Charge pump

Bootstapping? Now there's an idea I've never seen or tried. Have the Vdd for the 4093 provided by output of the voltage tripler. Start the

4093 off the available 5v and get it oscillating into the voltage tripler. With diode isolation, bring Vdd up to about 12v, which provides more current capacity from the 4093 and the voltage tripler. Once Vdd climbs to 12v, there should be more than enough current capacity in the single NAND section. Added cost would be the two isolation diodes and a filter capacitor.

Disclaimer: No warranty expressed or implied for this kludge.

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Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558
Reply to
Jeff Liebermann
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problem: the tripler takes three times the current in as you get out and that current comes in through VDD.

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Reply to
Jasen Betts

Haha - except it will then try to make 36V - except it won't because the

4093 output still needs to switch to the 5V supply :(

piglet

Reply to
piglet

Shore

Reply to
bitrex

I can use a 74HC series no problem.

Send teh coedz?

Reply to
bitrex

One section of a 74HC132 or HC14 should make a mA.

Here's the thing I posted to s.e.b. to educate Jim.

formatting link

With HC gates, you probably don't need the three buffer sections. I think you said that you only have one gate section available.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

Here's one way it could be done as a micropower boost:

And before you fall off your chairs laughing I actually used sucessfully in the 1980s to get a 5V rail from 2.5-3V at low power!

piglet

Reply to
piglet

Spiced it...that should work great. Yeah I only have one section as the others are tied up in a flip-flop sorta deal.

Do they make "small signal" Schottky diodes of that type in a six-pack package? Because I could actually use two more of them for another part of the circuit.

Reply to
bitrex

The 4093 has Schmitt trigger inputs, so it actually doesn't need a capacitor to oscillate, right? ; )

Yep, I recall that whole deal...

Reply to
bitrex

Oops. I hadn't thought of that. With a tripler, to get 1ma out, the tripler would need to supply 3ma to Vdd. But, since the tripler outputs only 1/3 of its input current, the tripler would need to be supplied with 9ma by the 4093 oscillator. Each time I need some output current, I would need to supply 3 times that in input current. Bad idea. This isn't going to work.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558
Reply to
Jeff Liebermann

Use 74HC logic, then you don't need to buffer. CD4000 is too soft at 5V, no oomph.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

If not regulated John's Cockroft-Walton version will do it but not with a CD4093, should be more powerful 5V logic. If it has to be a CD4093 and you can't use its 5V brother (can't imagine why not) you could add a PNP/NPN follower.

If you need the 12V regulated you can make an oscillator plus boost and have a NPN or TLV431 throttle the input when it gets to 12V. Like a governor. That needs only one inverter section, ideally a Schmitt inverter.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

Sure--connect the 4093 output to the enable pin of an SMPS.

:-)

HTH, James Arthur

Reply to
dagmargoodboat

Did you use schottky diodes? What output voltage did you get at 1 mA out? Post the .asc please.

There are some dual-series-pair parts, and some 3-diode ones too. I use a lot of duals, like the BAT74. Take a look at CMXD3004SR and MMBD3004BRM and HN2D03F. Those are all high-voltage PN parts, but lower voltage/schottky parts are similar.

Given the ten passives hung on that gate to make 12 volts, there is probably a simpler way to do it.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

Right, if you don't care too much about the frequency. Just a resistor will oscillate with the gate input capacitance.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

I was thinking about using an analog mux instead of the diodes, to make the charge pump. It could run off its own +12 supply. Need more coffee to work out the details.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

Connect the 4093 output to the enable pin of a SMPS which drives the startup circuitry for the control system that runs a factory which manufacturers SMPSes to be used in the startup circuitry which runs other factories.

Reply to
bitrex

Other factories that manufacture...4093s.

Reply to
bitrex

Yep. That works. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
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Reply to
Jim Thompson

Naturally.

Here are a few more serious ideas:

Fig. 1 - OUTPUT BOOSTER +5V === | |/ .--[R1]--+------| Q1 | | |>. | |\ | | +--| >o--+--[R2]--+--- | |/ | | --- C1 | |. | | === ===

Fig. 3 - BOOTSTRAP CONVERTER Here's an adaptation of piglet's bootstrap

+12V +5V === === | D2 | .--------------+---|. --- [R1] | | C1 | | === === ===

Cheers, James Arthur

Reply to
dagmargoodboat

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