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I am trying to generate 12V from a 5V source, capable of supplying around 15mA with less than 0.5V of ripple. The circuit must use all through-hole components, and cost is an issue.

Having encountered issues with designing a boost circuit, I decided to try a charge pump instead:

schematic

It works perfectly in simulation, but when assembled on matrix board, the 12V drops to around 6V if I try to draw more than about 2mA. I have no idea why.

I am using schottky signal diodes and standard electrolytic capacitors. I don't have any datasheets because they came from Aliexpress. The inverters are a single SN74HC14N.

I have tried varying the frequency and the size of the capacitors. I also added an extra stage (the original design had one less stage than you see here, and worked in simulation).

What am I doing wrong?

(NOTE: I know charge pump ICs exist, and I plan to try one; but I'd still like to know why my home-grown solution is so bad.)

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    \$\begingroup\$ Does the simulation use the real model of the gate output drivers, or are they "ideal" gates? Obviously, all of the input current must flow through those gates, and if they aren't beefy enough, you won't get any output current. \$\endgroup\$ – Dave Tweed Jul 11 at 20:37
  • \$\begingroup\$ That's a good point. I could check the datasheet for the gate. I can't really measure the actual current draw though, because it changes with each oscillation. What should I do? \$\endgroup\$ – Sod Almighty Jul 11 at 20:38
  • \$\begingroup\$ Would it be worth assembling another version of the circuit with transistors between the gates and the capacitors? \$\endgroup\$ – Sod Almighty Jul 11 at 20:39
  • \$\begingroup\$ The "through hole only" really limits your options as there are plenty of "better" components available as SMD only. Like AO3400 (NMOS) and AO3401 (PMOS) which you could use to make a more poweful inverter. Add those inverters to boost the output current of the 74HC inverters. For through hole you could try 2N7000 and BS250 but these are really much worse performance wise. \$\endgroup\$ – Bimpelrekkie Jul 11 at 21:02
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    \$\begingroup\$ As the answers say, you need more drive from the gates. What works really well for charge pumps is a FET gate driver, even the lowest power ones deliver the best part of 1A. You'll get 2 or 4 low side drivers in a single package, ideal for the Dixon multiplier you've shown. \$\endgroup\$ – Neil_UK Jul 11 at 21:36
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2mA sounds about right.

The datasheet of the SN74HC14 says it has an output current of about 4mA.

You are doubling the voltage. That means twice as much current must go in on the low side as you take out on the high side.

4mA available on the input gets you 2mA available at the output.

You need to supply more current going in.


I don't think a simple transistor will do it - you need to drive the charge pump high and low.

You need something like a "totem pole" output stage.

Here's an example from the wikipedia article on TTL circuits:

enter image description here

You need the stuff around V2, V3, and V4. It could also be done using a PNP and an NPN transistor. You only need two transistors instead of three, then.

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  • \$\begingroup\$ Wow, that's pretty weak. So, add transistors then? \$\endgroup\$ – Sod Almighty Jul 11 at 20:47
  • \$\begingroup\$ Actually, I can't really add transistors between the gates and the capacitors, because the cathodes of the capacitors need to go both high and low. Any suggestions? \$\endgroup\$ – Sod Almighty Jul 11 at 20:49
  • \$\begingroup\$ How about this? Reckon that would work? \$\endgroup\$ – Sod Almighty Jul 11 at 21:01
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    \$\begingroup\$ Yep. That's the idea. I don't know if there are any "gotchas" that will cause it to destroy transistors, but it ought to work at least well enough to learn about how the charge pump works. You might add a resistor between the 5V and the totem pole - size to limit the current to less than the maximum rated current for your transistors, but still allow enough current for your load. \$\endgroup\$ – JRE Jul 11 at 21:06
  • \$\begingroup\$ Thank you, I will give it a go. \$\endgroup\$ – Sod Almighty Jul 11 at 21:10
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You need high and low buffer drivers.

Using emitter followers adds 0.6V drop either way so you get about 3.8V swing and the Schottky takes 0.3V into the cap so you get 3.5V "pump" per stage. So ...

5 + 3.5 = 8.5V.
Then 8.5 + 3.5 = 12V (Yeah. Right! - ie there will be losses.)
Then 12 + 3.5 = 15.5
3 stages should then work even with some extra losses.

Buffer = NPN + PNP bjt.
Join bases = input.
Join emitters = output.
NPN collector high.
PNP collector low.
NO resistors.
Go !

You can reduce transistor drop by using common emitter stages - drop is then just CE saturation.

Circuits later maybe if wanted.

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  • \$\begingroup\$ Actually, according to the simulation, it seems that running the transistors in the normal way - rather than emitter-to-emitter - gives the best results. Should I believe the simulation? \$\endgroup\$ – Sod Almighty Jul 11 at 21:39
  • \$\begingroup\$ @SodAlmighty As I noted, when done that way you get lower forward voltage - just CE saturation rather than a Vbe drop+. If you want high currents there will be a point where the emitter followers work better. And the common emitter method notionally needs care taken with "shoot through" when/if both transistors are turned on at once BUT it's usually trivial in circuits at this power level (except when Murphy occasionally has a really good day). \$\endgroup\$ – Russell McMahon Jul 12 at 6:48
  • \$\begingroup\$ Even with the transistors, I'm getting the same voltage problem in my circuit. I went with the traditional transistor arrangement. Would it have worked with your arrangement do you think, or is something else wrong there? \$\endgroup\$ – Sod Almighty Jul 12 at 18:08
  • \$\begingroup\$ @SodAlmighty Check that all points are doing in practice what you expect in theory. With transistors driven as per your simulation you should get near rail to rail swing of the connected collectors. || Energy per cycle is E = 0.5 x C x V^2 . Power is nominally E x freq. So here 0.5 x 2.2uF x (5V - 0.3V (Vdiode) ) - a bit more. Power per stage is notionally approaching a Watt so is in the order of right. 5V source needs to be able to take the current peaks (suitable filter cap) and Schottky diodes should be much gruntier than currents concerned. eg 1A parts should do very well. ... \$\endgroup\$ – Russell McMahon Jul 13 at 12:25
  • \$\begingroup\$ ... Try loading the 1st stage without stage 2 on connected to see what happens. | The 220R is probably not needed. If sc the zener will clip whatever is excessive. This MAY load the diode pump but tbd. \$\endgroup\$ – Russell McMahon Jul 13 at 12:25

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