Can someone explain in an easy to understand the way how this circuit works? I get the very basics in that the circuit takes a positive voltage on the input and provides an equal but negative version on the output

 

hi t90,
Do you have a link to the site where the circuit posted.?
E

 

Hello,

Have a look at the pages 6 and 7 of the datasheet.

Bertus

 

hi t90,
Do you have a link to the site where the circuit posted.?
E

https://circuitdigest.com/electronic-circuits/switched-capacitor-voltage-converter
Thanks for the response

 

Hello,

Have a look at the pages 6 and 7 of the datasheet.

Bertus

ok thanks, i was actually looking at the T.i datasheet here and if you look at page 7 figure 11 you get the same circuit it show on the "intersil" datasheet. I have a question both figures look identical except the T.I datasheet figure doesnt show a ground connected to (Cr) capacitor, but in the other datasheet there is one?

 

hi t90,
Did you read page #9 of the datasheet posted by @bertus , it explains the 'Detailed Description'.
If you have a specific question please ask.
E

 

hi t90,
Did you read page #9 of the datasheet posted by @bertus , it explains the 'Detailed Description'.
If you have a specific question please ask.
E

Yes i read the T.I and Intersil but one shows a ground connected to the positve side of the output capactior and the other doesnt, is this known as a virtual ground?
Heres what i understand,
When switches s1 and s3 are closed, C1 charges to the supply voltage during one-half of the wave, on the other cycle s2 and s4 are closed and s1 and s3 are open (here's where i get a little confused) S2 connects C1 to ground, the output capacitor C2 develops a negative voltage because C2 positive rail is connected to a virtual ground? and the negative side of C1 goes to the output?
The second half of the cycle gets me confused.
Can someone explain it in a easy to understand way?
Thanks

 

any help?

 

The circuitdigest.com link you posted in #4 above gives about as simple and straightforward an explanation as possible:

The IC contains four large switches (mainly MOSFETs). In the first half of the input switching wave, switches S1 and S3 are closed, so it charges the pump capacitor Cp to supply voltage V+. During the second half of the switching wave, switches S2 and S4 are closed, and S1 and S3 are opened. As S2 connects the pump capacitor to ground, the output capacitor Cr develops a voltage which is -V+/2. After a few switching cycles, the voltage at the output capacitor will be exactly equal to -V+. At this point, the output voltage is the negative of the input voltage, and the input current is approximately equal to the output current.

I can't see any way to improve on that. Read it carefully.

 

here is a simulation of a switched capacitor voltage doubler
I see two capacitors they are like two batteries. You can measure them.
The switch can be mechanical however this switch is electrical. It fills up the capacitors.
https://www.multisim.com/content/Mt...r-positive-dc-to-positive-dc-voltage-doubler/