it against our zero volt reference relative to some other "positive voltage" that also exists in our design. If there is no positive voltage to compare against, there is no negative voltage either (other than conceptual ideas of "up" vs "down" on a schematic).

In practical terms, a negative voltage rail is sometimes needed if we need a voltage "below zero" compared to our zero volt reference, but often this does not need to supply a lot of current, and so the charge pump idea is often sufficient for this purpose. If both positive and negative rails need to be equally powerful, for example in an amplifier, then a dual-polarity power supply is needed.

[Rick Law]

Voltage is not absolute, it is relative.  5V means 5V above what you define as 0V.

May be easier to think of an analogy to see it.

Let say you are in the 30th floor of a high rise building.
.. Joe is on 32nd floor, so Joe is +2 floors (2 above) you.
.. Jane is on 27th floor, so Jane is -3 floors(3 below) you.

Now if you are in the 25th floor
.. Joe is on 32nd floor, so Joe is +7 floors (7 above) you.
.. Jane is on 27th floor, so Jane is +2 floors (2 above) you.

The + or - for voltage is the same as + or - for the floors.  It is comparative.  It depends on what you call your floor (your ground).  So one can think of it theoretically that there is no such thing as negative voltage.

In practice, current is like water pipes in the building.  Water flows down.  Where the water (current) is flowing, you need the capacity to take on that kind of drain at where it is receiving that water (current).  Draining water from the 27th floor to the 30th floor requires some thought.