Charge pump

Is it possible to bounce 5 volts up to 12 volts to provide maybe a mA of output current using a single 4093 section?

Reply to
bitrex
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Yup. Schmitt oscillator, three parallel buffers, voltage multiplier riding on +5. Might be a bit over 12 volts.

I could draw it if requested.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

But I only have one section... :(

Reply to
bitrex

Feed the classic Cockcroft-Walton from the 4093

Should be ok for 1mA load. Look up old data sheets that has info on the CMOS RdSon, simulate or test

Cheers

Klaus

Reply to
Klaus Kragelund

Requested. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

OK, but a real CD4096 will be a little wimpy as a charge pump. Might work, might droop too much. An HC series part would be a tougher driver. HC132 or HC14, something like that.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Should I post it to sci.electronics.basics for you?

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Nope. See Pg 2 and 3. Typical output current with Vdd=5V is 0.88ma high and 2.21ma low at

25C. About half that at temperature extremes. If you follow the NAND with a voltage tripler or quadrupler, you'll get 1/3 or 1/4th of the current ratings in output current.

See "gated oscillator" on Pg 4 of the above datasheet.

However, you can buffer the output with an emitter follower and easily get the necessary current.

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Jeff Liebermann     jeffl@cruzio.com 
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Reply to
Jeff Liebermann

OOOOH SNAP

Reply to
bitrex

Yes, please. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Reply to
Jim Thompson

(Oops. I skipped a line on the data sheet. It should be 0.88ma both high and low. Still won't work.

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Jeff Liebermann     jeffl@cruzio.com 
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Reply to
Jeff Liebermann

A single emitter follower would need a very low value pulldown resistor. A PNP-NPN follower would work, but will cost about 1.5 volts p-p swing.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

oscillator driving a 3 stage cockroft-walton?

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Reply to
Jasen Betts

You're running at 5V. You can't upgrade it to 74HC? That will get you the needed current capacity. Or 74AC even (beware the risetime though).

Tim

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Reply to
Tim Williams

The low side of the C-W multiplier can be parked on +5, so it only needs 2 stages. The first 5 volts is free.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

I'm picturing an oscillator, drivers, into cap and then, two more sections or diode's and caps... all referenced to 5V.

George H. (But sure put a scribble pic on SEB. :^)

Reply to
George Herold

Y'er right. I thought that the multiplier was a half wave device and only drew current on the positive cycle. That way the load acts as a pull down. However, that's wrong as it conducts on both the positive and negative parts of the cycle. Sorry(tm).

Do you agree that the 4093 will NOT be able to supply 1ma output by itself?

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Jeff Liebermann     jeffl@cruzio.com 
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Reply to
Jeff Liebermann

Probably not, if it's one section of an original CD4093. They are pretty wussy, especially with a 5 volt supply.

It would work fine with a 12 volt supply!

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

OK, scrawls-r-us.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

There used to be a 74HC4093. Does that series still exist?

Reply to
Tom Del Rosso

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