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I have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The switches flip at the same time. In the position shown, SW1 is causing a loop that only connects to V1+ and never to ground, because SW2 is off. What will happen there? Will there be any current flow in the whole L1/C1 branch?

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  • \$\begingroup\$ This looks like a SMPS.. a boost? I think you need to assume that before the position shown the switches were in the other direction. So Current was flowing through L1. (to C1 and C2 and R1.) Then in the config shown the current flowing in L1 will charge up C1, and L2, C2 will keep current into the load (R1) \$\endgroup\$ Oct 6 '14 at 12:48
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Call the schematic as shown the "base state". There is no voltage drop across L1 or C1, so the top od SW2 is at V1. SW2 ensures there is no voltage across R1.
When the switches change... L1 C1 charge as a resonant circuit. This is made more complex, because L2 (and C2 in parallel with R1) are all in parallel with C1.
Depending on Q (ratio of Lto C) there may be some ringing, but eventually ALL of V1 should appear across R1. C1 and C2 should both charge to V1.

When the switches change back ? C1 discharges via L1. There may be ringing again, but you won't see it because nothing is connected.
C2 discharges via L2 and R1, until the voltage across R1 eventually returns to zero - the original state.

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Presuming the circuit is in steady state (currents and voltages are not changing), both caps will have 0 volts across them and both inductors wil have zero current flowing through them. When you flip the switches, then C1 will pull one end of L1 to ground and start to charge through L1. As the voltage on that node increases, current will also start to charge C2 through L2. You will likely get some oscillations as the curent flowing through L1 and L2 will overcharge C1 and C2, which will push current back in the other direction. Eventually the circuit will settle with both caps charged. If you flip the switch back at this point, the caps will discharge through the inductors and ring for a while. L1/C1 will theoretically ring forever, but parasitics will cause this to decay. R1 will cause the oscillation of L2/C2 to decay until it returns to the original steady state.

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