# Producing negative voltage.

I’m a little puzzled as to why the cap is thrown below 0v … why does it not stay 0v? Maybe it’s because I’ve barely slept and it’s going straight over my head…

An electrolytic capacitor is often aluminium foil, plastic foil and acid liquid. They are only build to allow a positive voltage, so you should avoid a negative voltage. They might blow, spreading the electrolyte all around. http://en.wikipedia.org/wiki/Electrolytic_capacitor

In the circuit the second transistor is on or off. The voltage is 1.5V via the 100R resistor or 0V via the transistor. The third transistor does the opposite. So there are two states: 1 ) Led is off. The + side of the capacitor is 0V, the - side is near 1.5V via the led. That is a negative voltage, but not a big deal, the led lowers the voltage. 2 ) Led is on. The + side of the capacitor is 1.5V via the 100R resistor. The - side is 0V via the third transistor. That is a positive voltage.

No, that's not how its supposed to work.

When 2nd transistor is off the capacitor terminals sit at 1.5V and ~0.1V, since the third transistor is saturated (on fully).

When the 2nd transistor turns on the capacitor terminals both fall 1.4V or so to 0.1V and -1.3V, since its a capacitor the voltage across it cannot suddenly change.

The capacitor will then start discharging as current flows through the LED until the LED is no longer forward biased (perhaps 1.8V is needed, so cap at 0.1V and -0.3V by that point.

Then the 2nd transistor turns off again, the cap charges up via 3rd transistor and the 100 ohm resistor.

At no point does the capacitor have to be reverse-charged.

Also the LED cannot pull the negative end of the cap up to 0V since it doesn't start conducting until its cathode is (about) 1.8V below the 1.5V rail - until then its effectively open circuit.

MarkT: When the 2nd transistor turns on the capacitor terminals both fall 1.4V or so to 0.1V and -1.3V, since its a capacitor the voltage across it cannot suddenly change.

You say that, the cap can /not/ suddenly change but it does lol, so when the positive side of the cap sees ground, what precisely is going on here?

1v positive side of cap 0v negative side (until the transistor turned off)

Is this just maths, deducting voltage?

When these is 1V across a capacitor and the 1V end is suddenly put to 0V, then the other end which was zero volts goes instantly to -1V. It has to do this because of the conservation of energy law. At the instance of change of voltage there is no change in the energy stored in the capacitor so in order to maintain 1V across it the other end has to go to -1V.