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I have a voltage doubler charge pump that is used to provide voltage over an IR LED for a short pulse, I also have a driver circuit for the LED to sink the right amount of current that I am requiring. Individually, these two circuits work fine, with no load over the charge pump, there is a nice pulse up to double the input voltage (which could be anywhere from 2-3VDC), and with a constant supply over the LED driver, it sinks the right amount of current as I am expecting. Below is a circuit diagram of what I am talking about to make it more clear:

schematic

simulate this circuit – Schematic created using CircuitLab

On the right is the charge pump and on the left is the driver.
Now this circuit does kind of work for a 3V input, ie. gives out the current I am looking for ~150uA, but there is no spike in voltage over the LED that I can see on a scope where there should be. I believe the problem lies within the Charge pump side of things as the driver does give the right current given the proper voltage over the LED. The thing is, I am running out of ideas as to how to get it to work with a voltage as low as 2V ao if anyone has any ideas, that would be great!


It is probably very worth noting that the +2VDC I have put going into the transistor bases, is in fact a pulsed signal rather than a constant 2V supply, sorry for any confusion.

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  • \$\begingroup\$ What type of capacitors? Those values look pretty large for ceramic... See electronics.stackexchange.com/questions/134919/… \$\endgroup\$ – MarkU Nov 5 '14 at 10:09
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    \$\begingroup\$ When you say ... "no spitke in voltage over the LED" ... what are you expecting to see? The only way you'll get a large spike above the LED's normal operating voltage is by destroying the LED. Normally, you may see a few millivolts for each doubling of LED current. \$\endgroup\$ – Brian Drummond Nov 5 '14 at 11:57
  • \$\begingroup\$ Not an answer to your question, but why are you doing this? If you want a large current through the LED (presumably for a short time), just use a big capacitor on the power, a fast-switching transistor with a low saturation (and drive it hard), and a suitable (low-value) resistor. I would probably choose a low-voltage FET over a bipolar transistor. \$\endgroup\$ – Wouter van Ooijen Nov 5 '14 at 12:58
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I don't see anything dynamic here. No positive feedback, no bistable circuit.

There is no AC input (except maybe ripple at R1, where it does no good). There is no transistor switching. The transistors Q1 and Q2 are assembled as a two stage amplifier, rather than as a bistable oscillator - why? Perhaps R6 should be tied only to R7 and C2...

There are no rectifier diodes, which I expect to find among doubling capacitors. Perhaps R7 should be replaced with a diode?

The two transistors Q3 and Q4, below the LED, are connected in parallel - which is a distracting complication, rather than a true example of a LED constant current sink. Although biased (as shown) by a constant 2VDC supply so the base current is constant, this pair will not limit LED current using feedback. Perhaps if Q3's base was tied to the others emitter, and Q3's collector was tied to the others base...

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I eventually found the solution to this problem, and I forgot to update this post with the fix I needed.
First it might be worth me explaining how this circuit actually works:


The capacitors charge in parallel through the respective resistors and once they are fully charged (or near enough) a voltage can be pulsed to the base of Q2 (as well as the other two NPNs) which will pull the base of Q1 down to zero volts effectively creating a short circuit from the anode (+ve pin) of C3 to the cathode (-ve pin) of C2 - this is putting the capacitors in series, or 'stacking' them on top of each other. This acts the same way as putting two batteries in series - doubles the voltage.
Enabling the LED driver at the same time allows the voltage stored in these two capacitors to discharge through the LED, thus making it pulse briefly.


Now for the solution to the issue I was having:
I am afraid to say that it was nothing complicated but actually something quite silly and trivial. The PNP (Q1) I was using - a BC857C - didn't have a high enough current rating to be able to deal with the ~150mA I wanted to pass through it and this subsequently caused a substantial voltage drop over the emitter - collector junction which obviously stopped my doubler actually doubling properly.
To fix this I replaced the PNP with a low resistance PMOS (~10m ohm) between the drain and source junctions. With this change the circuit works exactly as it should and I am embarrassed to say it actually took be almost a month to find that this was the problem...

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  • \$\begingroup\$ What you're describing sounds more like a bootstrap circuit than a regular capacitative inductor, in that you can't provide a constant doubled output voltage on this, only pulses. \$\endgroup\$ – Nick Johnson Dec 17 '14 at 13:49
  • \$\begingroup\$ @NickJohnson That would probably be correct, yes. Either way, this is the solution to my problem \$\endgroup\$ – MrPhooky Dec 18 '14 at 10:33

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