Charging capacitors through a bipolar switch

[bucefal]

Homework Statement

Hi all,
My question is: In what kind of device the enclosed schematic can find an application?
The bipolar switch has two positions U and u- which are for upper position and D and d- which are for down position. The main question is on what kind of equation the voltage on capacitor Cv is depending.
http://sdrv.ms/14wcylh
2. The attempt at a solution
My attempt was quite a success I've measured the voltage on Cv with the voltmeter while moving the key from upper to down position, as said in the question. It was equal to e (the small battery -1,5V) no matter from what is the E (the 3 cell battery on the schematic).
So the simple solution is:
When the switch is switched to down position:
The capacitor Ch is charged to E.
When the switch is switched to upper position:
The capacitors Ch and Cv are in series, also E and e are in series. So Voltage on cv will be = E+e-Vch=E+e-E=e=1,5V.
MY PROBLEM IS WHERE THIS CIRCUIT CAN FIND AN APPLICATION?

 

[rude man]

"enclosed schematic"?

 

[bucefal]

Ch- 470uF electrolitic capacitor
Cv-4700uF electrolitic capacitor

 

[gneill]

Is the small battery +1.5V or -1.5V? Your problem statement seems to state both at different points.

This looks like some sort of "charge pump" setup. Presumably the Double Pole Double Throw (DPDT) switch is meant to be operated repeatedly.

Note that when the switch is in the down position that Ch will quickly obtain a fixed charge each time: q = Ch*E. When the switch moves to the upper position, sources E and e are placed in series with the two capacitors, and the then-existing charges on both Ch and Cv will have to move to accommodate the new total potential (E+e). The values of the two capacitors come into play here. Can you work out how the charges will adjust?

 

[rude man]

Assume you start with D/d and Cv is discharged (V=0). What happens when the switches are switched from D/d to U/u?
Then back to D/d?
Then back to U/u?
Etc.?

Is there a point at which Cv no longer increases voltage? What point must that be?
Hint: In the D/d position, no charge across Cv is lost. Look for the point when ΔV across Ch is zero when switching from D/d to U/u. Then at that point you know that Ch is not surging any more current into Cv.

The circuit does remove the common voltage E and present the differential voltage e at the output. Removing a common voltage is often necessary. Transformers do the same thing (not autotransformers!). Baluns are another way. Also certain op amp configurations.

If the small battery is -1.5V instead of what is shown on the diagram (+1.5V) then V will asymptotically approach -1.5V as you have reported.

 

[bucefal]

In the problem statement it's said that I have to measure the voltage across Cv, while switching from D,d to U,u. The maximum voltage across Cv, which can be obtained is +1,5V which is equal to e. I can explain this, but as gneil said this is some kind of "charge pump". I was asked from a professor where I can apply this in the real world. So it's not something special. I think it can be used as an inverted op amp, does this sound realistic?
I am very thankful to all of you, who replied, till this moment to my problem.

 

[rude man]

In the problem statement it's said that I have to measure the voltage across Cv, while switching from D,d to U,u. The maximum voltage across Cv, which can be obtained is +1,5V which is equal to e. I can explain this, but as gneil said this is some kind of "charge pump". I was asked from a professor where I can apply this in the real world. So it's not something special. I think it can be used as an inverted op amp, does this sound realistic?
I am very thankful to all of you, who replied, till this moment to my problem.

I told you what the purpose of the circuit can be. It is to reject a common mode voltage to detect a differential-mode voltage.

I don't see how it can be used as an inverter.

In your case E is the common-mode voltage and e is the differential-mode voltage.

In order to use the circuit as a common-mode -rejecting circuit, the switching rate would have to be more than twice as fast as the highest frequency component in e.

 

[bucefal]

I told you what the purpose of the circuit can be. It is to reject a common mode voltage to detect a differential-mode voltage.

I don't see how it can be used as an inverter.

In your case E is the common-mode voltage and e is the differential-mode voltage.

