solving a deceptively simple-looking switched RC circuit (singular circuit ?)

First of all, your circuit is a simplification of a real circuit that is either impossible, in reality, or involves infinite current impulses. But as long as you realize what the simplifications are, you can work around them.

You mention that C3 may have an arbitrary initial voltage at the moment it is switched into the loop made up of the other components. Do C1 and C2 also have arbitrary initial voltages at the moment the voltage source is connected to them? Id that moment t=0?

Yes. Except for the small possibility that the voltage across C3 happens to match the total voltage across C1 and C2, there will be an infinite current impulse through the switch and the equality will be instantaneously enforced. You need to solve for the amount of charge that must redistribute in that moment that will produce that equality, and also to solve for what the resulting voltage will be. Start with a simpler case of two capacitors, each charged to an arbitrary voltage and then connected in a loop. Before the connection, each capacitor has a charge of Q=C*V. After the connection they must have a common voltage. But charge is conserved, so Q1+Q2 in the initial state must equal Q1+Q2 in the final state, as well. Signs are important. During this impulse time, you can completely neglect the effect of any resistance in the circuit, because no time elapses, so only infinite impulses can do anything in zero time.

Hi John,

The DC-charge / transfer procedure is cycled, so C1 and C2 are initially uncharge at t=0 (before the first cycle), but will have an initial charge on subsequent cycles (I realise that these initial voltages will impact the initial condictions at t=T+). For now, I set out to do the circuit anaysis for the first cycle only... **Tangent: for a complete solution (assuming the charge / transfer cycling is square-wave periodic) would I be best advised to use a Laplace approach? i.e. is there a simple methodology that I can use for solving the circuit for multiple charge/tranfer cycles or must I work out the initial conditions for each stage, and solve the transient to obtain the initial conditions for the subsequent stage, and so on...?

I was able to work out what the charge-equilibration is for two initially-charged capacitors Ca and Cb that are suddently connected together in a loop to create an instantaneous impulse current:

Referring to my original charge/transfer circuit now, can I reduce the series-connected C1 and C2 of my circuit (which have different charges) to an equivalent capacitor Ceq, and then apply the above result for instantaneous current flow when Ceq is connected to C3? My hesitation stems from the fact that when I've calculated equivalent capacitance for series-arranged capacitors in the past, the original capacitors always had equal charges ... not the case here ...

Is the charge for the equivalent capacitor of two series-connected capacitors (C1, C2) having **different charges** calculated as: Veq = V1 + V2 Qeq / Ceq = Q1/C1 + Q2/C2 Qeq / [ C1C2 / (C1+C2) ] = Q1/C1 + Q2/C2 (here the Q's would normally be cancelled to get the familiar relationship for series-connected capacitors that have equal charge - but not here....) Qeq = [ C1C2 / (C1+C2) ] * [ Q1/C1 + Q2/C2 ]

and this last equation is what i use to "map" the charges from Ceq to C1 and C2 in my circuit. Afterwards I determine the instantaneous transfer of charge by doing a (finalCharge-initialCharge) calculation. Correct?

Thanks, Dan

Beats me. I would have to think about alot morethan I have.

I am not sure I follow all that.

But doing it my way (which may or may not help you)

I would say: Veq = V1 + V2

Ceq = 1/(1/C1 + 1/C2) (to be in the general form for any number of series capacitors)

Qeq = Veq*Ceq = (V1+V2)/( 1/C1 + 1/C2)

After the parallel connection with the third capacitor, the final Q1 and Q2 can be solved by

Qchange is the same for all 3 capacitors in the series loop, except for the sign:

Q1initial - Qchange = Q1final

Q2initial - Qchange = Q2final

Q3initial + Qchange = Q3 final

Then the voltages will be:

V1final = Q1final/C1

V2final = Q2final/C2

V3final = Q3final/C3

I think I would build a spread sheet with this sequence of formulas (across rows, including the RC decays between events), repeated the down the columns, so I could graph all variables for of an arbitrary number of cycles.

By that time, perhaps the formula for the equilibrium result would become obvious to me.

I didn't read all the equations but ...the basic principals when "cap smashing" (cap smashing = attaching caps together that have different voltages which ideally produce infintie currents) are that there is a conservation of CHARGE before and after the switch throws. so the charge is the same before and after the switch. But....there is NOT a conservation of energy. Energy is lost during the switch in the form of spark radition or EM radiation.

You may be able to simulate this circuit in PSPICE by putting small resistors in series with the switches.. The resistors will dissiapte the energy that is lost during the switching..you can change the value of the resistors to see the effect... but the sim probably will probably fail to work if you use 0 Ohms...

In a real circuit, there is always some resistance or a spark or something to dissiapte the lost energy. When connecting caps together of different voltages, energy is always lost, but charge is conserved. Note energy stored in a cap = 1/2 CV^2

Mark

snipped-for-privacy@yahoo.ca wrote: (snip)

Of course the ultimate state is the C3 voltage that matches the sum of the volts across C1 and C2 before the output switch closes. In other words, the final voltage will be one that allows no charge transfer (since there is nothing else to drain charge from C3).

Insert a resistor into the circuit between the old circuit and the new caps. Solve the circuit. Then, take the limit as that resistor goes to 0.