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Dumb question about voltage multipliers

  • Thread starter Electronic Techniciai
  • Start date
E

Electronic Techniciai

Jan 1, 1970
0
I have a dumb question.


Someone posted the following voltage multiplier::


------------o----O +12VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+7.5 to +12.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+8VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (+3.5 to +8.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+4VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (-0.5 to +4.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/|74HC04 | V
| ---
| / \ CR
--- ---
\ / GND |
V |
---
\ / GND
V



But wouldn't this do the same thing with fewer parts?



------------o----O +13VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+8.5 to +13.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|\ |
| \ U C | | (+4.0 to +9.0) |
---| O-------| |-------------------o
| / | | |
|/ 74HC04 |
---
/ \ CR
---
|\ |
| \ U C | | (-0.5 to +4.5) |
---| O-------| |-------------------o
| / | | |
|/|74HC04 |
| ---
| / \ CR
--- ---
\ / GND |
V |
 
J

John Larkin

Jan 1, 1970
0
I have a dumb question.


Someone posted the following voltage multiplier::


------------o----O +12VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+7.5 to +12.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+8VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (+3.5 to +8.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+4VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (-0.5 to +4.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/|74HC04 | V
| ---
| / \ CR
--- ---
\ / GND |
V |
---
\ / GND
V



But wouldn't this do the same thing with fewer parts?


For starters, the bottom 5 parts serve only to turn +5 volts into +4
volts.

John
 
R

Robert C Monsen

Jan 1, 1970
0
Electronic Techniciai said:
I have a dumb question.


Someone posted the following voltage multiplier::


------------o----O +12VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+7.5 to +12.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+8VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (+3.5 to +8.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+4VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (-0.5 to +4.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/|74HC04 | V
| ---
| / \ CR
--- ---
\ / GND |
V |
---
\ / GND
V



But wouldn't this do the same thing with fewer parts?



------------o----O +13VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+8.5 to +13.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|\ |
| \ U C | | (+4.0 to +9.0) |
---| O-------| |-------------------o
| / | | |
|/ 74HC04 |
---
/ \ CR
---
|\ |
| \ U C | | (-0.5 to +4.5) |
---| O-------| |-------------------o
| / | | |
|/|74HC04 |
| ---
| / \ CR
--- ---
\ / GND |
V |
---
\ / GND
V

What am I missing here?

With your design, the middle diodes will simply oscillate up and down with
the input gates, I think. The missing diodes and capacitors allow the charge
to be stored, and passed on to the next stage.

Look up 'charge pump' using google for a more reasonable way to do this. You
only need one gate.

Regards,
Bob Monsen
 
M

Mr. Civility

Jan 1, 1970
0
Electronic said:
I have a dumb question.


Someone posted the following voltage multiplier::


------------o----O +12VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+7.5 to +12.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+8VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (+3.5 to +8.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
|
o------------ (+4VDC)
| |
| | C
--- -----
/ \ CR
--- -----
|\ | |
| \ U C | | (-0.5 to +4.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/|74HC04 | V
| ---
| / \ CR
--- ---
\ / GND |
V |
---
\ / GND
V



But wouldn't this do the same thing with fewer parts?



------------o----O +13VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | (+8.5 to +13.5) | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|\ |
| \ U C | | (+4.0 to +9.0) |
---| O-------| |-------------------o
| / | | |
|/ 74HC04 |
---
/ \ CR
---
|\ |
| \ U C | | (-0.5 to +4.5) |
---| O-------| |-------------------o
| / | | |
|/|74HC04 |
| ---
| / \ CR
--- ---
\ / GND |
V |
---
\ / GND
V

What am I missing here?

Go back to understanding what happens in a simple one-stage charge pump
and make the simplifying assumptions that the inverters output 0-5
volts, the diodes are ideal and the output is unloaded.

------------o----O +VDC OUT
| |
+5VDC IN | | C
O --- -----
| / \ CR
| --- -----
|\| | |
| \ C | | | |
---| O-------| |-------------------o ---
| / | | | \ / GND
|/ 74HC04 | V
---
/ \ CR
---
|
___
\ / GND, but call it VREF later!
V

When the inverter output is low, the output series capacitor is
discharged to 0 volts across it. When the inverter output goes high,
charge is dumped into the output capacitor, but the output voltage won't
go any higher than the voltage across the series C (5 volts) plus the
reference voltage on the right side of that capacitor, which is zero,
because of the series diode. If the series C is small compared to the
output C, it will take a long time (in terms of number of cycles) for
the output to build up to 5 volts. If C is large, it will take fewer
cycles. (Of course, we're speaking asymtotically here.)

Your second schematic is really the equivalent of the above, except that
the series C is really 3C. 5 volts develops across each series
capacitor, but each is referenced to ground on their right sides when
the inverter outputs switch to "1," so they can not whack the output cap
above (5 volts across capacitors plus (VREF, which equals ground in this
case)).

If, instead of connecting the lower diode to GND you were able to
connect it to some other positive voltage, say VREF, you'd be able to
dump the 5 volts across the series C(s) onto the output capacitor from a
starting point of VREF. Ultimately, the output voltage would get up to
5 volts plus VREF.

The first circuit consists of 3 stages of the above circuit. Each
intermediate capacitors develops a VREF for the stage above it to
ultimately stuff 3 X 5 volts (ideal case) onto the output capacitor.

I could probably make that clearer, but that was hard enough :)

In the real world, the diode drops, inverter output voltage drops and
real load current drop the available output voltage to something around
the 12 volts shown.
 
B

Ben Bradley

Jan 1, 1970
0
I have a dumb question.


Someone posted the following voltage multiplier::
What am I missing here?

One thing in both the circuits, those inverters don't have anything
connected to their inputs. I presume they are all connected to a
square wave source, but are they all in phase, or should the phase
alternate between successive stages? It appears the top one doesn't
matter, but the bottom one DOES matter.
 
G

GPG

Jan 1, 1970
0
|\ |\
-| >O|-------------| >O-
|/ A |/ B




+5V ->|-----o------>|-----o----->|-------o----- +V
| | |
--- --- ---
--- --- ---
| | |
A B GND

A B A GND
| | | |
--- --- --- ---
--- --- --- ---
| | | |
|-----|<------o------|<------o-----|<----o----- -V
|
|
|
V
-
|
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
 
K

Ken Smith

Jan 1, 1970
0
One thing in both the circuits, those inverters don't have anything
connected to their inputs. I presume they are all connected to a
square wave source, but are they all in phase, or should the phase
alternate between successive stages? It appears the top one doesn't
matter, but the bottom one DOES matter.
Or:
When running a voltage multiplier from logic, you can use a flip-flip to
divide the frequency by two. The clock of the flip-flop can be run to the
even stages and the output of the flip-flop to the odd stages.
 
G

GPG

Jan 1, 1970
0
I see that my andy's posting has suffered.
The top bit should have 2 inverters in series, and (assumed) they
would be driven with a signal
approximating a square wave. Only useful for low currents because
of the losses. Add the Vf of the diodes and take into account
the output capability of the inverters.
If you have a hex inv parallel them for better drive.
The point is that you only need 2 inverters and somewhat
fewer caps and diodes to do this,compared with the first (asking for
advice)
posting and will generate both - and + simultaneuosly
 
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