Charge pump

I would model it but there's a problem. If I mange to get Vdd up to

Every time I look at this simple problem, I see if from a different angle. The oscillator starts using the +5v line, eliminating the startup problem. However, once started, Vdd will creep up to +15v as the circuit is 100% efficient and there are no losses. Well, there's a really big loss in the 12v 1ma load. Without the 5v to power Vdd, the circuit is powered by its own output. In other words, a perpetual motion machine. Rumor has it that such things don't work.

Also, it's not my problem. It belongs to Mr Bitrex.

The old data sheets show a problem. At high junction temperatures, the output impedance gets large-ish: it'll give 1.7 mA, but only into a 2.5V load, at 4.6V out it only promises a third of a milliamp.

Because of duty cycle, you need an order of magnitude better drive than that. Paralleling three sections, and keeping it cool, gives just about enough drive, barely.

There's no way that will work with VCC > 2Vbe

To fix split R2 into two resistors. R2 and the new R3 must be big enough that the gate can supply/sink current to one base without being pulled Vbe from the rail

IIRC with a 5V 4093 that means 10K or so, still it should be possible to meet the target if you can get enough gain from the trasistors.

protection diode as rectifier :)

Eliminate C1 and R2!

You're right of course, sloppy on my part. ASCII drawing mode is right-brain, electronic design & checking mode is left-brain, and n'er the twain shall meet.

We just need Rx= hfe * (5V-0.6V) / i.out. If hfe=100, i.out=5mA, Rx=112K.

Thanks for the catch.

Cheers, James Arthur

I'll give you C1, but R2 is useful--it helps reduce the Vbe loss.

Maybe bootstrap *both* Q1 and Q2? That would actually work.

I started ASCII drawing a whole regulated SMPS boost around a single 4093 gate, but quickly realized that's just plain silly.

Cheers, James Arthur

No, p-channel output transistor as synchronous rectifier :>

piglet

Which works quite well. ...Jim Thompson

Ahh, Sorry for being thick, but how does that work? (Where's the power come from?) Not through the 100 k ohm resistor.... ?

George H.

output of gate does the job of the switch and diode

-Lasse

OK I know that, but (maybe I reading the schematic wrong....) piglet has the 5V in going into the output gate through the inductor... Is that charging up the "power input" to the '4093 through some protection diodes on the output? Zenering the output transistors? How does the power get from the output gate to the power supply of the '4093?

George H.

All power flows through the inductor. During initial power up the output stage parasitic diode will help raise Vdd but once oscilation starts the output stage is acting as a synchronous boost convertor and the parastic diodes are out of the loop. The 4093 is literally pulling itself up by its own bootstraps :)

If you are tempted to build it I would suggest a bigger inductor, I sketched all those values off the cuff - I have forgotten the actual values I used way back in '88-89.

piglet

Oh I think I see. (It's Cmos) the output stage is a p-channel fet to the power rail when the output is high.. and that lets current flow up from the inductor to the 12V cap.

George H.

I actually built one. :( but it wasn't to provide continuous current, just for charging a capacitor...wasn't really regulated, just throttled so the drive shut off when the cap was fully charged.

I thought the 4093 logic gate sine wave generator was pretty cool though:

Using a 555, and "regulated" >:-}...

This is a discrete implementation of an ancient (42 years ago) chip design of mine, "FlybackDC-DC-Converter.pdf" on the S.E.D/Schematics Page of my website.

The current sensing could be made more efficient, but I figured I'd already used enough transistors that the whiners would be close to attacking >:-} ...Jim Thompson