Why won't this charge pump supply more than a couple of milliamps?

by Sod Almighty   Last Updated July 11, 2019 21:25 PM

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.)



Answers 2


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.

JRE
JRE
July 11, 2019 20:43 PM

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.

Russell McMahon
Russell McMahon
July 11, 2019 21:13 PM

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