for the op amp to work correctly the -V supply on the op amp should be -9.0V right? since the +V is equal to +9V

 

One way to determine the answer is to get the data sheet for an ICL7660.

 

Isn't there a misdrawn component there?

 

Hello,

I see no problems with the circuit.
Look at the two positions of the 7660, to see how the capacitors are switched.

Bertus

 

Hello,
Look at the two positions of the 7660, to see how the capacitors are switched.

Bertus

I'm switching between a datasheet and the circuit on this page now and looking at the capacitor connected to pins 3 and 5 of the chip I believe there is a misdrawn component on this circuit. Specifically the cap is backwards.

 

See the datasheet for the charge pump IC, or, draw the 2 states of the internal switches as separate circuits. Pay attention to what charges are on the capacitor with what polarity. Compare the two circuits (one for switches in state 1, one for switches in state 2). What happens?

 

http://www.intersil.com/data/fn/fn3072.pdf

circuit as drawn on page 6 of the pdf is what I have referenced as being different from this one. Is it a different circuit, that still gives a negative voltage? One of these two has a cap backwards. Who wants to build both and find out which is wrong?

 

on page 6 of the PDF, it states

VOUT = 2VIN

so Vout = +18V?

 

Hello,

When you look at the circuit, the capacitors are now in series.
In the other position they are parallel.
So what will be the negative voltage?

Bertus

 

Non-polarized caps (no + - signs). The curves on the caps, when used correctly, show which winding of a rolled cap is on the outside. On low-voltage, non-polarized caps, it is not an issue.

But this is a very non-standard way of using a 7660 (see data-sheet. Really, look closely). I won't elaborate yet because this is homework, but you need to not just believe the schematic, or that the designer knew what he was doing (as is much too often the case).

Does it work this way? Maybe.
Should it ever be used this way in a real circuit? No!

It will require knowing how the 7660 works, but somewhere in the world, someone will find this in their notes from when they were in school, and design this into something important.

Not on my watch!

 

so when the switches are in the other position, the capacitors will be in series. i see that, but i am getting lost trying to follow that out...

the power comes in the AC source, then flows through the 150k resistor, then down to terminal 3 or 5. im confused

 

so when the switches are in the other position, the capacitors will be in series. i see that, but i am getting lost trying to follow that out...

the power comes in the AC source, then flows through the 150k resistor, then down to terminal 3 or 5. im confused

The signal/AC Source is not connected to the charge pump, the lines simply cross. There isn't a "\(\bullet\)" (standard) or "\(\circ\)" (in this schematic) to indicate a connection. The signal only goes to the non-inverting input of the amp.

Follow the path of +9v, Ground, and the -V output.

 

9V goes to the V+ of the op amp and terminal #8 of the IC. terminal #8 on the IC is the Vin for the IC. that is going to be there no matter which way the switch is flipped.

Ground goes to terminal #5 on the IC which is also the Vout

the -V input goes to terminal #3 on the IC, which based on the IC datasheet should be a Ground.

so when the switches are in the other position, would power be flowing into terminal #5?

 

9V goes to the V+ of the op amp and terminal #8 of the IC. terminal #8 on the IC is the Vin for the IC. that is going to be there no matter which way the switch is flipped.

Ground goes to terminal #5 on the IC which is also the Vout

the -V input goes to terminal #3 on the IC, which based on the IC datasheet should be a Ground.

so when the switches are in the other position, would power be flowing into terminal #5?

Don't look at the datasheet labels for the IC pins. Look at the "internal picture" in your schematic above.

+9V goes to one side of a switch, then back out to a capacitor, and continues.

 

Don't look at the datasheet labels for the IC pins. Look at the "internal picture" in your schematic above.

+9V goes to one side of a switch, then back out to a capacitor, and continues.

yes i can follow that one. +9V comes in #8, then to #2, through a cap, back in #1, up to #3, then up to V- on the op amp

now when the switch is in the other position

+9v comes up to #8 and just sits there. the two capacitors are now in parallel and are supplying the V- of the op amp. the 2 caps in series are equal to 4.4uF, but how do I know what kind of voltage they are supplying? would both of them be storing a full 9V charge?

 

Hello,

Your calculation of the total capacitance for the series of the capacitors is wrong.
Take a look here how to calculate it:
http://www.allaboutcircuits.com/vol_1/chpt_13/4.html

Bertus

 

Hello,

Your calculation of the total capacitance for the series of the capacitors is wrong.
Take a look here how to calculate it:
http://www.allaboutcircuits.com/vol_1/chpt_13/4.html

Bertus

im sorry i meant to say "the two capacitors in parallel". but still the C total would be equal to 4.4uF.

 

Hello,

How will the voltage distribution be when the capacitors are in series?
Try to follow the voltage path from V+ to ground.

Bertus

 

i drew arrows in where i think the current path is. where does it go from where i left off?

 

As I see it, the way the schematic is drawn, V- will equal +4.5V. The way the 7660 datasheet is drawn, it will yield -9V.
I think either pins 3 and 5 are erroneously swapped, or it is a trick question. I am leaning toward the latter. It can sort out the student who is thinking, and understands the concept, vs one who just looks at the datasheet and parrots the answer.