Hi I played with some basic circuits in lushprojects.com simulator and came up with this one that surprised me - when the capacitor C1 is charged for the first time by pressing the SW1 switch the oscilloscope in the simulator displays that its final charge as 8.36V - well above the 5V source voltage.

My question hence is - is it just an inaccuracy in the simulator, perhaps caused by a missing resistor in the charging loop, or can the capacitor in fact charge to a higher voltage then the source?



simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Try putting a large (10 M ohm to 1 G ohm) resistor across the cap. \$\endgroup\$ – WhatRoughBeast Jul 29 '15 at 12:14

I think the simulator is wrong. I would expect the capacitor to be charged to about 4.3 volts, allowing for an 0.7 Volt base to emitter drop in the transistor. (Although the transistor may fail due to excessive base current when the switch is first closed.)

  • \$\begingroup\$ I would expect it to charge to very near 5 V, given enough time. Even though it's small there will be a leakage current through the b-e junction when it has even a very small positive bias. \$\endgroup\$ – The Photon Jul 29 '15 at 2:17
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    \$\begingroup\$ Using an inductor, it's possible to charge a capacitor to a higher voltage than source. But my gut feeling is that something other than inductance is going on in the O.P.'s simulation. \$\endgroup\$ – Nick Alexeev Jul 29 '15 at 3:06
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    \$\begingroup\$ See "charge pump" for example of capacitors boosting voltage. \$\endgroup\$ – vicatcu Jul 29 '15 at 3:30

The voltage across the terminals of C1 in your set-up can never rise above 5V unless your are able to change the physical properties of the capacitor after charging. This is however not realistically possible with a regular 10uF electrolytic or tantalum capacitor.

Watch demonstration


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