In order to use the circuit as a common-mode -rejecting circuit, the switching rate would have to be more than twice as fast as the highest frequency component in e.

Today, I said that this can be used as a rejecting circuit to reject common voltage and to detect differential-mode voltage, to my professor. And his reply was: "Of course it's not. Keep working. The man who developed the application of this circuit has became very rich."
Is there some relation that, when the DPDT switch is in position U,u the two capacitors are in series, and they can act as a voltage divider?

 

[rude man]

Today, I said that this can be used as a rejecting circuit to reject common voltage and to detect differential-mode voltage, to my professor. And his reply was: "Of course it's not. Keep working. The man who developed the application of this circuit has became very rich."
Is there some relation that, when the DPDT switch is in position U,u the two capacitors are in series, and they can act as a voltage divider?

Well, your prof is wrong about not rejecting common mode, but he could be referring to something else like the sampling theorem, digitization, Nyquist, all that stuff.

No comment about any capacitors being in series etc. Don't see it.

 

[bucefal]

I think Cv and Ch are in series in the first half of the image, which represents the U,u position of DPDT switch.

 

[rude man]

I think Cv and Ch are in series in the first half of the image, which represents the U,u position of DPDT switch.

I guess, but so what?

 

[NascentOxygen]

My question is: In what kind of device the enclosed schematic can find an application?

I suspect that you have mixed up the switching arrangement, and your schematic is not accurately representing what the original question intended. bucefal, can you post a photo of the question from your textbook (or from wherever this puzzle originated)?

 

[bucefal]

I suspect that you have mixed up the switching arrangement, and your schematic is not accurately representing what the original question intended. bucefal, can you post a photo of the question from your textbook (or from wherever this puzzle originated)?

Actually I can, but the problem statement is in Bulgarian, but the schematic is the same, as shown in the statement. The prof said this is used in device which is used for 50 years till today. The schematic is this I've just redraw it with software.
First: I have to measure the voltage drop on the capacitor Cv, which is 4700uF.
Second: Have to change E, by placing different combinations of batteries- 1,5V (one AA battery,) 3V (two batteries), 4,5 V and 0V (no batteries in this holder).
Third: I have to say what is the practice apply of this schematic.
I have one more chance to say the correct answer. Thinking to propose charge pump?!

 

[rude man]

Well, it is a charge pump, but so what? Charge pumps can be used to greatly multiply a voltage, but that is not the case here.

The only purpose I see continues to be common-mode voltage rejection.

 

[CWatters]

If it's not that the only thing I can suggest is a negative supply rail generator (eg http://en.wikipedia.org/wiki/MAX232) ? However the polarity of the output capacitor is wrong for that.

 

[rude man]

If it's not that the only thing I can suggest is a negative supply rail generator (eg http://en.wikipedia.org/wiki/MAX232) ? However the polarity of the output capacitor is wrong for that.

Thing is, though, the e supply already has to be a floating supply. So you can make a + or a - supply just by grounding the appropriate terminal (to the output ground).

 

[CWatters]

Good point.

Will be interesting to hear what the official answer is.

 

[rude man]

Good point.

Will be interesting to hear what the official answer is.

Yes. Made the inventor rich, by the OP's acount! A bit hard to see right now, though ...

 

[bucefal]

This is some kind of chopper. link. But what physics problem does the chopper is a solution for?

 

[rude man]

This is some kind of chopper. link. But what physics problem does the chopper is a solution for?

It can be a synchronous, phase-sensitive demodulator. If e = sin(wt) and the switch is 'up' for nπ/w < t < (n+1)/w and 'down' for (n+1)π/w to (n+2)π/w, n = 0, 2, 4, ... the output is a +dc voltage = (2/π) volts.

If e = -sin(wt) then the output is a negative dc voltage = -2/π.

If e = sin(wt +/- π/2) then the output is zero. That makes it a phase-sensitive demodulator, also called a phase-sensitive voltmeter or lock-in amplifier.

In any case it also rejects the common-mode voltage E.